text:
home > epc > summer 2002 > financing within the smo industry
Financing Within the SMO Industry
In the early 1990s a new phenomenon in the conduction of clinical trials began to emerge. Up until then physicians conducted clinical trials within their own practices. GPs and specialists recruited trial patients from their own databases. However, Good Clinical Practice (GCP) guidelines subsequently became more stringent and physicians involved in day-to-day health care became increasingly reluctant to concentrate on clinical trials.
The concept evolved of an organisation dedicated to conducting clinical trials with specialist physicians and nurses. The organisations were initially called trial management organisations and later became known as site management organisations (SMOs).
They were company-owned sites which had the best control over the specific trial circumstances. Because of this, an SMO had to cover a number of fixed costs such as physicians' and nurses' salaries. As SMOs have to at very least cover their costs in order to stay in business, I have tried to summarise the cost factors and income in this article. This summary is based on a practical evaluation.
By Dr Henk Mulder of Osteosupport
Dr Henk Mulder achieved his PhD in Internal Medicine and Endocrinology. He wrote his thesis 'Calcitonine concentration in blood in several disease sates' in 1980 at the University of Amsterdam.
After two decades of working as an Internist/Endocrinologist for the university hospital in Leiden and at a local hospital in Rotterdam, he founded the first European site management organisation named Good Clinical Practice (GCP). He participated in over 100 clinical trials as a Principal Investigator.
In early 2001 he left GCP and in June 2001 he founded Osteosupport - a hybrid company with integrated medical, paramedical and clinical trial activities. Osteosupport has been successfully involved in conducting three studies in the areas of arthritis, hypertension and male hormone replacement therapy.
Dr Henk Mulder
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If you asked 100 random people today who in their opinion is the most known investor, probably 70 would mention Warren Buffet, the icon of investing in stocks. The legend that has been built around the "Oracle of Omaha" over the last 40 years is indeed impressive.
While I very often agree with his views regarding, for example, the level of cash in portfolio or migration from growth to value stocks, I absolutely can not agree with what he wrote in the letter to shareholders about gold, once again showing how badly it performs in comparison to the US shares.
What is the probability that such a level will last for a long time? Close to zero. In order to show you how owning such overvalued one asset class in a portfolio ends, I will show you the results of Buffett's strategy during several periods that the management of Berkshire Hathaway would probably like to forget.
In 1973 (beginning of research) – 1980, Buffett's strategy suffered real losses (after taking inflation into account) in the amount of 4.78% per annum for 7 consecutive years, reaching a cumulative loss of 38.6%… a lot.
The following years are a period of very good results. First, because in 1981, the US stock exchange had been extremely cheap. Secondly, because in the next decade (1990-2000), capital from Japan (after the bubble had burst there) had moved to the USA. Ultimately, two decades of decent growth had ended in a bubble on technology stocks.
The story, as you know, likes to repeat itself and the years after 2000 have not been good for Buffett's strategy for the same reason as three decades earlier. Very high valuations!!! Well, owning US shares for the first decade of the 21st century has generated a real loss of 2.91% per annum. Apparently it is low, but the compound interest for 10 years has made its job. We can not forget that in 2008, a significant part of the companies of the Buffet's portfolio had been given 800 billion dollars of non-returnable assistance at the expense of taxpayers in the form of TARP without which many companies would have disappeared from the market.
Absolutely not, especially nowadays. First of all, because of the volatility, which is higher when the concentration of one asset class in our portfolio is bigger, by owning 90% of US shares in our portfolio, we risk losing almost 50% of our capital in just one year (max drawdown). How many of us after such a big loss would be willing to buy cheap stocks? Hardly anyone. Such a big loss will mentally exhaust many people. What is even worse, if knowledge and common sense prevail in some of us to buy these cheap stocks, we still only have 10% of the capital that we have kept in cash for special occasions.
What I absolutely want to emphasize is the fact that the 1970s belonged to commodities and precious metals, not shares. As you can see below, virtually all portfolios based on shares or bonds have been hurt heavily. The model based on the Buffett strategy has lost an average of 3.5% annually, 60/40 (stocks / bonds) over 4%. Bonds has lost the most – over 5%, as a result of drastic increase in inflation and interest rates.
b) At the beginning of the 1970s, we also had very undervalued commodities (below you can see their comparison against stocks).
At last, a very important difference between the actions of the legend and the strategy attributed to it is the fact that he is the insider with particularly good access to information and politicians, which greatly encourages investment in both shares of individual companies and acquisitions of them.
I really respect Buffett's approach to investment. His results, especially in the last decade, are very good, but his portfolio is too dependent on the US stock exchange, which has been demonstrated by 48% decreases after Lehman's bankruptcy, and in that time, shares in the US have not been as overvalued as it is now.
Note: This article appeared at ValueWalk.com. It was contributed by Independent Trader.


