text:
\section{Main Statement}

	Let $\mu$ be a Borel probability measure on the circle $\mathbb{T}
	:= \mathbb{R}/\mathbb{Z}\equiv [0,1)$ identified with the interval $[0,1)$ and let $(\ell_n)_{n\ge 1}$ be a sequence of
	positive numbers  with $0<\ell_n <1$.
	 Assume that $(\xi)_{n\ge 1}$ is  a sequence of i.i.d. random variables  having $\mu$ as probability law. Then, for each $n\ge 1$, we consider the random interval
	 $$I_n : = (\xi_n -\ell_n/2, \xi_n +\ell_n/2)$$
	  of length $\ell_n$ and centered at $\xi_n$. Sometimes, we say that $I_n$
	is the ball $B(\xi_n, r_n)$ centered at $\xi_n$ and of radius
	$r_n := \ell_n/2$. Under what condition on $\mu$ and on $(\ell_n)$ have we
	$$
	    P\big( \mathbb{T}= \limsup_{n\to \infty} I_n\big) =1 ?
	$$

	If the answer is affirmative, we say that $\mathbb{T}$ is covered
	for the {\em $\mu$-Dvoretzky covering}.
		We can also ask if a given compact set is covered or not.
		Without loss of generality, we always assume that $(\ell_n)$ is decreasing.

	\medskip

	When $\mu$ is the Lebesgue measure on $\mathbb{T}$, it is the
	classical Dvoretzky covering problem \cite{Dvoretzky1956}(1956), to which a necessary and sufficient condition for $\mathbb{T}$ to be covered is
	\begin{equation}\label{cond:Shepp}
	    \sum_{n=1}^\infty \frac{1}{n^2}\exp(\ell_1+\cdots +\ell_n)=\infty.
	\end{equation}
	This is called {\em Shepp's condition}, which was obtained by L. Shepp \cite{Sp}(1972). A compact set $F \subset \mathbb{T}$ is covered if and only if
	\begin{equation}\label{cond:Kahane}
	{\rm Cap}_{\Phi}(F)=0
	\end{equation}
	where
	$$
	        \Phi(t, s) = \exp \sum_{n=1}^\infty (\ell_n -|t-s|)_+.
	$$
	This is called {\em Kahane's condition}, which was obtained by J-P. Kahane \cite{Kahane1998}(1987). Recall that ${\rm Cap}_{\Phi}(F)=0$ means that
	for any Borel probability measure $\sigma$ supported by $F$ we have
	$$
	    \int_{\mathbb{T}} \int_{\mathbb{T}} \Phi(t,s) d\sigma(t) d\sigma(s) =+\infty.
	$$
	We refer to \cite{19kahane1985some} for the theory of capacity.
	The above cited results due to Shepp and Kahane
	will be our basic useful facts.
	Kahane's book \cite{Kahane1985} contains references on the study of classical Dvoretzky covering before 1985. For later works, let us only cite \cite{BarralFan,
		Durand2010, EP2018, Fan1995, Fan2002, Fan2004,  FK1993, FK2001,FLL2010, FW2004, FJJS2018, JJKLSX2017, JS2008, OSW2017,Seuret2018,Yu2003}. A first study on  $\mu$-Dvoretzky covering problem for Gibbs measures $\mu$ was  made in \cite{tang}, where the author used the method taken from \cite{A-Sh-T} in order to find the optimal covering exponent $t$ for $\ell_n = a/n^t, (a>0, t>0)$.

\medskip

	In this paper, we will study the $\mu$-Dvoretzky covering problem when $\mu$ admits a density. In the following, $f$ will always denote a density function and the measure of density $f$ will be
	denoted by $\mu_f$.  The solution to the problem depends on the
	essential infimum of the density function.
	\begin{defi}
		For a given Borel function $f:\mathbb{T} \rightarrow \mathbb{R}$, the {\em essential infimum} of $f$ is  denoted by $ m_f$, i.e.
		$$
		m_f:= \essinf_{\mathbb{T}}f=\sup \{a \in \mathbb{R}: a\le f(x) \ a.e. \}

		$$
		where "a.e." refers to the Lebesgue measure.
	\end{defi}
We can define the local essential infima function $\underline{E}_f(x)$ and the local essential supprema function $\overline{E}_f(x)$
(see  Section \ref{sect:essential-inf} for definitions).
 The set of essential infimum points of $f$, denoted by $K_f$ is defined to be the set of those $x$ such that  $\underline{E}_f(x) = m_f$.  It will be proved that $K_f$ is a non-empty compact set (Proposition \ref{prop:EI}).
 \medskip

 The flatness of the density will also play a role. It is considered as a regularity of the density.

\iffalse
	Let $B(x, r)$ be a ball and  $a>0$,  we denote $B(x, ar)$ by $a B(x, r)$. So, for a non-empty  open interval $U$,   $a U$ stands for the interval of the same center as
	$U$ and of $a$ times the length of $U$. This is not a standard notation, but it is convenient for us in this note.

Let $f$ be a density function on the unit circle $\mathbb{T}$. We denote by $\mu_f$ the absolutely continuous measure with the density $f$. For the ball $B(x, r_n) \subset \mathbb{T}$ we denote by $\mu_f(B(x, r_n))$ the measure of the ball.

For the given density function $f$ we denote by $K_f$  its set of essential infimum points (see definition \eqref{essinfset}). See Section \ref{prep} for definition of the notations $\overline{E}_f({x})$, $\underline{E}_f({x})$ etc.

The following property of the density function $f$ at a point will be important for us.
\fi

\begin{defi}\label{flat}
A point $x \in \mathbb{T}$ has the
{\em flatness property} for the measure $\mu_f$ and the sequence $\{\ell_n\}_{n \geq 1}$ if
\begin{equation}\label{bound1}
\sum_{n=1}^\infty |\mu_f(B(x, r_n))-m_f \ell_n|<\infty.
\end{equation}
The set of all flat points will be denoted by $F_f(\{\ell_n\}_{n \geq 1})$ or simply by $F_f$.
\end{defi}

	Here is another regularity of the density function $f$: there exists a sequence of points $\{x_n\}_{n \geq 1}\subset \mathbb{T}$ such that
	\begin{equation}\label{dom_1}
	\lim_{n \rightarrow \infty}\overline{E}_f(x_n)=m_f.
	\end{equation}
In is easy to see, that if $f$ is continuous at least at one point of the set $K_f$, then the condition \eqref{dom_1} is fulfilled.

We are now ready to state one of the main theorems in this paper.
\begin{thm}\label{thm1}
Let $\mu_f$ be a Borel probability measure on $\mathbb{T}$ with the density function $f$ and $\{\ell_n\}_{n=1}^\infty$ a sequence of positive numbers. Assume that the condition \eqref{dom_1} is fulfilled.
We distinguish two cases.

$(1)$ Assume
$m_f = 0$ and $K_f$ is countable. Then, the circle is covered for the $\mu_f$-Dvoretzky covering if and only if the following two conditions are satisfied:
\begin{equation}\label{cond0}
\forall x \in K_f,\quad \sum_{n=1}^\infty \mu_f(B(x, r_n))=\infty;
\end{equation}

\begin{equation}\label{cond01}
\forall a> 0, \sum_{n=1}^\infty \frac{1}{n^2}e^{a(\ell_1+\dots + \ell_n)}=\infty.
\end{equation}

$(2)$ Assume $m_f>0$ and there exists a sequence $\{a_n\}_{n \geq 0}$, with $a_0 =0$, so that the set $K_f \setminus \cup_{n \geq 0}(a_n + F_f \cap K_f)$ is at most countable. Then, the circle is covered for the $\mu_f$-Dvoretzky covering if and only if the following two conditions are satisfied:
\begin{equation}\label{cond1}
\forall a> m_f, \sum_{n=1}^\infty \frac{1}{n^2}e^{a(\ell_1+\dots + \ell_n)}=\infty; \quad {\rm Cap}_{\Phi^{(m_f)}}(F_f \cap K_f)=0,
\end{equation}
where the capacity refers to the kernel
$$
      \Phi^{(m_f)}(t,s) = \exp \sum_{n=1}^\infty m_f (\ell_n - |t-s|)_+.
$$
\end{thm}

\iffalse
\begin{remark}
In the above theorem it is sufficient to require the property \eqref{bound1} only for a subset $K_0 \subset K_f$, for which there exists $\{a_n\}_{n \geq 1}$, $a_n \in [0,1), n \geq 1$ so that
$$
\sigma\left(K_f \setminus \bigcup_{n \geq 1}\{K_0 + a_n\}\right)=0.
$$
for arbitrary measure $\sigma$.
\end{remark}
\fi

We have the following corollaries of Theorem \ref{thm1}.
\begin{cor}\label{cor1}
	Assume that for all $x \in K_f$, there is an open neighborhood  $U$ of $x$ and a Lipschitz function $g_x \in \lip(U)$, such that $f(t) \leq g_x(t)$ for almost every $t \in U$ and $f(t)=g_x(t)$ on $K_f \cap U$. \\
	\indent {\rm (1)}  If $m_f=0$ and $\sum_{n=1}^\infty \ell_n^2<\infty$,  there will be no $\mu_f$-Dvoretzky covering. \\
		\indent {\rm (2)}
	If $m_f>0$,  the circle is covered if and only if the following two conditions are satisfied
\begin{equation}
\forall a> m_f, \sum_{n=1}^\infty \frac{1}{n^2}e^{a(\ell_1+\dots + \ell_n)}=\infty; \quad {\rm Cap}_{\Phi^{(m_f)}}(K_f)=0,
\end{equation}

\begin{cor}\label{cor2}
Assume there exists $U \subset \mathbb{T}$, so that $f(x)=m_f$, for a.e. $x \in U$. Then the circle is covered for the $\mu_f$-Dvoretzky covering if and only if
$$
\sum_{n=1}^\infty \frac{1}{n^2}e^{m_f(\ell_1+\dots + \ell_n)}=\infty.
$$
\end{cor}

	\iffalse
Assume for $\forall x \in K_f$ there is an open set $U$, with $x \in U$, and a function $g \in \lip(U)$, such that $f(t) \leq g_x(t)$ for almost all $x \in U$ and $f(x)=g(x)$ on $K_f \cap U$. Then the following two condition are necessary and sufficient for covering the circle:
\begin{equation}\label{cond1c}
\sum_{n=1}^\infty \frac{1}{n^2}e^{a(l_1+\dots + l_n)}=\infty, \hbox{ for all }a>m_f,
\end{equation}
and for any measure $\sigma$ supported on $K_f$, we have
\begin{equation}\label{cond2c}
\int_{K_f} \int_{K_f} \exp{\left( m_f \sum_{k=1}^\infty (\ell_k - |t-s|)_+\right)}d\sigma(s)d\sigma(t)=\infty.
\end{equation}
\fi
\end{cor}

In the special case of $\ell_n = \frac{c}{n}$ ($c>0$), we don't require
any regularity of $f$.
\begin{thm}\label{thm_seq}
Suppose $\ell_n = \frac{c}{n}$ ($c>0$). Let $\mu_f$ be an arbitrary
Borel probability measure with density $f$.  A necessary and sufficient condition for covering the circle in the $\mu_f$-Dvoretzky covering is
$
cm_f\geq 1.
$
\end{thm}

The $\mu$-Dvoretzky covering problem is subtle
and is not treated in this paper when $\mu$ is singular. But we have the following
local comparison principle. This principle will serve us as a tool
in our present study and it has its own interests.

\begin{thm}\label{thm3} Consider two Dvoretzky covering respectively defined by two Borel probability measures $\mu$ and $\nu$. Assume that
	$\mu|_U \le \nu|_U$ for some non-empty open set $U \subset \mathbb{T}$. Let $K\subset U$ be a compact set in $U$. If $K$    is covered for the $\mu$-Dvoretzky covering, then it is covered for the $\nu$-Dvoretzky covering.
\end{thm}

We organize the rest of the paper as follows. In Section 2, we prove that the set of essential infimum is a non-empty compact set. Section 3 contains two basic results which are respectively qualified local Billard criterion and local Kahane criterion. In Section 4, we prove the comparison principle (Theorem \ref{thm3}) which is our third basic result. After these preparations,  we  find sufficient conditions for covering $\mathbb{T}$ in Section 5, and necessary conditions for covering $\mathbb{T}$ in Section 6. Then Theorem \ref{thm1} together with its corollaries and Theorem \ref{thm_seq}
are proved in Section 7.