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text:
Q: Why List return empty when execute 2nd event Why List return empty when execute 2nd event?
List<string> ErrorList = new List<string>();

protected void Page_Load(object sender, EventArgs e){}

protected void btnFirst_Click(object sender, EventArgs e)
{
   for (int i = 0; i < 5; i++)
       {
          ErrorList.Add(i);
       }
   txtResult.Text = "Length of list: " + ErrorList.Count;
 }

protected void btnSecond_Click(object sender, EventArgs e)
{
   txtResult.Text = "Length of list: " + ErrorList.Count;
 }

When click btnFirst: txtResult.Text = "Length of list: 5"
When click btnSecond: txtResult.Text = "Length of list: 0"

A: Try this code but it only work if cookies enabled on client side
The session is a container located server side. In the cookie on client (browser) is a id to make a relation with the session. It is also possible to save more informations in a cookie but i would not recommend it. All other informations infos are lost every request.
protected void Page_Load(object sender, EventArgs e){}

protected void btnFirst_Click(object sender, EventArgs e)
{
   List<string> ErrorList = new List<string>();

   for (int i = 0; i < 5; i++)
   {
        ErrorList.Add(i);
   }
   Session["Errors"] = ErrorList;
   txtResult.Text = "Length of list: " + ErrorList.Count;
 }

protected void btnSecond_Click(object sender, EventArgs e)
{
   List<string> ErrorList = (List<string>)Session["Errors"];
   txtResult.Text = "Length of list: " + ErrorList.Count;
}

redpajama_set_name: RedPajamaStackExchange


You can import any file, choose the settings for samplerate, bitdepth, sign, channel number and byte order.
You can also select an offset to skip the first n bytes of the file.
You need the lame_enc.dll to be present in the Wavosaur folder (or system folder).
This feature exports all the data regions delimited by markers in separated files.
Go to menu Files -> Export -> Export as text This feature exports the current files in text format, for using in application such as Excel, Matlab, PSpice etc.
It's not recommended to export long files as text!