\section{Set of essential infimum}\label{sect:essential-inf}
 We leave the Dvoretzky covering problem for a while.
In this section, we study the set of points where a measurable function attains its "minimal" value.
\medskip

Let $f:\mathbb{T} \rightarrow \mathbb{R}$ be a Borel measurable function and $I \subset \mathbb{T}$ an interval.
The {\em essential infimum} of the function $f$ on the interval $I$ is defined as follows
$$
\essinf_{I}f := \sup\{a \in \mathbb{R}: a \le f(x) \ {\rm for\ almost \ all } \ x\in I\}.
$$
We will denote $\essinf_{\mathbb{T}}f$ by $m_f$.
Let $x_0 \in \mathbb{T}$ be fixed. The {\em  essential infimum} at  $x_0$ of $f$ is defined  to be the following limit
$$
\underline{E}_f(x_0) : = \lim_{n \rightarrow \infty}\essinf_{B(x_0,\frac{1}{n})}f.
$$
The {\em set of essential infimum points},
denoted $K_f$, is defined by
$$
K_f	:=\{x \in \mathbb{T}: \underline{E}_f(x) =m_f\}.
$$

Similarly  we  define the essential suprema $\esssup_{I}f$
and $\overline{E}_f(x_0)$.
\iffalse
$$
\esssup_{I}f = \sup_{a \in \mathbb{R}}\{P(f(x) > a|x \in I)>0\},
$$
and respectively
$$
\underline{E}_f(x_0)= \lim_{n \rightarrow \infty}\esssup_{B(x_0,\frac{1}{n})}f.
$$
\fi
Clearly
$$
\underline{E}_f(x_0) \leq \overline{E} _f(x_0).
$$
It is also cleat that $\underline{E}_f(x_0) = \overline{E} _f(x_0)$ if $f$ is continuous at $x_0$.
\iffalse
and since $f$ is positive, then there is equality if and only if $f$ is continuous at $x_0$ in an almost sure sense, i.e. for arbitrary $\epsilon>0$ there is $\delta>0$, so that
$$
|f(x)-f(x_0)|<\epsilon,
$$
for almost all $x \in (x_0-\delta,x_0+\delta )$.

\begin{defi}\label{essinfset}
For a non-negative function $f$ defined on $\mathbb{T}$ we denote by $K_f$ the set of all essential infimum points, i.e.
$$
K_f	=\{x \in \mathbb{T}: \underline{E}_f(x) =m_f\}.
$$
\end{defi}
\fi

\begin{prop}\label{prop:EI}
For every Borel measurable function $f: \mathbb{T} \to \mathbb{R}$, $K_f$ is a non-empty compact set.
\end{prop}
\begin{proof}
We prove the non-emptyness of $K_f$  by a dissection argument using the fact
$$\essinf_{I \cup J} f = \min (\essinf_I f, \essinf_J f).
$$
Indeed, cut $\mathbb{T}$ into $\mathbb{T} = I \cup J = [0,1/2) \cup [1/2, 1)$. Then we have
$$
\essinf_{\mathbb{T}} f = \min(\essinf_I f,\essinf_J f).
$$
We continue this process and construct a sequence of nested intervals
$$
I_1 \supset I_1 \supset \dots \ I_n \supset \dots,
$$
so that $I_{n+1}$ is one of the two halves of the interval $I_n$ and such that
$$
\essinf_{I_n} f = \essinf_{I_{n+1}} f=\essinf_{\mathbb{T}} f.
$$
Since $|I_{n+1}|=|I_n|/2$, then $\bigcap_{n=1}^\infty I_n$ is a single point, say $\{x_0\}$.
\iffalse Let
$$
\{x_0\}=\bigcap_{n=1}^\infty I_n.
$$
\fi
We claim that $x_0 \in K$. Indeed, for arbitrary $n \in \mathbb{N}$ there is $m \in \mathbb{N}$ such that
$$
B(x_0, 1/n)\supset I_m,
$$
which implies
$$
\essinf_{B(x_0, \frac{1}{n})}f\leq  \essinf_{I_m}f=\essinf_{\mathbb{T}}f.
$$
Let $n\to \infty$. Thus we get $\underline{E}_f(x_0) = \essinf_{\mathbb{T}}f$.

Now show that $K_f$ is closed. Assume $\{x_n\}_{n \geq 1}\subset K_f$ and $\lim_{n \rightarrow \infty}x_n = x_0$.
For any $n \in \mathbb{N}$ there exists $m \geq 1$ such that
$$
x_m \in B(x_0, \frac{1}{n}),
$$
which implies that
$$
\essinf_{{B(x_0,\frac{1}{n})}}f\leq \underline{E}_f(x_m)= m_f.
$$
It follows that
$
\underline{E}_f(x_0)=m_f
$, i.e. $x_0\in K_f$.
\end{proof}

We now show that the regularity condition (\ref{dom_1}) is not always fulfilled.

\medskip
Let $A \subset [0,1]$ be a set, which is constructed in the same way as the classical Cantor set, but at each step, we remove the central interval of shorter length. This results in a compact set with positive Lebesgue measure, which  contains no intervals and is nowhere dense. Consider the characteristic function of the complementary of $A$:
$$
f_A(x) = 1 - \chi_A(x), \quad x \in [0,1).
$$
Observe that
\begin{equation}\label{Cantor}
\forall x \in [0,1],\ \overline{E}_{f_A}(x) = 1;  \quad
\forall x \in A, \ \underline{E}_{f_A}(x) = 0.
\end{equation}
Then, we define the following function
$$
   f(0)= 0; \quad  \forall x \in  [2^{-n-1}, 2^{-n}),
f(x) =
f_A(2^{n+1}x - 1)+\frac{1}{n+1}  \ \ \ (n\ge 0).
$$
From (\ref{Cantor}), we get that for each $n\ge 0$ we have
$$
\essinf_{[\frac{1}{2^{n+1}}, \frac{1}{2^n}]}f=\frac{1}{n+1}; \quad
\forall x \in  [2^{-n-1}, 2^{-n}),
\overline{E}_{f}(x)= 1 + \frac{1}{n+1}.
$$
Finally we get
$$
K_f=\{0\}; \quad
\underline{E}_{f}(0) = 0;  \quad \forall x \in [0,1), \overline{E}_{f}(x) \geq 1.
$$
Thus, the Condition \ref{dom_1} is not fulfilled by this function $f$.
However,  recall that the condition
(\ref{dom_1}) is satisfied for a function $f$ which is continuous at one point of $K_f$.
\medskip

\section{Two basic results due to Billard and Kahane}

The proof of Theorem \ref{thm1} will be based on Theorem \ref{thm3} and on the following two criteria, which have their own interests.

\subsection{Local Billard necessary condition}
The following
theorem gives us a necessary condition for covering the circle.
The idea of  second moment used in the proof  came from P. Billard \cite{Billard1965}.
Therefore the condition will be refered to as (local) Billard condition.

	\begin{thm}[local Billard criterion]\label{thm:B}
		  	Let $F \subset \mathbb{T}$ be a non-empty compact set.
		  	Suppose that $\mu$ is a probability measure on $\mathbb{T}$ such that
		  	\begin{equation}\label{eq:cond-noncover1}
		  	       \sup_{t \in F} \sum_{n=1}^\infty \mu(B(t, r_n))^2 <\infty.
		  	\end{equation}
		  	Then $F$ is not covered for the $\mu$-Dvoretzky covering if there exists a probability measure	$\sigma$ supported by $F$ such that
		  	\begin{equation}
		  	       \label{eq:cond-noncover2}
 \int_{F} \int_{F} \exp\sum_{n=1}^\infty \mu(B(t, r_n)\cap B(s,r_n))d\sigma(t)d\sigma(s) <\infty.
		  	\end{equation}
	\end{thm}

\begin{proof}
	 Consider the martingale
$$
M_n = \int_{F} Q_n(t) d\sigma(t)
$$
where
$$
Q_n(t) = \prod_{j=1}^n \frac{1- \chi_{B(\xi_n, r_n)}(t)}{1 - \mu(B(t, r_n))}.
$$
Notice that $Q_n(t) =0$ means that $t$ is covered by one of the intervals $I_j$ ($1\le j \le n$). If $M_n$ doesn't tend to zero, then some point in $F$ is not covered.
By Fubini theorem and simple computations, we get
$$
\mathbb{E} M_n^2 =
\int_{F}  \int_{F}  \prod_{j=1}^n \frac{1-\mu(B(t, r_n) - \mu(B(s, r_n)) + \mu (B(t, r_n) \cap B(s, r_n))}{(1-\mu(B(t, r_n))(1- \mu(B(s, r_n)))}dtds.
$$
Using $1-x = e^{-x + O(x^2)}$ and the conditions (\ref{eq:cond-noncover1}) and (\ref{eq:cond-noncover2}), we get $\mathbb{E}M_n^2 = O(1)$ which implies that the limit of $M_n$ is not almost surely zero.
\end{proof}

The condition (\ref{eq:cond-noncover2}) means that $F$ has a positive capacity.
The condition (\ref{eq:cond-noncover1}) can be relaxed to the pointwise finiteness, because $F$ can be approximated by
compact sets on each of which the sum is uniformly bounded.

\subsection{Local Kahane criterion}
The next result can be considered as a local version of Kahane's theorem. For $a>0$, define the kernal
$$
     \Phi^{(a)}(t, s) := \exp{\left( a \sum_{k=1}^\infty (l_k - |t-s|)_+\right)}.
$$
As we see below, this criterion is not perfect because we need the assumption that the density is constant around the set to be covered  in question. But it will be one of our basic tools, because we can approximate our density function by functions which are locally constant.

\begin{thm}[local Kahane criterion]\label{thm:K} \label{kh-loc}Consider a probability measure $\mu_f$
	having its density function $f$.
Let $I \subset \mathbb{T}$ be an open interval. Suppose that $f(x)=a$ for almost all $x \in I$. Then
a compact subset  $F$ of $I$ is covered for the $\mu_f$-Dvoretzky covering if and only if ${\rm Cap}_{\Phi^{(a)}}(F) =0$.
\iffalse
 for arbitrary measure $\sigma$ supported on $F$ we have
\begin{equation}\label{kah_thm}
\int_F \int_F \exp{\left( a \sum_{k=1}^\infty (l_k - |t-s|)_+\right)}d\sigma(s)d\sigma(t)=\infty.
\end{equation}
\fi
\end{thm}
\begin{proof} The idea of proof is to compare the $\mu_f$-Dvoretzky
	covering with a classic Dvoretzky covering.
Let $M$ be the distribution function of $\mu_f$, i.e.
$$
     M(x) = \mu_f([0,x]) = \int_0^{x} f(t) dt \quad (0\le x\le 1).
$$
Let $X$ be a random variable which is uniformly distributed on $[0,1)$. Define
$$
           Y= M^{-1}(X)
$$
where the inverse function is defined by
$$
      M^{-1}(x) = \inf \{t\in [0,1]: M(t) \ge x\}.
$$
It is well known that the probability law of $Y$ is $\mu_f$.
So, take a sequence $(\omega_n)$ of i.i.d. random variables uniformly distributed on $[0,1)$ and use $(\xi_n)$ to model a $\mu_f$-Dvoretzky
covering, where $\xi_n = M^{-1}(\omega_n)$.

 Observe that the restriction $M: I \to M(I)$ is affine and invertible. It follows that  for any interval  $J \subseteq I$,  its image $M(J)$ is an interval of length
$$|M(J)|=\int_J f(t)dt=|J|a$$
and the center of $M(J)$ is  the image of the center of $J$.
So,  if $I_n = (\xi_n-\ell_n/2, \xi_n +\ell_n/2) \subset I$, then   $M(I_n)$ is centered at $(\omega_n)$ and of length  $a\ell_n$.

Now choose a proper subset $J \subset I$ such that $F \subset J$. Then choose $N \in \mathbb{N}$ so large that $\ell_n \leq {\rm dist}\{J, \partial I\}$ for all $n \geq N$.
Assume $F \subset \limsup I_n(\xi_n)$. Then for any $x\in F$,
$x\in I_n(\xi_n)$ for an infinite number of $n's$ with $n\ge N$ (these $n$'s depend on $x$). These $\xi_n$'s must fall into the interval $J$.   For such $n$, $M(x) \in M_n(I_n(\xi_n))$.
It follows that
$$
P(F \subset \limsup_{n \rightarrow \infty} I_n)=1
\Longrightarrow
P(M(F) \subset \limsup_{n \rightarrow \infty} M(I_n))=1.
$$
The converse implication can be similarly proved, because $M: I \to M(I)$
is invertible.
In other words,  $F \subset J$ is covered for the $\mu_f$-Dvoretzky coverring   by intervals of length $\{\ell_n\}_{n \geq 1}$  if and only if $M(F)$ is covered for the classic Dvoretzky covering by intervals of length $\{a \ell_n\}_{n \geq 1}$.  But, by Kahane's theorem, the set $M(F)$ is covered for the classic Dvoretzky covering if and only if
${\rm Cap}_{\Phi^{(a)}}(M(F)) =0$, which is equivalent to
${\rm Cap}_{\Phi^{(a)}}(F) =0$, because $M$ is affine around $F$.
\end{proof}

\subsection{Two elementary facts}
Recall that we always assume that $(\ell_n)$ is decreasing.
The following fact is known when $F=\mathbb{T}$. The general case is not trivial.
The proof given below involves both Shepp's condition and Kahane's condition.

	\begin{lem}\label{lem-1}
Let $F$ be a compact set with $|F|>0$ and $a>0$ a positive number.
Then the following two conditions are equivalent:
\begin{equation}\label{subs}
{\rm Cap}_{\Phi^{(a)}}(F)=0,
\end{equation}
\begin{equation}\label{whole}
\sum_{n=1}^\infty \frac{1}{n^2} \exp(a(\ell_1+\cdots +\ell_n))= \infty.
\end{equation}
\end{lem}
\begin{proof}
A simple calculation shows that there exists a constant $C_a>0$
such that
$$
\int_F \int_F \exp{\left( a \sum_{k=1}^\infty (\ell_k - |t-s|)_+\right)}ds dt
\leq C_a
\int_\mathbb{T} \int_\mathbb{T} \exp{\left( \sum_{k=1}^\infty (a \ell_k - |u-v|)_+\right)}du dv.
$$
(First replace $F$ by $\mathbb{T}$ and then make a change of variable). Then (\ref{whole}) is implied by (\ref{subs}) because the last double integral equals to the infinity if and only if
(\ref{whole}) holds.
\iffalse
While the last expression is known to be equivalent to the following condition
$$
\sum_{n=1}^\infty \frac{1}{n^2} \exp(a(\ell_1+\cdots +\ell_n))= \infty.
$$
\fi

Now assume (\ref{whole}).
It is nothing but Shepp's condition for the classic
Dvoretzky covering with the sequence of lengths $\{a\ell_n\}_{n \geq 1}$. Then any non-empty compact set $K' \subset \mathbb{T}$
is covered.  If $F'$ has positive Lebesgue measure, it supports  the restriction of Lebesgue measure. Thus, by Kahane's condition, we have
\begin{equation*}
\int_{F'} \int_{F'} \exp{\left( \sum_{k=1}^\infty (a \ell_k - |u-v|)_+\right)}du dv=\infty.
\end{equation*}
Take $F'=aF$, where $aF$ is the scaling of $F$ by a coefficient $a$.
Then, by making a change of variable,  we get
\begin{equation*}
\int_{F} \int_{F} \exp{\left( \sum_{k=1}^\infty a( \ell_k - |t-s|)_+\right)}dt ds=\infty.
\end{equation*}

\end{proof}

The following fact shows that $\sum_{n=1}^\infty \ell_n^2 =\infty$ is a strong condition for the Dvoretzky covering problem.

\begin{lem}\label{2-summable}
	The condition
	$
	\sum_{n=1}^\infty \ell_k^2 = \infty
	$
	implies
	$$
	\forall a>0, \ \ \ \sum_{n=1}^\infty \frac{1}{n^2} \exp(a(\ell_1+\cdots +\ell_n))= \infty.
	$$
\end{lem}

\begin{proof
This is known, however we will include a proof for completeness.	The condition
	$
	\sum_{n=1}^\infty \ell_n^2=\infty
	$
	implies that 	$
	\ell_n > \frac{1}{n^{2/3}}
	$ holds for infinitely many $n$. Then, for such a $n$, by the monotonicity of $(\ell_n)$ we have
	$$
	\sum_{k=1}^n \ell_k > \frac{n}{n^{2/3}}>n^{1/3}.
	$$
	Therefore the general term of the series in question doesn't tend to zero. So, the series diverges.

\end{proof}

	 \section{A comparison principle: Proof of  Theorem \ref{thm3}}

	 Intuitively, if $\mu|_U\le \nu|_U$, then sets contained in an open set
	 $U$ are easierly covered for the $\nu$-Dvoretzky covering than
	 for the $\mu$-Dvoretzky covering, because there are more chances
	 to get points in $U$ for the $\nu$-Dvoretzky model.
	 We give here a rigourous proof of this intuition which is stated
	 as Theorem \ref{thm3}. The proof benefits from our discussion with Meng Wu.

	 \subsection{Model of $\mu$-Dvoretzky covering}

Let $\mu=\mu_0 + \mu_1$ be a decomposition of a probability measure $\mu$ on $\mathbb{T}$. Then we set
$$
\widetilde{\mu}_0 := \frac{\mu_0}{\mu_0(\mathbb{T})}, \quad
\widetilde{\mu}_1 := \frac{\mu_1}{\mu_1(\mathbb{T})}.
$$
For simplicity, we write $\alpha_0= \mu_0(\mathbb{T})$ and $\alpha_1= \mu_1(\mathbb{T})$, so that $\mu = \alpha_0  \widetilde{\mu}_0
+  \alpha_1  \widetilde{\mu}_1$.

Let $\Omega= \{0,1\}^\mathbb{N} \times \mathbb{T}^\mathbb{N}$. A point in $\Omega$  is denoted by
$$
\omega := (\epsilon; \xi):= (\epsilon_1, \cdots, \epsilon_n, \cdots; \xi_1,\cdots, \xi_n, \cdots) \in \Omega.
$$
Using the decomposition $\mu = \alpha_0  \widetilde{\mu}_0
+  \alpha_1  \widetilde{\mu}_1$, we define a Borel probability measure $P$ on $\Omega$ as follows.
For any integer $n\ge 1$, any $(a_1, \cdots, a_n) \in \{0, 1\}^n$
and any $(A_1, \cdots, A_n) \in \mathcal{B}(\mathbb{T})^n$, define
\begin{equation}\label{def:P}
P(\epsilon_1=a_1, \cdots, \epsilon_n =a_n; \xi_1\in A_1, \cdots, \xi_n \in A_n)
:= \prod_{j=1}^n \alpha_{a_j} \widetilde{\mu}_{a_j}(A_j)
= \prod_{j=1}^n \mu_{a_j}(A_j).
\end{equation}
Notice that $P$ is not exactly a tensor product. But it has a nice
desintegration as we show below.

Let $\pi$ be the
$(\alpha_0, \alpha_1)$-Bernoulli measure on  $\{0,1\}^\mathbb{N}$.
For $\epsilon$ fixed,
let
$$
\Lambda^{(\epsilon)} =\{n \in \mathbb{N}: \epsilon_n =1\},
$$
and we define two measures respectively
on $\mathbb{T}^{\Lambda^{(\epsilon)}}$ and  $\mathbb{T}^{\mathbb{N}\setminus \Lambda^{(\epsilon)}}$ where
$$
P_{\Lambda^{(\epsilon)}} = \bigotimes_{n \in \Lambda^{(\epsilon)}} \widetilde{\mu}_{1},
\quad
P_{\mathbb{N}\setminus\Lambda^{(\epsilon)}} = \bigotimes_{n \in \mathbb{N}\setminus \Lambda^{(\epsilon)}} \widetilde{\mu}_{0},
$$
Let $P^{(\epsilon)}=  P_{\Lambda^{(\epsilon)}} \otimes P_{\mathbb{N}\setminus\Lambda^{(\epsilon)}}$. Notice that
$P^{(\epsilon)}$ is a version of the conditional probability
$P(\cdot | \epsilon)$.

\begin{lem}\label{lem:Lem1} The measure $P$ has the following desintegration
	\begin{equation}\label{eq:desint}
	P =
	\int P^{\epsilon} d\pi(\epsilon) =\int P_{\Lambda^{(\epsilon)}} \otimes
	P_{\mathbb{N}\setminus\Lambda^{(\epsilon)}} d \pi(\epsilon).
	\end{equation}
	Consequently, the following statements hold:\\
	\indent {\rm (1)} All the random vectors $(\epsilon_j, \xi_j)$ ($j =1,2,\cdots$) are i.i.d..\\

	\indent {\rm (2)} Each $\epsilon_j$ obeys the $(\alpha_0, \alpha_1)$-Bernoulli law;\\
	\indent {\rm (3)} Each $\xi_j$ admits $\mu$ as probability law;\\

	\indent {\rm (4)}
	For  $\epsilon$ fixed, with respect to $P^{(\epsilon)}$, $\{\xi_n\}_{n\in \Lambda^{(\epsilon)}}$ are i.i.d. random variables with probability law $\widetilde{\mu}_1$ and $\{\xi_n\}_{n\in \N\setminus\Lambda^{(\epsilon)}}$ are i.i.d. random variables  with probability law $\widetilde{\mu}_0$.
\end{lem}
\begin{proof}
	Using the fact
	$
	{\mu}_{0}+  {\mu}_{1}
	= \mu
	$, we check that the measure $P$ is well defined.  The desintegration (\ref{eq:desint}) is nothing but the definition (\ref{def:P}). It follows the other properties.
\end{proof}

Since $(\xi_n)$ is a sequence of i.i.d. random variables with $\mu$
as probability law, the random intervals
$I_n(\xi_n) = (\xi_n - \ell_n/2, \xi_n + \ell_n/2)$  define a model of $\mu$-covering. Recall that $\mu = \mu_0 + \mu_1$

\begin{lem}\label{lemma comparison1}
	Let $U \subset \mathbb{T}$ be a non-empty open interval.
	Suppose ${\mu}_0(U)=0$. Then for any compact set  $K\subset U$, for $P$-a.e. $(\epsilon,\xi)\in \Omega$, we have
	\begin{equation}\label{eq. lem comp1}
	K\subset \limsup_{n\in \mathbb{N}} I_n(\xi_n) \Longleftrightarrow  K \subset \limsup_{n\in \Lambda^{(\epsilon)}} I_n(\xi_n).
	\end{equation}
\end{lem}
\begin{proof}

	Recall that conditioned on $\epsilon\in \{0,1\}^\N$, the sequence $\{\xi_n\}_{n\in \Lambda^{(\epsilon)}}$ are i.i.d. random variables with probability law $\widetilde{\mu}_1$ and $\{\xi_n\}_{n\in \N\setminus\Lambda^{(\epsilon)}}$ are i.i.d.  with probability law $\widetilde{\mu}_0$. Since $\widetilde{\mu}_0(U^c)=1$,  for $P^{(\epsilon)}$-a.e. $\xi$ we have $\xi_n\in U^c$ for all $n\in \N\setminus\Lambda^{(\epsilon)}$. Since $K$ is compact, it has positive distance from the boundary of $U$, we must have $$
	P^{(\epsilon)}(K\cap \limsup_{n\in \N\setminus\Lambda^{(\epsilon)}} I_n(\xi_n)=\emptyset)=0.
	$$ From this we deduce \eqref{eq. lem comp1}, with the aid of the desintegration (\ref{eq:desint}).
\end{proof}

However, notice that $\epsilon_j$ and $x_j$ are not independent.

\subsection{Proof of Theorem \ref{thm3}}
\iffalse
\begin{thm}\label{dom_thm} Consider two Dvoretzky covering respectively produced by Borel probability measures $\mu$ and $\nu$. Assume that
	$\mu|_U \le \nu|_U$ for some non-empty open set. If any compact set  $K\subset U$  is a.s. covered in the $\mu$-covering, it is  a.s. covered in the $\nu$-covering.
\end{thm}
\fi

	We can model the covering as above, according to the following decompositions
	\begin{eqnarray*}
		\mu &=& \mu_0+\mu_1 \ \ \ {\rm with}\ \ \   \mu_1=\mu|_U, \mu_0=\mu|_{U^c} \\
		\nu &=& \nu_0+\nu_1 \ \ \ \ {\rm with}\ \ \  \nu_1= \mu|_U,  \nu_0 = (\nu|_U-\mu|_U)+ \nu|_{U^c}.
	\end{eqnarray*}
	The corresponding measures will be denoted respectively by $P_\mu$ and $P_\nu$.

	Suppose $K$ is a.s. covered in the $\mu$-covering. Then by Lemma \ref{lemma comparison1}, we have

	$$
	\pi\!-\!a.e.\ \epsilon, \ \ \ P_\mu^{(\epsilon)}\left(K \subset \limsup_{n\in \Lambda^{(\epsilon)}}
	I_n(\xi_n)\right)=1.
	$$
	But $(\xi_n)_{n \in \Lambda^{(\epsilon)}}$ has the same probability law under
	$P_\mu^{(\epsilon)}$ and under $P_\nu^{(\epsilon)}$, because of $\mu_1 = \nu_1$ (see Lemma \ref{lem:Lem1} (4)). Thus
	$$
	\pi\!-\!a.e.\ \epsilon, \ \ \ P_\nu^{(\epsilon)}\left(K \subset \limsup_{n\in \Lambda^{(\epsilon)}}
	I_n(\xi_n)\right)=1.
	$$
	It follows obviously that
	$$
	\pi\!-\!a.e.\ \epsilon, \ \ \ P_\nu^{(\epsilon)}\left(K \subset \limsup_{n\in \mathbb{N}}
	I_n(\xi_n)\right)=1.
	$$
	Finally using once more the desintegration we get
	\begin{eqnarray*}
		P_\nu\left(K \subset \limsup_{n\in\N} I_n(\xi_n)\right)=
		\int
		P_\nu^{(\epsilon)}\left(K \subset \limsup_{n\in \mathbb{N}}
		I_n(\xi_n)\right)
		d\pi (\epsilon) =1.
	\end{eqnarray*}

\subsection{Comparison with the classic Dvoretzky covering}
\begin{cor}\label{f-0}
	Let us consider a probability measure $\mu_f$ on $\mathbb{T}$ with density $f$.
	Let $F$ be a non-empty compact set contained in an open set $U$.

	1) Assume $f(x) \geq a$ for almost all $x \in U$. The set $F$
	is covered for the $\mu_f$-Dvoretzky covering if
	${\rm Cap}_{\Phi^{(a)}}(F)=0$.
	\iffalse
	\begin{equation}\label{f-1}
	\forall \sigma\int_{K} \int_{K} \exp{\left( \sum_{k=1}^\infty a( l_k - |t-s|)_+\right)}d\sigma(s) d \sigma(t)=\infty.
	\end{equation}
	then $F$ is $\mu_f$-Dvoretzky covered almost surely.
	\fi

	2)
	Assume $f(x) \leq a$ for almost all $x \in U$.
	The set $K$
	is not covered for the $\mu_f$-Dvoretzky covering if
	${\rm Cap}_{\Phi^{(a)}}(F)>0$.
	\iffalse
	 $\subset \mathbb{T}$, and for some $\sigma$
	\begin{equation}\label{f-2}
	\int_{K} \int_{K} \exp{\left( \sum_{k=1}^\infty a( l_k - |t-s|)_+\right)}d\sigma(s) d \sigma(t)<\infty.
	\end{equation}
	then $K$ will not be $\mu_f$-Dvoretzky covered.
	\fi
	\begin{proof}
		Let $h$ be a density function on $\mathbb{T}$ so that $h(x)=a$ for all $x \in U$.