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The 8.3 earthquake that shook Samoa on September 29 triggered a tsunami that devastated the south-coast of Upolu island, the south-coast of Tutuila in American Samoa and the Niueas group in Tonga. People on the coast had around 5- 10 minutes warning to evacuate – many did not make it.
The districts of Aleipata and Falealili on the south-east coast of Upolu have been the most extensively damaged with entire villages and all hotels and small resorts annihilated along the coast.
The mounting death toll stands now at 178 with many people not yet accounted for. A total of 40 villages have been affected and the homeless population is estimated now over 10,000. This represents 6% of the total population of Samoa being affected by the tsunami.
The Aleipata District is where CI is most active in Samoa in partnering and supporting local communities marine protected areas (MPAs), as part of the Coral Reef Initiative in the Pacific (CRISP) program. We are also the only International NGO in Samoa, so we are liaising closely with the relief effort agencies, the Red Cross, the Disaster Management Unit, SPREP and UNDP.
Beyond the immediate relief work, there will be a much larger longer-term need for reconstruction, community income-generation initiatives and ecosystem restoration as the reef, we suspect, has been extensively damaged. We also hope to be able to fully capitalize the MPA Trust Fund set up by the local communities to ensure sustainability of the reconstruction and conservation initiatives. Communities income in these affected areas is derived mainly from coastal tourism and subsistence fishing, both heavily affected by the tsunami.
CI, as the only international NGO in Samoa, and with our long-established partnerships with communities of the south-coast, is well positioned to provide long-term support. Our technical staff, Sue Taei (marine) and James Atherton (terrestrial) are both involved in the Government of Samoa mandated Tsunami Environmental Impact Assessment and the team is seconded to UNEP, jointly with MNRE, UNESCO and SPREP to conduct the first rapid field assessment of all coastal villages impacted by the tsunami.
This is a real human tragedy at the scale of our small island, so we do thank all of CI staff and our larger CI family who have expressed their condolences and have shared their thoughts with our small team in Samoa.
Francois Martel is the Executive Director of the Pacific Islands Program at Conservation International. The program's headquarter is located in Apia, Samoa.


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text:
\section{Introduction}

A recurring theme in descriptive set theory is that of analyzing the descriptive (or definable) complexity of naturally occurring sets from other areas of mathematics.  In the present work, we consider certain sets which arise in ergodic theory.

Suppose that $T: [0,1] \rightarrow [0,1]$ is a Borel map which preserves Lebesgue measure.  It is of interest to consider those points $x \in [0,1]$ which exhibit ``chaotic'' or random behavior with respect to $T$ and its iterates.  For instance, one might consider those points $x$ such that $\{ T^n (x) : n \in \NN\}$ is dense in $[0,1]$.  (Here $T^n (x)$ denotes the $n$-fold iterate of the map $T$, applied to $x$.)  Given such a $T$, there is another type of chaotic behavior that is related to uniform distribution.  Specifically, one considers those points $x$ such that the sequence $x , T(x) , T^2(x) , \ldots$ is {\em uniformly distributed}, that is, for each subinterval $I \subseteq [0,1]$,
\[
\lim_{n \rightarrow \infty}\frac{\mathrm{Cardinality}( \{ i < n : T^i (x) \in I \} )}{n} = \mathrm{length} (I).
\]
If one considers the transformation $T(x) = bx \, (\mathrm{mod}\ 1)$, for some fixed integer $b \geq 2$, then an $x \in [0,1]$ exhibiting this type of chaotic behavior is called {\em normal to base $b$}.  It can be shown that $x \in [0,1]$ is normal to base $b$ if, for each integer $k \geq 1$ and $m < b^k$,
\[
\lim_{n \rightarrow \infty}\frac{\mathrm{Cardinality}( \{ i < n : m/b^k \leq T^i (x) < (m+1)/b^k \} )}{n} = 1/b^k.
\]
In turn, this is equivalent to the combinatorial statement that every finite string, $\sigma$, of numbers $0$ through $b-1$ occurs in the $b$-ary expansion of $x$, with frequency (in the limit) $b^{-\mathrm{length} (\sigma)}$.  It is a consequence of the Birkhoff Ergodic Theorem that, for each integer $b \geq 2$, the set of $x \in [0,1]$ which are normal to base $b$ has Lebesgue measure $1$.

Next we introduce some notions from descriptive set theory.  Recall that a {\em pointclass} is a family of sets which can be described in any complete separable metric space, e.g., the classes of $F_\sigma$, $G_\delta$, or analytic sets are all pointclasses.  In the present work, the pointclasses we consider are $\mathbf \Pi^0_3$, $\mathcal D_2 (\mathbf \Pi^0_3)$ and $\mathcal D_\omega (\mathbf \Pi^0_3)$.  The class $\mathbf \Pi^0_3$ is that of sets which have the form $\bigcap_{m\in \NN} \bigcup_{n\in \NN} F_{m,n}$, with each $F_{m,n}$ a closed set.  The class $\mathcal D_2 (\mathbf \Pi^0_3)$ is that of sets having the form $A\setminus B$, with $A,B \in \mathbf \Pi^0_3$.  Finally, sets in $\mathcal D_\omega (\mathbf \Pi^0_3)$ have the form $\bigcup_k A_{2k+1} \setminus A_{2k+2}$, where $A_1 \supseteq A_2 \supseteq \ldots$ are $\mathbf \Pi^0_3$ sets.