		1) Assume 	${\rm Cap}_{\Phi^{(a)}}(F)=0$. Then, according to the local Kahane criterion (Theorem \ref{thm:K}), the set $F$
		is covered for the $\mu_h$-Dvoretzky covering. Then, by Theorem \ref{thm3},  the set $F$
		is covered for the $\mu_f$-Dvoretzky covering for
		$f|_U \ge h|_U$.

			2) Assume 	${\rm Cap}_{\Phi^{(a)}}(F)>0$. Then, according to the local Kahane criterion (Theorem \ref{thm:K}), the set $F$
		is not covered for the $\mu_h$-Dvoretzky covering. Then, by Theorem \ref{thm3},  the set $F$
		is not covered for the $\mu_f$-Dvoretzky covering for
		$f|_U \le h|_U$..
	\end{proof}
\end{cor}

\section{Sufficient conditions for covering $\mathbb{T}$}\label{sct-5}
In this section we discuss several sufficient  conditions for covering the circle, which is decomposed into $K_f$ and $\mathbb{T}\setminus K_f$.

\begin{prop}\label{l-1}
The set $\mathbb{T}\setminus K_f$ is covered for the $\mu_f$-Dvoretzky covering under the condition
\begin{equation}\label{eq12}
		    \forall a> m_f \,, \quad
		    \sum_{n=1}^\infty \frac{1}{n^2} \exp(a(\ell_1+\cdots +\ell_n))= \infty.
\end{equation}
If, additionally, $f$ satisfies the property \eqref{dom_1}, then the condition \eqref{eq12} will also be necessary for covering the set $\mathbb{T}\setminus K_f$.
\end{prop}

\begin{proof} Let $K_f^c =\mathbb{T}\setminus K_f$.
Let us assume \eqref{eq12}. Notice that  $K_f^c$ is an open set,
 by Proposition \ref{prop:EI}.  By definition,
$
\underline{E}_f(x) > m_f
$ for all $x \in K_f^c$.
\iffalse
This means that there exists $n \in \mathbb{N}$, so that
$$
\essinf_{{B(x,\frac{1}{n})}}f>m_f.
$$
But this, in its turn, implies that there is a number $a > m_f$, so that
$$
f(t)\geq a, \hbox{ for almost all }t \in B(x,\frac{1}{n}).
$$
\fi
Therefore, for every $x \in K_f^c$ there is an integer $n_x \in \mathbb{N}$ and a number $a_x>m_f$ such that
$$
f(t)\geq a_x, \hbox{ for almost all }t \in B(x,\frac{1}{n_x}).
$$
Thus, by Corollary \ref{f-0}, the interval $(x-\frac{1}{n_x}, x+\frac{1}{n_x})$ is covered for the $\mu_f$-Dvoretzky covering.
However, the open set $K^c_f$ is a union of countably many
such intervals $B(x,\frac{1}{n_x})$. So, $K_f^c$ is covered.

Now assume  the condition \eqref{dom_1}, which implies that for any
$\varepsilon>0$, there is $N \in \mathbb{N}$ such that
$
\overline{E}_f(x_N) < m_f + \varepsilon,
$
which implies that there is $k \in \mathbb{N}$ such that
$$
f(x)< m_f + \varepsilon, \hbox{ for almost all }x \in B(x_0, 1/k).
$$
Let $a = m_f + \varepsilon$.
Notice that $\varepsilon$ is arbitrary. By Corollary \ref{f-0}, the condition \eqref{eq12} must be necessary for covering $K_f^c$.
\end{proof}

\begin{cor}\label{cont}
If $f$ is continuous at some point of $K_f$, then the condition \eqref{eq12} is necessary and sufficient for covering the set $\mathbb{T}\setminus K_f$.
\end{cor}
\begin{proof}Assume $f$ is continuous at $x_0 \in K_f$.
	Then $
	\overline{E}_f(x_0)=m_f,
	$
	which clearly implies \eqref{dom_1} if we take $x_n=x_0$
	for all $n\ge 1$.

\end{proof}

\subsection{Sufficient conditions for covering $\mathbb{T}$}

\begin{prop}\label{suff}
Assume $m_f>0$.	The circle is covered for the $\mu_f$-Dvoretzky covering if
	\begin{equation}\label{suff-cond1}
	\forall a >m_f, \ \ \ \sum_{n=1}^\infty \frac{1}{n^2}e^{a(\ell_1+\dots + \ell_n)}=\infty;
	\quad {\rm Cap}_{\Phi^{(m_f)}}(K_f)=0.
	\end{equation}
\end{prop}
\begin{proof} The first condition in (\ref{suff})
	implies that $\mathbb{T}\setminus K_f$ is covered,  	Proposition \ref{l-1}. The covering of $K_f$ follows from
	the second condition in (\ref{suff}), Corollary \ref{f-0} and Kahane's result.
\end{proof}

\begin{prop}\label{c-2}
	If we have $m_f>0$, then the condition
	$$
	\sum_{n=1}^\infty \frac{1}{n^2} \exp(m_f(\ell_1+\cdots +\ell_n))= \infty,
	$$
	is sufficient for covering the circle.
\end{prop}
\begin{proof} It follows from  Corollary \ref{f-0} and Shepp's result, because
	$f(x)\geq m_f$ for almost all $x \in \mathbb{T}$.
\end{proof}

We now show that each of the conditions in \eqref{cond1} does not imply the other.

Let us recall the two conditions in \eqref{cond1} :
\begin{equation*}
{\rm (C1)} \ \ \ \forall a> m_f, \sum_{n=1}^\infty \frac{1}{n^2}e^{a(\ell_1+\dots + \ell_n)}=\infty; \qquad \qquad
{\rm (C2)} \ \ \ {\rm Cap}_{\Phi^{(m_f)}}(K_f)=0.
\end{equation*}
Consider the following density function
$$
f(x)=|x| + \frac{3}{4}, \hbox{ for } x \in \left[-\frac{1}{2},\frac{1}{2}\right).
$$
Observe that $m_f=\frac{3}{4}$, $K_f=\{0\}$  and $f$ is Lipschitz at $x=0$.  By Corollary \ref{cor1} (2), the conditions (C1) and (C2) are necessary and sufficient for covering the circle. Since $K_f=\{0\}$, the condition (C2) is equivalent to
$
\sum_{n=1}^\infty \ell_n = \infty,
$
which does not imply (C1). Indeed, when $\ell_n = \frac{3}{4n}$, we have
$
\sum_{n=1}^\infty \ell_n = \infty,
$ but
$$
\forall a<4/3, \ \ \ \sum_{n=1}^\infty \frac{1}{n^2}e^{a(\ell_1+\dots + \ell_n)}\approx\sum_{n=1}^\infty \frac{1}{n^{(2-\frac{3a}{4})}}<\infty.
$$
Observe however that in this case (C1) implies (C2).

We now show that (C1) does not imply (C2). To see this, let us look at the sequence
$
\ell_n = \frac{2}{n}-\frac{4}{n\ln n}
$
and the density function
$$
f(x)=
\frac{1}{2} 1_{[0,1/2)}(x) +
\frac{3}{2} 1_{[1/2, 1)}(x).
$$
We have that $m_f=1/2$ and $K_f = [0,1/2]$. From Theorem \ref{thm_seq}, the conditions (C1) and (C2) are necessary and sufficient for covering the circle.
Note that for all $a>1/2$
$$
\sum_{n=1}^\infty \frac{1}{n^2}e^{a(\ell_1+\dots + \ell_n)}\approx\sum_{n=1}^\infty \frac{1}{n^2}e^{(2a\ln n - 4a \ln \ln n)}=\sum_{n=1}^\infty \frac{1}{n^{(2-2a) } \ln^{4a} n}=\infty.
$$
Hence, (C1) is satisfied. Since $|K_f|>0$, according to Lemma \ref{lem-1} the condition (C2) is equivalent to \eqref{whole}, which reads as
$$
\sum_{n=1}^\infty \frac{1}{n^2}e^{\frac{1}{2}(\ell_1+\dots + \ell_n)}=\infty.
$$
However
$$
\sum_{n=1}^\infty \frac{1}{n \ln^{2} n}<\infty.
$$
Thus, the condition (C1) does not imply (C2) for all density functions $f$ and all sequences $\{\ell_n\}_{n \geq 1}$.
\medskip

\section{Necessary conditions for covering $\mathbb{T}$}

In this section we discuss  necessary conditions for covering the circle. The proof of the next proposition is based on Lebesgue's
differentiation theorem and Billard's criterion.

\begin{prop}\label{cond_seq}
Suppose that  the sequence $\{\ell_n\}_{n=1}^\infty$ satisfies the condition
\begin{equation}\label{s2}
\limsup_{n \rightarrow \infty}\frac{n \ell_n}{ \ell_1 + \cdots + \ell_n}<1.
\end{equation}
If $\mathbb{T}$ is covered for the $\mu_f$-Dvoretzky covering, then
\begin{equation}\label{suff-cond1-}
\forall a >m_f, \ \ \ \sum_{n=1}^\infty \frac{1}{n^2}e^{a(\ell_1+\dots + \ell_n)}=\infty.
\end{equation}
\end{prop}
\begin{proof} For  $\epsilon>0$, let
	$$
	A = \left\{ x\in \mathbb{T}:f(x)<m_f+ \frac{\epsilon}{2}\right\}.
	$$
By the definition of $m_f$,   $A$ has positive Lebesgue measure.
Consider
$$
S_n(x):=\frac{\mu_f(B(x, r_n))}{\ell_n}=\frac{1}{\ell_n}\int_{x-r_n}^{x+r_n}f(t)dt.
$$
By  Lebesgue's differentiation theorem,
$
S_n(x)$ converges to $f(x)$ almost everywhere. Then,
by Egoroff's theorem,   there is a compact set $A' \subset A$ with $|A'|>0$ such that $
S_n(x)$ converges to $f(x)$ uniformly in $A'$. Therefore, for all large enough $n$, we have
$$
\forall x \in A', \ \ \ |S_n(x)- f(x)|<\frac{\epsilon}{2}.
$$
It follows that for large $n$ and for  $x \in A'$ we  have
\begin{equation}\label{eqn:A}
\mu_f (B(x,r_n))\leq \Big(f(x) + \frac{\epsilon}{2}\Big) \ell_n \leq (m_f + \epsilon)\ell_n.
\end{equation}
Notice that
$
B(t, r_n)\cap B(s,r_n) \subset B(s, r_n)
$
and
$
f(x)\geq m_f$ for almost all $x \in \mathbb{T}$.
It follows that for all $s, t \in \mathbb{T}$
\begin{equation}\label{eqn:A2}
\mu_f(B(t, r_n)\cap B(s,r_n))
\leq m_f(\ell_n - |t-s|)_+ +(\mu
_f(B(s, r_n)-m_f \ell_n).
\end{equation}
Indeed, this is trivial if $|t-s|>\ell_n$. Otherwise the inequality reads as
$$
   \mu_f(B(t, r_n)\cap B(s,r_n)) \le \mu_f(B(s,r_n))- m_f |t-s|,
$$
which is  true, because $B(s, r_n)$ is the union of $B(s, r_n)\cap B(t, r_n)$ and an interval of length $|t-s|$.
Assume now $s\in A'$. From (\ref{eqn:A2}) and (\ref{eqn:A}) we get
\begin{equation} \label{eqn:A3}
\mu(B(t, r_n)\cap B(s,r_n)
\leq m_f(\ell_n - |t-s|)_+ +\epsilon\ell_n.
\end{equation}

We claim that there exists $0<\delta<1$ such that for all
$(t, s)\in A'\times A'$ with $|t-s|$ small enough we have
\begin{equation} \label{eqn:A4}
 \sum_{n=1}^\infty \mu_f(B(t, r_n)\cap B(s, r_n)
\le \Big(m_f +\frac{\epsilon}{1-\delta}\Big)\sum_{n=1}^\infty (\ell_n -|t-s|)_+.
\end{equation}
Indeed, assume $\ell_{N+1}<|t-s|\le \ell_N$ we have
\begin{equation*}\label{estim}
\sum_{n=1}^\infty \mu_f(B(t, r_n)\cap B(s,r_n)) = \sum_{n=1}^N \mu_f(B(t, r_n)\cap B(s,r_n)).
\end{equation*}
By the estimation (\ref{eqn:A3}), for $(t, s)\in A'\times A'$ we have
\begin{equation}\label{EST}
\sum_{n=1}^\infty \mu_f(B(t, r_n)\cap B(s,r_n)) \leq \sum_{n=1}^N (\ell_n-|t-s|)_+ +\epsilon \sum_{n=1}^N \ell_n.
\end{equation}
However $$
\sum_{n=1}^N (\ell_n-|t-s|)_+= \sum_{n=1}^N \ell_n-N|t-s|\ge
\sum_{n=1}^N \ell_n-N\ell_N.
$$
\iffalse
Consider now the following ratio
$$
\frac{(\ell_1-u)_+ +\cdots +(\ell_n - u)_+}{\ell_1 + \cdots + \ell_n}.
$$
Let $\ell_{n+1}<u \leq \ell_n$. Then
$$
\frac{(\ell_1-u)_+ +\cdots +(\ell_n - u)_+}{\ell_1 + \cdots + \ell_n}=\frac{\ell_1 + \cdots
 +\ell - nu}{\ell_1 + \cdots + \ell_n}
$$
$$
=1 - \frac{nu}{\ell_1 + \cdots + \ell_n}.
$$
But
$$
\frac{nu}{\ell_1 + \cdots + \ell_n}\leq \frac{n\ell_n}{\ell_1 + \cdots + \ell_n}.
$$
From the condition \ref{s2}, there should exist $0<\epsilon'<1$, so that for all large $n$
$$
\frac{nu}{\ell_1 + \cdots + \ell_n}\leq \epsilon'.
$$
\fi
By the assumption (\ref{s2}), for some $0<\delta <1$, when $N$ is large enough (i.e. $|t-s|$ is small enough), we have
\begin{equation}\label{inter}
\sum_{n=1}^N (\ell_n-|t-s|)_+ \ge (1- \delta)\sum_{n=1}^N \ell_n.
\end{equation}
Therefore, by
(\ref{EST}) and (\ref{inter}), we get
$$
\sum_{n=1}^\infty \mu(B(t, r_n)\cap B(s,r_n))
\leq \left(m_f+\frac{\epsilon}{1-\delta}\right)\sum_{n=1}^N (\ell_n - |t-s|)_+.
$$
This is what we have claimed for (\ref{eqn:A4}) because
$(\ell_n -|t-s|)_+=0$ for $n>N$.