\begin{definition}
If $\Gamma$ is a pointclass, we say that $X \in \Gamma$ is {\em $\Gamma$-complete} iff every $Y \subseteq \{0,1\}^\NN$, with $Y \in \Gamma$, is a continuous preimage of $X$.
\end{definition}

\noindent In a sense, a $\Gamma$-complete set ``encodes'' all $\Gamma$ subsets of $\{0,1\}^\NN$.

There are many well-known $\mathbf \Pi^0_3$- complete sets.  For instance, the set
\[
\{ x \in \NN^\NN : \lim_{n \rightarrow \infty} x(n) = \infty\}.
\]
(See \S23 of Kechris \cite{KECHRIS.dst}.)  In 1994, Haseo Ki and Tom Linton \cite{ki.linton} published an interesting completeness result related to the set of numbers normal to base $b$.

\begin{theorem}[Ki-Linton]
The set of real numbers which are normal to base $b$ is $\mathbf \Pi^0_3$-complete.
\end{theorem}

More work in this vein has been done subsequently by others.  For instance, Ver\'onica Becher, Pablo Heiber and Ted Slaman \cite{becher.heiber.slaman.2} showed that the set of real numbers which are normal to all bases is also $\mathbf \Pi^0_3$-complete.  In other work, Becher and Slaman \cite{becher.slaman.sigma4} have shown that the set of numbers normal to at least one base is $\mathbf \Sigma^0_4$-complete.

In general, however, there are not many known completeness results for difference classes, e.g., $\mathcal D_2 (\mathbf \Pi^0_3)$ and $\mathcal D_\omega (\mathbf \Pi^0_3)$.  In what follows, we shall prove completeness results for the classes $\mathcal D_2 (\mathbf \Pi^0_3)$ and $\mathcal D_\omega (\mathbf \Pi^0_3)$.  Before stating our result, we introduce some more terminology.

For the present work, we mostly restrict attention to the case of $b=2$, as this will simplify our notation somewhat.  As a weakened form of normality, one may consider those $x \in [0,1]$ which are {\em order-$k$ normal} in base 2.  That is, such that, for each $j < 2^k$,
\[
\lim_{n \rightarrow \infty}\frac{\mathrm{Cardinality}( \{ i < n : T^i(x) \in [j/2^k , (j+1)/2^k) \} )}{n} = 2^{-k},
\]
where $T:[0,1] \rightarrow [0,1]$ is the map $x \mapsto 2x \, (\mathrm{mod} \ 1)$.  Let $N_k$ denote the set of numbers which are order-$k$ normal in base $2$.  Note that a real number $x$ is normal iff it is order-$k$ normal, for each $k \geq 1$.

Examining the proofs in Ki-Linton \cite{ki.linton}, one may deduce the following theorem.

\begin{theorem}[Ki-Linton]
The sets $N_1$ and $N_2$ are $\mathbf \Pi^0_3$-complete.
\end{theorem}

Inspired by this observation, we proved the following result.

\begin{theorem}\label{T1}
The set $N_1 \setminus N_2$ is $\mathcal D_2 (\mathbf \Pi^0_3)$-complete.
\end{theorem}

\begin{corollary}
The set $N_1 \setminus N_2$ is properly $\mathcal D_2 (\mathbf \Pi^0_3)$.
\end{corollary}

The method of our proof is somewhat different from that of Ki-Linton.  Specifically, we employ a permitting structure to construct a reduction of an arbitrary $\mathcal D_2 (\mathbf \Pi^0_3)$ set to $N_1 \setminus N_2$.  Our task is necessarily complicated by the fact that there are not many combinatorially tractable sets which are known to be $\mathcal D_2 (\mathbf \Pi^0_3)$-complete.