\iffalse
Since $\epsilon$ was arbitrary and $\epsilon'$ was fixed, then by taking $N$ sufficiently large and $\epsilon$ sufficiently small we will have that
$$
d+\frac{\epsilon}{1-\epsilon'}<d + \varepsilon.
$$
Thus, for $N$ large enough, we will arrive at the following estimate
$$
\sum_{n=1}^N \mu(B(x, r_n)\cap B(x,r_n))\leq (d+\varepsilon)\sum_{n=1}^N (\ell_n-u)_+.
$$
\fi

We are ready to check (\ref{suff-cond1-}). By Lemma \ref{2-summable},
we can assume that
$
\sum_{n=1}^\infty \ell_n^2<\infty.
$
Then, by (\ref{eqn:A}), we have
$$
\sup_{t \in A'}\sum_{n=1}^\infty \mu(B(t, r_n))^2 =(m_f + \varepsilon)^2\sum_{n=1}^\infty \ell_n^2<\infty.
$$
Since the compact set $A'$ is covered and it has positive Lebesgue measure, we can apply Billard's local criterion (Theorem \ref{thm:B}) to confirm that
$$
\int_{A'} \int_{A'} \exp\sum_{n=1}^\infty \mu(B(t, r_n)\cap B(s,r_n))dt ds =\infty.
$$
Then, by the estimation (\ref{eqn:A4}) we get
$$
\int_{A'} \int_{A'} \exp \left((m_f+ \frac{\varepsilon}{1-\delta})\sum_{n=1}^\infty (\ell_k - |t-s|)_+ \right) dt ds  =  \infty.
$$
Notice that $0<\delta<1$ is fixed and $\epsilon >0$ is arbitrary.
So, we conclude for (\ref{suff-cond1-}) from the last equality and
Lemma \ref{lem-1}.

\end{proof}

\begin{prop}\label{t-1}
Suppose that $\mu_f$ admits a density $f$.
If  $K_f\cap F_f$ is covered, then ${\rm Cap}_{\Phi^{(m_f)}}(K_f\cap F_f)=0$.
\iffalse
: for any measure $\sigma$ supported on $K$ we have
\begin{equation}\label{cond3}
\int_{K_f} \int_{K_f} \exp{\left( m_f \sum_{k=1}^\infty (l_k - |t-s|)_+\right)}d\sigma(s)d\sigma(t)=\infty.
\end{equation}
\fi
\end{prop}

\begin{proof}
	If $K_f \cap F_f =\emptyset$, there is nothing to prove. Suppose then $K_f \cap F_f\not =\emptyset$.
Assume  $m_f=0$. Then by  the Borel-Cantelli lemma, any fixed point $x\in K_f$ is only finitely covered. Since $K_f\cap F_f$ is covered, we must have  $m_f>0$.
We can assume that $\sum_{n=1}^\infty \ell_n^2 <\infty$. Otherwise,
we get immediately ${\rm Cap}_{\Phi^{(m_f)}}(K_f)=0$
from Lemma \ref{2-summable} and Corollary \ref{c-2}.

Notice that $x \mapsto \mu_f(B(x, r_n))$ is continuous. So
the function
$\sum_{n=1}^\infty |\mu_f(B(x, r_n))-m_f \ell_n|$ is lower semi-continuous. Thus  the following sets are compact:
$$
K_m : =K_f\cap \left\{x\in \mathbb{T}: \sum_{n=1}^\infty |\mu_f(B(x, r_n))-m_f \ell_n|\le  m\right\}.
$$
\iffalse
Indeed, assume $x_0 \in \overline{K_m}$ and $x_0 \notin K_m$. Then there is $N_{K_m} \in \mathbb{N}$, so that
$$
\sum_{n=1}^{N_{K_m}}|\mu_f(B(x_0, r_n))-d \ell_n|>m.
$$
But from the continuous of the sum below in $x_0$
$$
\Phi(x_0)=\sum_{n=1}^{N_{K_m}}|\mu_f(B(x_0, r_n))-d \ell_n|,
$$
it follows that for $x' \in K_m$, sufficiently close to $x_0$,  we should have
$$
\sum_{n=1}^{N_{K_m}}|\mu_f(B(x', r_n))-d \ell_n|>m,
$$
which contradicts the assumption $x' \in K_m$. Thus,
 $K_m=\overline{K_m}$.
 \fi
 It is clear that $K_m$ is increasing and it tends to $K_f\cap F_f$. Therefore
 we have only to prove that ${\rm Cap}_{\Phi^{(m_f)}}(K_m)=0$
 for every $m\ge 1$.

By the inequality (\ref{eqn:A2}), for all $t, s \in \mathbb{T}$ we have
\iffalse
The following inclusion is straightforward
$$
B(t, r_n)\cap B(s,r_n) \subset B(s, r_n).
$$
It is not difficult to see that the following estimate holds
		  	$$
		  	\mu(B(t, r_n)\cap B(s,r_n)) \leq m_f (\ell_k - |t-s|)_+ + |\mu(B(s, r_n))-m_f \ell_n|.
		  	$$
From here
\fi
$$
\sum_{n=1}^\infty \mu(B(t, r_n)\cap B(s,r_n))\leq \sum_{n=1}^\infty m_f (\ell_k - |t-s|)_+ + \sum_{n=1}^\infty |\mu(B(s, r_n))-m_f \ell_n|.
$$
Assume $s\in K_m$. Then the last sum is bounded by $m$.
\iffalse
But for the second sum on the right, for $\forall s \in K_m$ we have
\begin{equation}\label{c-0}
\sum_{n=1}^\infty|\mu(B(s, r_n))-d \ell_n|< m.
\end{equation}
\fi
Therefore, for all $ s \in K_m$ and all $t\in \mathbb{T}$ we have
\begin{equation}\label{N1}
\sum_{n=1}^\infty \mu(B(t, r_n)\cap B(s,r_n))\leq \sum_{n=1}^\infty m_f (\ell_k - |t-s|)_+ + m.
\end{equation}
On the other hand, letting $a_n=|\mu(B(s, r_n))-m_f \ell_n|$, we have
\begin{equation}\label{N2}
\sup_{s \in K_m}\sum_{n=1}^\infty \mu(B(s, r_n))^2\le  \sup_{s \in K_m} \sum_{n=1}^\infty (a_n^2 + 2 a_n m_f\ell_n +(m_f\ell_n)^2)<\infty
\end{equation}
because $\sum \ell_n^2 <\infty$ and $\sum a_n \le m$.
\iffalse
In view of the Remark \eqref{bound} and  \eqref{c-0} we can assume
$$
\sum_{n=1}^\infty \ell_n^2 < \infty\hbox{ and } \sum_{n=1}^\infty a_n < m.
$$
From which it will follow
$$
\sum_{n=1}^\infty \mu(B(s, r_n))^2<\infty.
$$
\fi
By the local Billard criterion, for any probability
measure $\sigma$ on $K_m$ we have
\begin{equation*} \label{eq1}
\int_{K_m} \int_{K_m} \exp\sum_{n=1}^\infty \mu(B(t, r_n)\cap B(s,r_n))d\sigma(t)d\sigma(s) = +\infty,
\end{equation*}
\iffalse
\begin{equation}
\int_{K_m} \int_{K_m} \exp(\sum_{n=1}^\infty d (l_k - |t-s|)_+ + m)d\sigma(t)d\sigma(s).
\end{equation}
From which
\fi
which, together with (\ref{N1}), implies
\begin{equation}
\int_{K_m} \int_{K_m} \exp\sum_{n=1}^\infty m_f (\ell_k - |t-s|)_+d\sigma(t)d\sigma(s)=\infty.
\end{equation}
Thus we have proved ${\rm Cap}_{\Phi^{(m_f)}}(K_m)=0$.
\end{proof}

We now discuss the condition \eqref{s2} and  its relation to \eqref{suff-cond1-}.

\medskip

(A) {\em
The condition \eqref{s2} is not always fulfilled.}

 We will construct a decreasing sequence $(\ell_n)$ such that $
\sum_{n=1}^\infty \ell_n^2 < \infty
$ and
\eqref{s2} is not satisfied.
Let $1 = n_0 < n_1 < \cdots < n_k < \dots$ be a sequence of positive integers such that for all $k\geq 1$
\begin{equation}\label{d11}
\frac{\ln n_k}{n_k}\geq \frac{\ln n_{k+1}}{n_{k+1}}.
\end{equation}
Define $(\ell_n)$ as follows
$$
 \ell_n := \frac{\ln n_{k+1}}{n_{k+1}}, \ \ {\rm when}\ \
n_1 + \dots + n_k <n \leq n_1 + \dots + n_k + n_{k+1}.
$$
Clearly $(\ell_n)$ is decreasing and if $n_k$ grows fast enough, then
$$
\sum_{n=1}^\infty \ell_n^2 < \sum_{k=1}^\infty n_k \cdot \frac{\ln^2 n_k}{n_k^2}= \sum_{k=1}^\infty\frac{\ln^2 n_k}{n_k}< \infty.
$$
For $n = n_1 + \dots + n_k$, we have
$$
\frac{n \ell_n}{ \ell_1 + \cdots + \ell_n}=\frac{(n_1 + \dots + n_k) \frac{\ln n_k}{n_k}}{ \ln n_1 + \cdots + \ln n_k}.
$$
If $n_k$ grows fast enough, then
$$
{\overline{\lim_{n \rightarrow \infty}}}\frac{n \ell_n}{ \ell_1 + \cdots + \ell_n} \geq \lim_{k \rightarrow \infty}\frac{\frac{n_1}{n_k}\ln n_k+ \dots + \frac{n_{k-1}}{n_k}\ln n_k+ \ln n_k}{\ln n_1 + \cdots + \ln n_k}=1.
$$

(B) \eqref{s2} is not necessary for {\em  \eqref{suff-cond1-}}.

For this we modify the construction above. For all even $k$s, we alter the construction above as follows. For all $n \in \mathbb{N}$, with
$$
n_1 + \dots + n_k \leq n \leq n_1 + \dots + n_{k+1},
$$
we define
$$
\ell_n = \frac{c}{n},
$$
where $c>1$.
Note that
$$
\frac{\ln n_k}{n_k}\geq \frac{c}{n_1 + \dots + n_k}.
$$
Next we take $n_{k+1}$ so large that
$$
\sum_{n=1}^{n_1 + \dots  + n_{k+1}}\frac{1}{n^2}e^{a(\ell_1 + \dots +\ell_n)}>k.
$$
This is doable, since $ac \geq 1$.
As a result, we will get a sequence $(\ell_n)_{n \geq 1}$ for which \eqref{s2} fails, but for all $a\geq c$ the series in \eqref{suff-cond1-} diverges.

\iffalse
\section{Proof of Theorem. \ref{thm:non-cover}}

	  {\bf Proof of Theorem \ref{thm:non-cover}.}   Let
	  $$
	       Q_n(t) = \prod_{j=1}^n \frac{1- \chi_{B(\xi_n, r_n)}(t)}{1 - \mu(B(t, r_n))}.
	  $$
	  Notice that $Q_n(t)=0$ means $t$ is covered by one of the intervals $I_j$ ($j=1, 2,\cdots, n$). Let
	  $$
	          M_n =  \int_{\frac{1}{2}U} Q_n(t) dt.
	  $$
	  This is a martingale. Observe that $\mu(B(t, r_n)) \sim f(t) \ell_n$ if $t \in \frac{1}{2}U$.  We can estimate
	  $$
	     \mathbb{E} M_n^2 \approx
	     \int_{\frac{1}{2}U} \int_{\frac{1}{2}U} \exp\left(
	     f(\frac{t+s}{2})\sum_{n=1}^\infty (\ell_n -|t-s|)_+\right) dt ds.
	  $$
	  Since $f(x) \le a$ over $U$. The assumption implies that
	  the last double integral is finite. So, the martingale
	  $M_n$ converges in $L^2$-norm. Then almost surely,
	  there exists a point in $\frac{1}{2}U$ which is not covered.

	\fi

\section{Proofs of Theorem ~\ref{thm1}  , Corollaries ~\ref{cor1},~\ref{cor2} and Theorem \ref{thm_seq}}
\subsection{Proof of Theorem ~\ref{thm1}}

{\em Case $m_f=0$}.
By assumption, we have \eqref{dom_1}. Hence, by Proposition \ref{l-1} the condition \eqref{cond01} is necessary and sufficient for covering the set $\mathbb{T}\setminus K_f$.
Obviously, the condition \eqref{cond0} is necessary for covering $K_f$. Otherwise for some $x_0 \in \mathbb{T}$ we will have
$$
\sum_{n=1}^\infty \mu_f(B(x_0, r_n))<\infty,
$$
a contradiction to the fact that $x_0$ is covered.
Since $K_f$ is countable, then the condition \eqref{cond0} will also be sufficient for covering the set $K_f$.

{\em Case $m_f>0$}.
The sufficiency of the
first conditions in \eqref{cond1} follows from Proposition \ref{suff}. We now show its necessity.
As above, the necessity of the first condition in \eqref{cond1} follows from the assumptions \eqref{dom_1}.

The second fact in \eqref{cond1} implies that $K_f\cap F_f$
and $K_f \cap (a_n + K_f\cap F_f)$ are all covered by Kahane's result and
Corollary \ref{f-0}. The points in $K_f$, which are not covered by the
translates of $K_f\cap F_n$ are countable, these countable set are covered
according to the condition \eqref{cond0}.
\iffalse
According to Propositions \ref{l-1} and \ref{cond_seq}, both of the conditions \ref{l-1} and \ref{cond_seq} imply the first condition in \eqref{cond0}, i.e.
$$
\forall a> m_f, \sum_{n=1}^\infty \frac{1}{n^2}e^{a(\ell_1+\dots + \ell_n)}=\infty.
$$
But this, in view of Proposition \ref{l-1}, implies covering for the set $\mathbb{T}\setminus K_f$. Observe that from Proposition  \ref{l-1}, this condition is also sufficient for covering the set $\mathbb{T}\setminus K_f$. Thus, both of the condition \eqref{cond0} and \eqref{cond1} are necessary and sufficient for covering the circle for $m_f=0$.
\fi
Observe that when $m_f>0$, the condition \eqref{cond0} is automatically fulfilled. This is because $\mu_f(B(x, r_n)) \ge m_f \ell_n$
and $\sum \ell_n =\infty$. Actually we get more
$$
    \forall x \in \mathbb{T}, \quad \sum_{n=1}^\infty \mu_f(B(x, r_n)) =\infty.
$$
\iffalse
Indeed, by the Lebesgue differentiation theorem, for almost all $x \in \mathbb{T}$ we have
$$
\lim_{n \rightarrow \infty}\frac{\mu_f(B(x, r_n))}{\ell_n}= f(x).
$$
Hence, for any $\epsilon>0$, for almost all $x$ we have
$$
\sum_{n=1}^\infty \mu_f(B(x, r_n)) \leq C_x\sum_{n=1}^\infty(f(x) + \epsilon)\ell_n
$$
for some $C_x>0$. But $m_f>0$ implies $\sum_{n=1}^\infty \mu_f(B(x, r_n))=\infty$ for all $x \in \mathbb{T}$.  So, we must have $\sum_{n=1}^\infty \ell_n=\infty$. From this and in view of the estimate below, the condition \eqref{cond0} should hold for arbitrary $x \in \mathbb{T}$
$$
\sum_{n=1}^\infty m_f\ell_n \leq \sum_{n=1}^\infty \mu_f(B(x, r_n)).
$$
Observe, however, that the condition \eqref{cond0} is not automatic when $m_f = 0$.
\fi
So, we have only to check the  necessity of the second condition in \eqref{cond1}
which, in turn, follows from Proposition \ref{t-1}.
\iffalse
 Since for all $x \in F_f \cap K_f$ we have the flatness condition \ref{flat}, then from Proposition \ref{t-1}, it follows
\begin{equation}
\int_{F_f \cap K_f} \int_{F_f \cap K_f} \exp{\left( m_f \sum_{k=1}^\infty (\ell_k - |t-s|)_+\right)}d\sigma(s)d\sigma(t)=\infty.
\end{equation}
This proves the necessity of the second condition in \eqref{cond1}. The necessity of the first condition in \eqref{cond1} again follows from the assumptions \eqref{s2} and \eqref{dom_1}. To see that we also have covering for $K_f$, we recall that if there is covering for $F_f \cap K_f$, then there is also covering for $F_f \cap K_f + a$, for any fixed $a>0$. Thus, the existence of a $\mu_f$-Dvoreczky covering for the set $F_f \cap K_f$ implies covering for the set $\cup_{n \geq 1}\{K_0 + a_n\}$, where $a_n \in \mathbb{T}$, $n \geq 1$. By assumption
$$
K_f \subset \cup_{n \geq 1}\{F_f \cap K_f + a_n\} \cup K_0
$$
where $K_0$ is a countable set. Thus, we will have covering for $K_f$.
The sufficiency of the conditions \eqref{cond0} and \eqref{cond1} follows from Proposition \ref{suff}.
\fi

\subsection{Proof of Corollary ~\ref{cor1}}
\begin{proof}
By assumption, the set of points where we do not have a Lipschitz dominant $g \in \lip(U)$, is countable, hence the condition \eqref{cond0} will be necessary and sufficient for covering this set.
As for all $x \in K_f \cap F_f$, there is a set $U$ with $x \in U$ and a function $g_x=g \in \lip(U)$, so that $f(t)\leq g(t)$ for almost all $t \in U$ and $f(t)=g(t)$, for $t \in K_f \cap U$. Since $g$ is continuous at $x$ and $f(x)=g(x)=m_f$, then $f$ is continuous at $x$ too, so according to Corollary \ref{cont} the first condition in \ref{cond1} is necessary and sufficient for covering the set $K_f^c$. To finish the proof, we need to show that for $\forall x \in K_f$ we have the flatness condition.

Fix $x\in K_f$. Let $C$ be the  Lipschitz constant of $g$.
Hence
$$
\mu(B(x, r_n))\le\int_{x-r_n}^{x+r_n}g(y)dy \leq \int_{x-r_n}^{x+r_n}(C|x-y|+|g(x)|)dy
\le C r_n^2 + \ell_n f(x).
$$
Since $x \in K_f$, then $f(x)=m_f$. Hence
$$
|\mu(B(x, r_n))-m_f \ell_n|\leq C r_n^2.$$
From which
$$
\sum_{n=1}^\infty |\mu(B(x, r_n))-m_f \ell_n|\leq C \sum_{n=1}^\infty r_n^2.
$$

\indent {\rm (1)}
Let $m_f=0$ and $\sum_{n=1}^\infty \ell_n^2<\infty$, then
$$
\sum_{n=1}^\infty \mu(B(x, r_n)) < \infty.
$$
Hence, there will be no $\mu_f$-Dvoretzky covering.

\indent {\rm (2)}
If $m_f>0$, then we can assume
$$
\sum_{n=1}^\infty \ell_n^2 < \infty,
$$
from which
$$
\sum_{n=1}^\infty |\mu(B(x, r_n))-m_f \ell_n|< \infty.
$$

Thus, at $x$ we have the flatness property. To finish the proof we will need to apply Theorem \ref{thm1}.

\end{proof}

\subsection{Proof of Corollary ~\ref{cor2}}
\begin{proof}
Let $K = K_f \cap F_f$. By assumption, $K$  is of positive Lebesgue measure. Suppose $\mathbb{T}$ is covered. Then so is $K$ which supports the Lebesgue measure restricted on $K$,  and by Proposition \ref{t-1},  we have
\begin{equation}
\int_K \int_K \exp{\left( m_f \sum_{k=1}^\infty (\ell_k - |t-s|)_+\right)}d t ds=\infty,
\end{equation}
which is equivalent to the condition \eqref{whole}
according to Lemma \ref{lem-1}. Thus, the condition \eqref{whole} is necessary for covering the circle.
The sufficiency of the condition \eqref{whole} for covering the circle follows from  Proposition \ref{c-2}.
\end{proof}

\subsection{Proof of Theorem \ref{thm_seq}}
\begin{proof}
Observe that $\ell_1 + \dots + \ell_n = c\log n + O(1)$. Hence
\begin{equation}\label{c/n}
\sum_{n=1}^\infty \frac{1}{n^2}e^{a(\ell_1 +\cdots +\ell_n)}= \infty
\Leftrightarrow ac \geq 1.
\end{equation}

Assume
$
cm_f \geq 1.
$
\iffalse
We have
$$
\ell_1 + \dots + \ell_n = c \ln n + o_{n \rightarrow \infty}(1).
$$
Then
$$
\sum_{n=1}^\infty \frac{1}{n^2} \exp(m_f(\ell_1+\cdots +\ell_n))\approx C \sum_{n=1}^\infty \frac{1}{n^2} \exp(cm_f \ln n)=\sum_{n=1}^\infty \frac{1}{n^{2-cm_f}}.
$$
The last series will be convergence if and only if $cm_f \geq 1$.
\fi
By   Corollary \ref{c-2} and (\ref{c/n}) applied to $a=m_f$, the circle is covered.

Conversely assume that the circle is covered.
Remark that  the condition (\ref{s2})
is satisfied. Indeed,
\iffalse now that the sequence $\ell_n = \frac{c}{n}$ satisfies the conditions of Theorem \ref{cond_seq}. Indeed
\fi
$$
\frac{n \ell_n}{\ell_1 + \dots + \ell_n}=\frac{c}{c \ln n + O(1)}=o(1).
$$
Then by Proposition \ref{cond_seq}, we have
$$
\forall \epsilon >0, \ \ \ \sum_{n=1}^\infty\frac{1}{n^2}e^{(m_f+\epsilon)(\ell_1 + \dots + \ell_n)}=\infty,
$$
which, according to (\ref{c/n}), implies
$c(m_f + \epsilon)\geq 1$ for all $\epsilon>0$. Therefore
$
cm_f \geq 1.
$
\end{proof}

\medskip
\textit{Acknowledgement.}
The second author would like to thank the School of Mathematics and Statistics
of the Central China Normal University for its hospitality during his visit in the Spring semester of 2018, where the work was started.

\bibliographystyle{plain}

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text:
PAC → Archive → Vol. 83 (2011) → Issue 12 → Preface
Pure Appl. Chem., 2011, Vol. 83, No. 12, pp. iv
Jung-Il Jin
IUPAC's vision statement declares that the Union advances the worldwide role of chemistry for the benefit of Mankind. And one of its long-range goals states "IUPAC will utilize its global perspective and network to contribute to the enhancement of chemistry education, the career development of young chemical scientists, and the public appreciation of chemistry". In pursuit of this spirit, the Union established in 2000 the IUPAC Prize for Young Chemists and has been honoring since then outstanding young research chemists at the beginning of their careers by making annual awards.
The prizes are given for the most outstanding Ph.D. theses in the area of the chemical sciences, as described in 1000-word essays. For details, please refer to <www.iupac.org/news/prize.html> (IUPAC Prizes for Young Chemists). IUPAC awards up to five prizes annually, each comprising USD 1000 and travel expenses to the next IUPAC Congress.
As immediate Past President of IUPAC, I was honored to chair the prize selection committee of eminent chemists, who enjoyed reading essays of 47 applicants from 19 countries. The committee was extremely pleased to note the increase in the number of applicants to 47 from 29 last year. This increase is in excellent accord with IUPAC's worldwide celebration of the International Year of Chemistry throughout this year. After critical evaluation of the originality and excellence of the essays and research results, the committee decided unanimously to award 2011 Prizes to the following six chemists:
Rubén D. Costa, University of Valencia, Valencia, Spain: "Theoretical and experimental study of light-emitting electrochemical cells based on ionic transition-metal complexes: From the molecule to the device"
William Donald, University of California, Berkeley, CA, USA: "Ion nanocalorimetry: Measuring absolute reduction potentials and investigating effects of water on electron solvation and ion fluorescence"
Matthew Macauley, Simon Fraser University, Burnaby, BC, Canada: "Insight into O-GlcNAc protein modification using chemical and biochemical tools"
Bozhi Tian, Harvard University, Cambridge, MA, USA: "Design, synthesis, and characterization of novel nanowire structures for photovoltaics and intracellular probes"
Sheng Xu, Georgia Institute of Technology, Atlanta, GA, USA: "Oxide nanowire arrays for energy sciences"
Chengqi Yi, University of Chicago, Chicago, IL, USA: "Probing the AlkB family DNA/RNA repair enzymes with a chemical disulphide cross-linking approach"
All the awardees were invited to present posters on their research at the 43rd IUPAC World Chemistry Congress, San Juan, Puerto Rico, 30 July-6 August 2011. Upon IUPAC's invitation, five of the six winners offered review papers on their research topics for consideration as publications in Pure and Applied Chemistry (PAC). And the five refereed articles appear in this issue of PAC.
Finally, it is an honor and a pleasure to congratulate each of the winners and their supervisors for winning the 2011 IUPAC Prize for Young Chemists. It is hoped that each of them will continue to contribute to a bright future for chemical sciences and technologies and to take active roles in IUPAC bodies in the future.
IUPAC Immediate Past President and Chair of the IUPAC Prize Selection Committee
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The Chicago Syndicate: Is the Polish Government Under Attack from Organized Crime Groups?
Is the Polish Government Under Attack from Organized Crime Groups?
The United States remains "a very important ally" to Poland and that will not change despite the leak of disparaging remarks by the country's top diplomat, the president said Monday.
President Bronislaw Komorowski spoke a day after the magazine Wprost released a transcript in which Foreign Minister Radek Sikorski said Poland's alliance with the U.S. is worthless and even harmful for the eastern European country because it created a false sense of security. "The United States is our very important ally and partner," Komorowski told reporters in Warsaw.
Transcripts of bugged restaurant conversations between top officials are rocking Prime Minister Donald Tusk's center-right government and critics have urged that it resign.
Earlier Monday, Tusk said he will not dismiss officials whose compromising conversations were caught on tape in what he called a "criminal" action by "ill-intentioned people."
Sikorski asserted Monday that organized crime was behind the secret recordings. "The government was attacked by an organized crime group," Sikorski said in Luxembourg. "We don't yet know who stands behind it."
He offered no proof for his statement.
The magazine says the recordings came from a "businessman" who did not do the taping and were made in the private VIP rooms of Warsaw restaurants.
In the transcript released by Wprost, Sikorski used vulgar terms while telling the former finance minister, Jacek Rostowski, the Polish-U.S. alliance is not helping Poland. "The Polish-American alliance isn't worth anything. It is even harmful because it gives Poland a false sense of security," Sikorski said. "(We are) suckers, total suckers."
In Washington, State Department spokeswoman Marie Harf wouldn't comment directly on the tapes, but said "the United States and Poland have an incredibly strong relationship ... based on shared values."
"It's a key part of our alliances in that part of the world. And the crisis in Ukraine, I think, has made that even more the case, where we're confronting a shared threat together," Harf said.
Earlier, Wprost released a conversation between Central Bank head Marek Belka and Interior Minister Bartlomiej Sienkiewicz where they discussed how the bank could help the governing party win re-election in 2015, a seeming violation of the bank's independence.