In response to a question posed, in conversation, by Su Gao, we extended the method used to prove Theorem~\ref{T1} and obtained the following result.

\begin{theorem}\label{T2}
The set $\bigcup_k N_{2k+1} \setminus N_{2k+2}$ is $\mathcal D_\omega (\mathbf \Sigma^0_3)$-complete.
\end{theorem}

\section{Preliminaries and notation}

We now introduce some notation which largely follows Kechris \cite{KECHRIS.dst}, our principal reference for descriptive set theory.

Let $\langle \cdot , \cdot \rangle : \NN^2 \rightarrow \NN$ be a fixed bijective pairing function such that, for fixed $m \in \NN$, the sequence $\langle m , 0\rangle , \langle m , 1 \rangle , \ldots$ is increasing.  Likewise, we fix a bijective ``triple function'' $\langle \cdot , \cdot , \cdot \rangle : \NN^3 \rightarrow \NN$.

Let $\{0,1\}^n$ denote the set of finite binary strings of length $n$, $\{0,1\}^{\leq n}$ denote the set of binary strings of length not greater than $n$, and $\{0,1\}^{<\omega}$ denote the set of all finite binary strings (of all lengths).  Let $\{0,1\}^\NN$ denote the set of all infinite binary sequences, equipped with the product, over $\NN$, of the discrete topology on $\{0,1\}$.  For $x \in \{0,1\}^\NN$, let $x(n)$ denote the $n$th term of $x$ and let $x \upto n$ denote the finite string $(x(0) , \ldots , x(n-1))$.  For $\sigma \in \{0,1\}^{< \omega}$, let $[\sigma]$ denote the basic open set \[
\{ x \in \{0,1\}^\NN : \sigma \mbox{ is an initial segment of } x\}.
\]
For $\sigma , \tau \in \{0,1\}^{< \omega}$, let $\sigma \concat \tau$ denote the concatenation of $\sigma$ and $\tau$.  For $\alpha \in \{0,1\}^{< \omega}$, let $\alpha^n$ denote the $n$-fold concatenation of $\alpha$ with itself.  Similarly, let $\alpha^\infty$ denote the infinite binary sequence $\alpha \concat \alpha \concat \ldots$.  For $\sigma \in \{0,1\}^{< \omega}$, let $|\sigma|$ denote the length of $\sigma$.

For $a \in \ZZ$ and $x \in \{0,1\}^\NN$, let $a \, . \, x$ denote the number
\[
a + \sum_{n = 0}^\infty x(n) / 2^{n+1}.
\]
That is, $a \, . \, x$ is the least real number greater than or equal to $a$ whose fractional part has binary expansion $x$.

If $\alpha$ and $\sigma$ are finite binary strings, with $|\alpha \| \leq |\sigma|$, let $d_\alpha (\sigma)$ be
\[
\frac{\mathrm{Cardinality}\big(\{ i <  |\sigma| - |\alpha| : (\exists \beta \in \{0,1\}^{< \omega}) (\sigma \upto (i + |\alpha|) = \beta \concat \alpha)\}\big)}{|\sigma|}.
\]
In other words, $d_\alpha (\sigma)$ indicates the proportion of substrings of $\sigma$ which are equal to $\alpha$.

For the rest of this paper, we will use the following well-known equivalent definition of order-$k$ normality in base 2, rather than that introduced in the previous section.

\begin{definition}
A real number, $a \, . \, x$, is {\em order-$k$ normal} in base 2 iff, for each $\alpha \in \{0,1\}^k$, the sequence $(d_\alpha (x\upto s))_{s \in \NN}$ is convergent, with
\[
\lim_{s \rightarrow \infty} d_\alpha (x \upto s) = 2^{-k}.
\]
We let $N_k$ denote the set of all order-$k$ normal numbers in $[0,1]$.
\end{definition}

Proving that this definition is equivalent to the one given earlier is a relatively straightforward matter. (See Kuipers-Niederreiter \cite{kuipers.niederreiter}, Chapter 1, exercise 8.7.)