STAUB, L. M. (2000). The Correlation Between Eating Breakfast and School Performance. National Undergraduate Research Clearinghouse, 3. Available online at http://www.webclearinghouse.net/volume/. Retrieved April 25, 2019 .
The purpose of this study was to determine if there is a positive relationship between eating breakfast before class and school performance. Ninety-six freshmen undergraduate students from Loyola University New Orleans participated in the study. These students were given memory tests where they had to recall eight words in the correct sequence in which they were recited. After taking the memory test, the participants filled out a survey where they had to rate their school performance in areas such as alertness, attention, mood, test anxiety, etc. The survey also asked whether they eat breakfast or not. The results indicated that breakfast does seem to have some effect on school performance, but not as much as hypothesized. Eating breakfast before class seems to improve students' memory, but it does not appear to have any real influence on other components of school performance.
INTRODUCTION Many people have often wondered why some students have more difficulty than others with their academic performance. Some students are better able to function in the school environment and to perform in a more positive way than others. For instance, why is it that one student may seem to be very alert in class, take wonderful notes, participate in all classes, and do well on tests, while others are not capable of doing any of those things? Researchers try very hard to provide reasons why these particular students are lacking in some very important abilities. They still do not know every cause for a lack of academic functioning ability, but through their research they have formed a few hypotheses that have to do with the cognitive functioning and performance of young individuals. There appears to be a link between the lifestyles of students and their cognitive abilities, as reflected in their academic performance. One specific lifestyle habit that has been studied is breakfast consumption. Some scientists (e.g., Simeon, 1989; Grantham-McGregor, 1989; Worobey, 1999; Pollitt, 1981-83, etc.) think that eating breakfast before school will help students perform better on the cognitive level. Research has been done on this topic because so many people could be helped if eating breakfast truly does have a positive effect on students' learning abilities and functioning in their classes (academic performance). Eating breakfast before attending school is a fairly easy task to do, and it may have more than just physical health advantages. Past research (e.g., Simeon & Grantham-McGregor, 1989; Benton & Parker, 1998; Worobey & Worobey, 1999; etc.) has indicated that eating breakfast can be beneficial to many aspects of mental health as well. There are some studies, in particular, that seem to highlight this matter of interest, and the results of the studies discussed below demonstrate the supposed positive relationship between eating breakfast and academic success. In 1989, an experiment was conducted that demonstrated the effects of breakfast on cognitive functions of three categories of children: stunted, nonstunted controls, and malnourished. The stunted children were those whose linear growth was retarded, which meant that the lack of proper nutrition hindered the complete growth of their bodies. The nonstunted controls, in this experiment, were the children who had no growth impairment. The malnourished were the children who were undernourished (lacking the essential vitamins, minerals, and nutrition) in early childhood because they came from poor households. The children, who were ages 9-10.5, were divided into three groups of 30. In this experiment, an abundance of tests were done to examine many aspects of cognitive performance. The first of these cognitive tests were the arithmetic, the digit-span, and the coding tests, which were all subtests of the Wechsler Intelligence Scale for Children. These tests were useful because they determined important aspects of cognitive functioning such as attention, math skills, short-term memory, speed, and accuracy. Fluency and listening comprehension tests were taken from the Clinical Evaluation of Language Functions, and these were also very helpful. The fluency test was used to see the children's capabilities to generate ideas, and the listening comprehension test was a good way of measuring attention, auditory short-term memory, and comprehension. Another test that was given was the Matching Familiar Figures Test (MFFT). This test examined the number of errors made and the time it took children to respond. These are good indicators of problem-solving ability and speed of response. The last of the tests was the Hagen's Central-Incidental task (HCI), which contained a central task and an incidental task, and it measured visual short-term memory and attention to task-irrelevant information. Although these tests made up a large portion of the experiment, there were some other components, as mentioned below, that were also significant because they may have had an effect on cognitive functioning as well. It was necessary for the experimenters to go to the homes of the children to obtain even more information (e.g., eating habits, social background, economic status) regarding the youngsters. After receiving all the necessary information and consent from the children's parents, the participants were randomly assigned to different groups: those receiving the breakfast and those not receiving the breakfast, and they took the same cognitive tests again to see what, if any, effects the breakfast had. It appeared that missing breakfast had more unfavorable effects on the cognitive functioning of the poorly nourished children (stunted and malnourished). Missing breakfast did not have a significant effect on the other participating children. The results of this study indicate that skipping breakfast could have a detrimental effect on the achievement of the undernourished children (Simeon & Grantham-McGregor, 1989). This study emphasized the impact that breakfast seemed to have on undernourished children; however, a review of the research on the effects of breakfast on cognition, which was conducted by Pollitt (1995) demonstrated a wider range of influence. In this study, the participants were children and teenagers who had to fast, and then take cognitive tests to see the influence of not eating breakfast on cognitive performance. The results of the studies showed that fasting had negative effects on the students' cognitive functioning, especially in those students who were already undernourished, as found in the previous study. The range of influence, however, was increased as shown in the studies that took place in Cambridge, Massachusetts (Pollitt, 1981) and Houston, Texas (Pollitt, 1982/1983), which demonstrated that breakfast does not only have beneficial effects on the cognitive performance of the under-nourished, but it also has been a positive influence on well-nourished, middle-class 9- 11-year-old children's problem-solving performance. The children who were in the experimental group were given waffles and syrup, margarine, orange juice, and milk for their breakfast, but those in the control group were only given a placebo (noncaloric, non-caffeinated drink). The next week, the groups switched, and the breakfast eaters were given the placebo and those who had received the placebo last week were given the breakfast instead. Both studies show that the children who received the breakfast made fewer mistakes on their tests than on the day they did not eat the breakfast. Although it is important to know that breakfast can have positive effects on students' performance, it is also essential that we know what specific aspects of breakfast make it beneficial. For example, another experiment exhibiting the effects of eating breakfast on cognitive processes dove deeper to examine the nutritional components of breakfast meals to see which ones had the most effect on people. This study focused more on the blood glucose levels of people who do eat breakfast, and whether or not that has something to do with cognitive functioning. In this study, three experiments were done where memory tests and intelligence tests were given to the participants, who were 33 university students, 16 men and 17 women, and the average age was 21.3 years. Word lists, spatial memory tests, the Wechsler Memory Scale, and abstract reasoning tests were also used. The data obtained from these experiments supported the notion that blood glucose does improve memory, based on the memory tests, but it does not have any real effects on intelligence. The results of one of the experiments did show that when people's blood glucose was manipulated by a glucose drink, their memory did improve (Benton & Parker, 1998). Glucose is the energy source for the brain and the nerves, which may account for some of the memory improvement (Morse & Pollack, 1988). The irony comes in; however, when the effects of carbohydrates are studied. Although carbohydrates are the source that glucose comes from, some studies (e.g., Spring, 1982/83) show that carbohydrates can have adverse effects on children's performance. A study was conducted that looked at the effects of specific types of breakfast meals on cognitive functioning. Spring, Maller, Wurtman, Digman, and Cozolino (1982/83) as cited in Logue (1986) observed that children who ate a meal high in carbohydrates did not perform as well on attention tasks as those who ate a meal high in protein. Spring and her colleagues supported the hypothesis that eating breakfast is positively related to success on cognitive tasks. They did, however, consider breakfast meals that consist of some protein to be better at improving cognitive functioning. The results of the mentioned studies, as well as others, have triggered people's interest in the relationship between breakfast and academic performance, which is why many school breakfast programs were started in hopes that they would benefit young individuals. There was an experiment that examined the impact of a two-year breakfast program in certain schools. The focus was on preschool-aged children, their intake of nutrients, and pre-academic performance. The main purpose was to find out the differences between breakfast eaten at home and breakfast eaten at school during the breakfast program. This experiment was broken down into two studies. The first study contained 20 children and the second study contained 19 children. One tested the children's cognitive performance by providing them with various memory and intellectual behavior tests, rather than general intelligence tests. The children would arrive at school, eat their breakfast, and, later that day, would be taken to a testing room to take the tests. They were only separated from their classmates during free play period, and they were always told that they were going to "play some games" to prevent them from getting nervous about the tests. The second study was similar to the first except for one factor. The second study had a control group, but the first one did not. The children in the first study were just compared to themselves on the days they ate at school and on the days they ate at home. However, the second study has a true control group that consisted of the students who were not participants of the school breakfast program. Both studies did come up with basically the same results: Calories from carbohydrates increased with those who ate the school breakfast, and the calories from refined sugar decreased, which is an added nutritional benefit. The performance on the pre-academic tasks also went up in those who ate what the school provided them with for breakfast (Worobey & Worobey, 1999). This is not the only experiment to involve the school breakfast program. Another study looking at the school breakfast program and academic and psychosocial success was conducted by Kleinman (1998) as cited on the website ("New Harvard Research Shows," 1998). The purpose of the research was to see if there was an improvement in children's school performance after participating in the breakfast program. In Philadelphia and two public schools in Baltimore, which had been using the school breakfast program for four months, twice as many students were eating breakfast. The study observed 133 students both before and after the start of the free breakfast program. The results showed that the children who participated in the breakfast program paid closer attention in class, received higher grades in math, and behaved in a more positive manner. There was also a big drop in the number of absences, tardiness rates went down, and depression, anxiety, and hyperactivity decreased when the students participated in the breakfast program. Much of the literature gained has supported the hypothesis that eating breakfast has a positive influence on students' performance in school. Some studies showed that breakfast had more of an effect on the cognitive performance of undernourished individuals than on students in general (Simeon & Grantham-McGregor, 1989). However, the studies in Cambridge (Pollitt, 1981) and in Houston (1982/83), which were reviewed by Pollitt (1995), demonstrated that breakfast also had positive effects on well-nourished people. Some of the studies also explained that the level of glucose had an effect on children's performance. For example, it appeared that those who consumed breakfast also raised their levels of glucose; therefore, it was thought that glucose might be the major contributor to academic success (Benton & Parker, 1998). Another important factor to be aware of is the type of breakfast consumed. For example, in the study of the impact of the two-year school breakfast program, there was a group who ate the school breakfast and the others who ate their breakfast at home. Everyone ate breakfast, but the school breakfast was the one that was more nutritional and well-balanced, as opposed to many of the breakfasts that the low-income families provided to their children at home (Worobey & Worobey, 1999). Hardly any studies have taken place that show the role breakfast plays in college students' lives. The current study examined the relationship between eating breakfast and academic performance of college students. It was hypothesized that there would be a positive correlation between eating breakfast and school performance. In the study, eating breakfast was defined as whether or not the children consume food (does not matter what type) in the morning before attending their classes. The other variable, school performance, was defined as alertness in class, participation in class, attention span in class, concentration in class, mood in class, the frequency of note taking in class, and the anxiety level in class. The objective was to persuade parents with the results of the research to instill good eating habits in their children, such as eating breakfast in the morning, so they might be able to better their school performance.
Ninety-six freshmen undergraduate students from Loyola University New Orleans participated in this study. The researchers from Loyola recruited these individuals. They got class credit or class participation points for participating in the study. The type of sampling used for this experiment was convenience sampling because these participants were easily accessible.
Two copies of the informed consent form (the participant's copy and the researchers' copy) were handed out before the study. The testing packages, which contained the survey and an extra sheet of paper with numbers on it, were also distributed. The numbered sheet of paper was provided so the participants could write down the words from the recited list that they could remember in the proper sequence. Scratch paper, which was used to write down the U.S. states, was distributed to anyone who did not have any at their disposal. There was also a list of words (fish, house, apple, light bulb, bird, leaf, dentist, snow), which was seen by only the researchers, that would be recited to the participants. The questionnaire consisted of two parts. Part A contained 7 questions asking the students to rate their academic performance in class. Some of the questions had to do with alertness in class, mood in class, test-taking anxiety, etc. Part B consisted of questions 1-4 regarding whether or not the individuals eat breakfast and, if they do, how often do they eat it and what does it consist of (see Appendix A for complete copy of questionnaire).
This was a correlational study that examined the relationship between healthy lifestyles and school performance. The variables in the study were healthy lifestyles and school performance. In the study, healthy lifestyles were operationalized as eating breakfast. The conditions were whether or not the students eat breakfast and how often they do. This was measured by asking the students questions regarding whether or not they consume breakfast, how often they do, and what it consists of. School performance was described as cognitive performance on a memory test that was measured by the scores they received on the test. It was also called academic performance as well, and it was measured by asking students to rate themselves in the following areas: alertness in class, attention span, mood in class, note taking frequency, test-taking skills, concentration in class, participation in class, and recall of information. Every participant was tested in a classroom setting, and the procedure took place in the same order for every group of participants. The experimenters also made sure that everyone took the memory test at the same time so no one would have an advantage. These were all means of control and helped to prevent confounds from hindering the accuracy of the study. Upon arrival at the testing site, the participants were asked to sign the informed consent forms that were given to them. There were two copies of the informed consent form and the participants were told to sign both because one was theirs to keep for their records and the other one was the researchers' copy. The researchers also told the participants to only fill in the address section of the form if they want the results of the group as a whole sent to them. Once these forms were completed, they were collected, and the participants were told to get out some scratch paper. If they did not have any, the researchers provided them with some. The participants were then given a memory test where the researchers recited a list of words in a certain order: fish, house, apple, light bulb, bird, leaf, dentist, snow. Before the words were recited, the participants were told to try to remember as many words as they could in the correct sequence. After hearing the list of words, they were told to write down as many of the U.S. states as they could recall for thirty seconds in order to distract their attention from the word list. After the thirty seconds were up, the the testing packages were handed out to the participants who were told to write down the words, in the correct order, that they could remember from the list that was recited to them. The experimenters told the students to write down the words on the numbered page that they had provided them with. Finally, after writing down the words, the instructions were given as to how to complete the survey. They simply told the students to read each set of instructions for each section of the survey because they were self-explanatory. They were then told that they would have ten minutes to complete the survey. They, then, instructed them to begin unless they had any further questions. Once the ten minutes were up, the testing packages were collected from the participants. They were debriefed verbally explaining that the purpose of the memory test and the survey was to see if there is a relationship between healthy lifestyles and school performance. They were also informed that if they have any further questions they could contact the researchers at the phone number given on the consent form. The participants were also thanked for taking part in the study.
RESULTS An independent groups t test was used to analyze the obtained data. The findings indicated that breakfast does have an impact on school performance, but not as much as previously thought. The results showed that there was a significant difference between the memory test scores of the breakfast eaters and those who did not eat breakfast before class. There were 54 students in the breakfast group and 42 students in the non-breakfast group. The mean memory test score of those who reported that they eat breakfast before class was 4.4, and the mean memory test score of those that do not eat breakfast was 3.4. See Table 1 to get a better view of the other important statistics such as standard deviation, t-value, and probability level. Although the findings indicate that there is a significant difference between the two samples' performance on the memory test. The effect size is only .13, which is rather low. Despite the significant results obtained for the memory test, there was no significant difference found between the breakfast group and the non-breakfast group on other areas of school performance including alertness, participation, attention span, concentration, mood, note-taking, and test-taking. Table 1 presents the descriptive and inferential statistics regarding these aspects of school performance.
DISCUSSION The present study was done to see if eating breakfast before class improves school performance. It was hypothesized that there is a positive relationship between eating breakfast before class and school performance. The results of the study supported the hypothesis partially but not completely. The findings indicated that there is a positive relationship between eating breakfast and memory. However, breakfast did not seem to impact the other types of school performance (alertness, participation, attention, concentration, mood, note taking, and test taking). These results are similar to those found in past research (e.g., Benton & Parker, 1998 and Pollitt, 1995) and they differ from the findings of other studies (e.g., Kleinman, 1998). The study conducted by Benton and Parker (1998) came up with similar results to the present study. In this study, it was found that breakfast did improve memory because those who ate breakfast before taking the memory tests did receive higher scores than those who did not eat breakfast. Like in the present study, breakfast was found to not have any real impact on any other types of school functioning. In the studies done by Pollitt (1981-83), children did better on cognitive tests, including memory tests, after eating breakfast than when they were only given a placebo (noncaloric, non-caffeinated drink). Both of these studies (Benton & Parker, 1998 and Pollitt, 1995) showed that breakfast has a positive impact on students' memory just like the present study concluded. However, there is another study (Kleinman, 1998) that reported findings that differ from those of the present research. In this differing study, students' school performance was measured both before and after the start of a school breakfast program, and the results showed that the students' school functioning improved after the start of the breakfast program. Unlike the present study, the students' attention increased and their anxiety and hyperactivity went down, among many other things. This study's findings indicated that breakfast really does positively effect other aspects of school performance besides just memory. The reason that the present study may not have obtained the same results as the one done by Kleinman (1998) could be due to the way that the variables were tested, and it could have limitations. One of the shortcomings of this present study was that the participants were not gathered randomly. Convenience sampling was used instead because the participants were easily accessed this way, and there was only a short amount of time for the research to be conducted. Another big limitation of the study was that self-report measures were used to rate school performance. In this case, it is not known whether or not the participants are actually telling the truth or if they are just making up answers. Because, it is not certain whether or not the students' rated their school functioning accurately, the self-reports are unreliable measures. For these reasons, more reliable tests that measure the same components of cognitive performance should be given in the future rather than self-reports, which give students room to lie and deceive. Random sampling should also be done in future research to allow for better representation of the true population. Although this study had many limitations, it also presented many strengths. The sample was very large, consisting of 96 college students. The larger the sample, the more representative of the true population it is. Another positive aspect of the study was that it examined the effects of breakfast on college students. Most of the past research has been done on younger students, which limits the representation of the population, but the present study broadens the representation. By looking at the relationship between breakfast and school performance in college students, one is able to see whether breakfast improves school performance in people in general or just young individuals. In addition to the strengths of the sample size and the participants that make up the sample, there was still another strong point of this study. The memory test that was given to the participants was valid because it really did measure what it was supposed to measure—memory. This study has many practical implications that can be very beneficial to society today, especially students. Because the breakfast group did do significantly better on the memory test than the non-breakfast group, it appears that eating breakfast before class can be very beneficial to students everywhere. For this reason, more school breakfast programs should develop and the government should provide more funding for these programs that will help their students both now and in the future. If these school breakfast programs are successful and it is proven that breakfast truly does have a positive effect on learning, there will be less reliance on medication (e.g., Ritalin and anti-sleep agents) to help people function. If people are aware of the positive effects of breakfast, they will be more likely to eat it more frequently, which would help them both mentally and physically. Not only does this study have many practical implications, but it also opens the door to a wide array of theoretical implications. This study contributes to the body of knowledge of health and school psychology. Every bit of data that we gain regarding this subject will help to further our understanding of the influence of breakfast. The more information collected through research, the more specific and representative the results will be. In the future, there will probably be much more detailed knowledge of this topic, and the true cause of breakfast's impact will eventually be known. Researchers will hopefully go further to test whether eating breakfast can have even more of an effect on people's lives besides just school performance. Although this study attempted to test students to see if there is a positive relationship between consuming breakfast before class and functioning in school, improvements should be made in the future. For example, although a significant difference was found between those that did eat breakfast and those that did not, the effect size (.13) was very small, which meant that other factors besides breakfast could have contributed to the scores of the memory tests. That is why future studies should use more reliable methods of testing for memory as well as for other components of school performance. In the future, the independent variable (whether or not the students eat breakfast) should be manipulated by the researchers to make sure that each person is placed in the correct category. More cognitive tests should be done repeatedly to see if the results are the same each time, which would determine if the tests were accurate and reliable. A variety of tests should be tried out to see which ones work the best and are the most valid. It appears that in the future, our body of research knowledge will be much larger, and we will be able to find out a lot more information about breakfast and learning. The present study, however, does provide a foundation and a start for further research to be built on.
REFERENCES Benton, D., & Parker, P.Y. (1998). Breakfast, blood glucose, and cognition. American Journal of Clinical Nutrition, 67, 772S-778S. Kleinman, R. (1998, March). New Harvard research shows school breakfast program may improve children's behavior and performance. Kidsource Online. Available: http://www.kidsource.com/kidsource/content4/breakfast.html Logue, A.W. (1986). The Psychology Of Eating And Drinking. New York: W.H. Freeman and Company. Morse, D.R., & Pollack, R.L. (1988). Nutrition, Stress, And Aging. New York: AMS Press, Inc. Pollitt, E. (1995). Does breakfast make a difference in school? Journal of the American Dietetic Association, 95, 1134. Simeon, D.T., & Grantham-McGregor, S. (1989). Effects of missing breakfast on the cognitive functions of school children of differing nutritional status. American Journal of Clinical Nutrition, 49, 646-653. Worobey, J., & Worobey, H.S. (1999). The impact of a two-year school breakfast program for preschool-aged children on their nutrient intake and pre-academic performance. Child Study Journal, 29, 113.
Directions:  Please list as many words that you can remember from the list in the space provided.  Try to list them in the order in which they were called out to you.
Part A: On each of the following questions, please circle the choice that best defines you.
2.  How often do you participate in your classes before lunch?       Very Often  Often  Sometimes  Very Little  Never      1         2        3          4          5                                              3.  How long is your attention span in class before lunch?
4.  How hard do you find it to concentrate in class before lunch?
7.  During tests, how much anxiety do you generally experience in classes before lunch?
Part B: Please present us with the following information about yourself.
2.  If you answered yes to question #1, how often do you eat breakfast?
Every day  Often  Sometimes  Very Little                    1       2        3           4                 3.  If you answered yes to question #1, when are you most likely to eat breakfast?
4.  For the following list, please put a check next to the item(s) that you tend to eat/drink for breakfast.  You may check as many that apply to you.  Also, please circle how often you tend to eat/drink the item(s) you check off.