\section{The proof of Theorem~\ref{T1}}

Let $L = \bigcap_m \bigcup_n L_{m,n}$ and $F = \bigcap_m \bigcup_n F_{m,n}$ be fixed $\mathbf \Pi^0_3$ subsets of $\{0,1\}^\NN$, with the $L_{m,n}$ and $F_{m,n}$ all closed sets.  With no loss of generality, we may assume that, for each $m$,
\[
L_{m,0} \subseteq L_{m,1} \subseteq \ldots \qquad \mbox{and} \qquad F_{m,0} \subseteq F_{m,1}\subseteq \ldots.
\]
Also, since we will be considering the difference set $L\setminus F$, we may assume that $L \supseteq F$.  Were this not so, we could replace $F$ with $L \cap F$.  We now proceed to define a continuous map $f : \{0,1\}^\NN \rightarrow \mathbb R$ such that $f^{-1} (N_1 \setminus N_2) = L\setminus F$.

In the first place, let
\[
\alpha_n = (0110)^n \concat (10)
\]
and
\[
\beta_n = (0110)^n \concat 0.
\]
Observe that
\[
\lim_{k \rightarrow \infty} d_{10} (\alpha_n^\infty \upto k) = (n+1)/(4n+2) > 1/4
\]
and
\[
\lim_{k \rightarrow \infty} d_0 (\beta_n^\infty \upto k) = (2n+1)/(4n+1) > 1/2.
\]
Also, if $y \in \RR$ is of the form $0\, . \, \alpha_{i_0}^{a_0} \concat \beta_{j_0}^{b_0} \concat \alpha_{i_1}^{a_1} \concat \beta_{j_1}^{b_1} \concat \ldots$, then $y \in N_1$ if $j_p \rightarrow \infty$, as $p \rightarrow \infty$, since the $\alpha$'s do not affect the density of 0's and 1's in the binary expansion of $y$.  In addition, if  both $i_p , j_p \rightarrow \infty$, as $p \rightarrow \infty$, then $y \in N_2$.  This follows from the fact that the real number $0.01100110\ldots$ is order-2 normal and inserting a density-zero set of digits does not affect normality.

Given $x \in \{0,1\}^\NN$, we will let
\[
f(x) = 0\, . \, \alpha_{i_0}^{a_0} \concat \beta_{j_0}^{b_0} \concat\alpha_{i_1}^{a_1} \concat \beta_{j_1}^{b_1} \concat \ldots,
\]
where $i_p$, $j_p$, $a_p$ and $b_p$ are natural numbers, defined as follows, for each $p \in \NN$:
\[
i_p = \begin{cases}
i_{p-1} + 1 & \mbox{if } p = \langle m,n \rangle, \ (\forall n' < n)([x \upto \langle m, n-1 \rangle] \cap F_{m,n'} \neq \emptyset \\ &\implies [x \upto p] \cap F_{m,n'} \neq \emptyset) \mbox{ and } [x \upto p] \cap F_{m,n} \neq \emptyset,\\