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Arsenal 4 - Dundalk 2 match report: Hale End FC
The Gunners cruised to victory in a match that didn't mean much for either side.
By Aaron Lerner@AaronCLerner Dec 10, 2020, 3:21pm EST
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Arsenal maintained their 100% winrate in the Europa League with a 4-2 win over Dundalk. The Gunners are the only team in European competition to still be unbeaten, un-drawn, by the way. It wasn't the best played match, but on a rainy night, sloppy pitch, and with nothing for either team to play for, what did you expect?
There were some definite highlights and a few lowlights from the match.
Mo Elneny's missile of a goal, top bins. The lad doesn't score many (it was his first for Arsenal in three years), but when he does, they are absolute bangers.
Emile Smith Rowe's strong play, generally. He made the most of his first senior start in a while. He provides mobility and makes runs that very few, if any, of the other guys on the roster make. Would love to see him start getting run-outs in the Premier League.
Joe Willock and Ainsley Maitland-Niles had solid games. They have so much more mobility than Granit Xhaka does that I think it may be time to give one of them a try in lieu of him. Xhaka is pretty clearly a better passes than both though, so I get why Mikel Arteta keeps picking him.
Eddie Nketiah keeps scoring goals. Today, he scored from 12 yards out, which is easily the longest goal he's scored in his Arsenal career. Goal-poaching is a skill, and he's got it.
Cedric Soares is aggressively mediocre. He's fine, I guess, as a Premier League backup, but he's not bringing anything to the table that a younger player at the same position would. Giving him a four-year deal will always be a head-scratcher for me. As for those of you who'd rather he start over Hector Bellerin, I just don't see it.
Nicolas Pepe had a quieter outing tonight. He was great in the other two Europa League matches and wasn't as good against Dundalk. That maddening inconsistency is what makes him so frustrating as a player.
Alex Runarsson could have and probably should have had both Dundalk goals. He's a, uh, let's say "work in progress" in goal. I really hope Bernd Leno stays healthy.
Great to see Pablo Mari and Calum Chambers back on the pitch after their lengthy injury layoffs.
Another appearance, another goal for Folarin Balogun. Yes, it's late in matches against Europa League competition, but I've really been impressed with him. Strong hold-up play, turns really well with the ball, and can finish. Arsenal need to do all they can to keep him at the club.
Massive congratulations to Ben Cottrell and Miguel Azeez on their senior debuts. I feel like we say "congrats" and move on, but stop to think about it for a second. I know they're young, but stepping onto the pitch tonight was, for both of them, what they've been working towards their entire lives. That must be a wonderful feeling, and I hope they and their families enjoy it. Both had nice little plays, too — Cottrell with a cheeky, gutsy chipped ball to bypass the press and start the break, Azeez with an ankle-breaker of a turn in midfield.
All told, it was a good evening for the Gunners. The knockout stage draw is on Monday, and we'll let you know who Arsenal are paired with. The Gunners are home to Burnley on Sunday.