m &\mbox{if } p = \langle m,n \rangle \mbox{ and } ((\exists n' < n)([x \upto \langle m,n-1 \rangle] \cap F_{m,n'} \neq \emptyset \\
&\mbox{and } [x \upto p] \cap F_{m,n'} = \emptyset) \mbox{ or } [x \upto p] \cap F_{m,n} = \emptyset)
\end{cases}
\]
We refer to the two cases in this definition as {\em Case 1} and {\em Case 2}.  The definition of $j_p$ is identical to that of $i_p$, except with the $L_{m,n}$ in place of the $F_{m,n}$.  We let
\[
a_p = \begin{cases}
1 &\mbox{if $p = \langle m,n \rangle$ and Case 1 from the definition of $i_p$ holds,}\\
k &\mbox{if $p = \langle m,n \rangle$ and Case 2 from the definition of $i_p$ holds,}\\ &\mbox{where $k$ is chosen to be large enough that}\\
&d_{10} (\alpha_{i_0}^{a_0} \concat \beta_{j_0}^{b_0} \concat\alpha_{i_1}^{a_1} \concat \ldots \concat \beta_{j_{p-1}}^{b_{p-1}} \concat \alpha_{i_p}^k) \geq (m+(3/4))/(4m+2).
\end{cases}
\]
We may alway find such a $k$ in Case 2, since $i_p = m$ and, therefore,
\[
\lim_{k \rightarrow \infty} d_{10} (\sigma \concat \alpha_m^\infty \upto k) = (m+1)/(4m+2),
\]
for any $\sigma \in \{0,1\}^{< \omega}$.  Similarly, we define
\[
b_p = \begin{cases}
1 &\mbox{if $p = \langle m,n \rangle$ and Case 1 from the definition of $j_p$ holds,}\\
k &\mbox{if $p = \langle m,n \rangle$ and Case 2 from the definition of $j_p$ holds,}\\ &\mbox{where $k$ is chosen to be large enough that}\\
&d_{0} (\alpha_{i_0}^{a_0} \concat \beta_{j_0}^{b_0} \concat\alpha_{i_1}^{a_1} \concat \ldots \concat \beta_{j_{p-1}}^{b_{p-1}} \concat \alpha_{i_p}^{a_p} \concat \beta_{j_p}^{k}) \geq (2m+(3/4))/(4m+1).
\end{cases}
\]
Again, we may always find such a $k$, since $j_p = m$ and, hence,
\[
\lim_{k \rightarrow \infty} d_0 (\sigma\concat \beta_m^\infty \upto k) = (2m+1)/(4m+1),
\]
for any $\sigma \in \{0,1\}^{< \omega}$.

\medskip

\noindent{\scshape Verification.}  The claims below will complete the proof.  In particular, they will establish that $f^{-1} (N_1 \setminus N_2) = L\setminus F$.

\begin{claim}
The map $f : \{0,1\}^\NN \rightarrow \RR$ is continuous.
\end{claim}

\begin{proof}[Proof of claim]
The continuity of $f$ follows from the observation that $i_0 , \ldots , i_p$, \newline $j_0 , \ldots , j_p$, $a_0 , \ldots , a_p$ and $b_0 , \ldots , b_p$ are all determined by the first $p$ terms of $x$.  In particular, at least the first $p$ digits of the binary expansion of $f(x)$ are determined by the first $p$ terms of $x$.  Thus, if $x , y \in \{0,1\}^\NN$ are such that $x \upto p = y \upto p$,
\[
|f(x) - f(y)| \leq 2^{-p}.
\]
Thus, $f$ is continuous.
\end{proof}

At this juncture, we introduce some convenient terminology.  For fixed $x \in \{0,1\}^\NN$, we say that $m \in \NN$ {\em acts infinitely often} for the $i_p$ (resp., $j_p$) iff there exist infinitely many $n$ such that Case 2 holds for $p = \langle m,n \rangle$, in the definition of $i_p$ (resp., $j_p$).  Otherwise, we say that $m$ {\em acts finitely often} for the $i_p$ (resp., $j_p$).  Also, note that $i_p \rightarrow \infty$, as $p \rightarrow \infty$, iff each $m \in \NN$ acts only finitely often for the $i_p$.  Likewise, for the $j_p$.  Although this terminology refers implicitly to a specific $x$, we will supress mention of $x$, since $x$ will always be fixed in what follows.

\begin{claim}
If $x \in F$, then $f(x) \in N_2$.
\end{claim}

\begin{proof}[Proof of claim]
As noted above, it will suffice to show that $i_p \rightarrow \infty$ and $j_p \rightarrow \infty$.  In turn, it will suffice to show that each $m$ acts only finitely many times for both the $i_p$ and the $j_p$.  Indeed, suppose $x \in F$ and fix $m \in \NN$.  Let $n_0$ be such that that $x \in F_{m,n_0}$.  It follows that $x \in F_{m,n}$, for each $n \geq n_0$, since $F_{m,0} \subseteq F_{m,1} \subseteq \ldots$.  Hence, $[x \upto p] \cap F_{m,n} \neq \emptyset$, for all $p \geq \langle m,n_0\rangle$ and $n \geq n_0$.  We may also assume $n_0$ is large enough that, if $n \geq n_0$ and $n' < n_0$, we have $[x \upto \langle m , n-1 \rangle] \cap F_{m,n'} = \emptyset$.  In particular, we are in Case 1 of the definition of $i_p$, provided $p = \langle m,n \rangle$, with $n \geq n_0$.  Thus, $m$ acts only finitely many times for the $i_p$.

We omit the corresponding argument for the $j_p$, as it is entirely analogous, using the fact that $x \in L \subseteq F$.  This completes the proof of the claim.
\end{proof}

\begin{claim}
If $x \in L \setminus F$, then $f(x) \in N_1 \setminus N_2$.
\end{claim}

\begin{proof}[Proof of claim]
Fix $x \in L\setminus F$.  As in the previous claim, each $m$ acts only finitely many times for the $j_p$.  It follows that $j_p \rightarrow \infty$ and hence $f(x) \in N_1$.  On the other hand, we shall see that some $m$ acts infinitely often for the $i_p$.  Indeed, since $x \notin F$, there exists $m$ such that $x \notin F_{m,n}$, for all $n$.  For each $n$, let $k_n$ be least such that $[x \upto k_n] \cap F_{m,n} = \emptyset$.  Note that $k_0 \leq k_1 \leq \ldots$, since $F_{m,0} \subseteq F_{m,1} \subseteq \ldots$.

We consider two cases.  In the first instance, suppose that there are infinitely many $r$ such that $k_r > \langle m,r \rangle$.  We may therefore select $r_0 < r_1 < \ldots$ and $n_0 < n_1 < \ldots$ such that
\begin{itemize}
\item $(\forall e) (r_e < n_e)$ and
\item $(\forall e) (\langle m , n_e -1 \rangle < k_{r_e} \leq \langle m, n_e \rangle$.
\end{itemize}
Thus, for each $p = \langle m , n_e \rangle$, there will be an $n' < n_e$ (namely $n' = r_e$) such that $[x \upto \langle m, n_e -1 \rangle] \cap F_{m,n'} \neq \emptyset$ (since $\langle m,n_e -1 \rangle < k_{r_e}$), but $[x \upto \langle m, n_e \rangle] \cap F_{m,n'} = \emptyset$ (since $\langle m,n_e\rangle \geq k_{r_e}$).  It follows that, for each $p = \langle m,n_e\rangle$, we will be in Case 2 of the definition of $i_p$.

In the second case, we assume that $k_n \leq \langle m,n \rangle$, for all but finitely many $n$.  Thus, by the definition of the $k_n$, we have that $[x \upto \langle m,n\rangle ] \cap F_{m,n} = \emptyset$, for all but finitely many $n$.  Hence, for cofinitely many $n$, if $p = \langle m , n \rangle$, we are in Case 2 of the definition of $i_p$.

It now follows that $f(x) \notin N_2$, since each time $p = \langle m,n \rangle$ is in Case 2 of the definition of $i_p$, we have
\[
d_{10} (\alpha_{i_0}^{a_0} \concat \beta_{j_0}^{b_0} \concat\alpha_{i_1}^{a_1} \concat \ldots \concat \beta_{j_{p-1}}^{b_{p-1}} \concat \alpha_{i_p}^{a_p}) \geq (m+(3/4))/(4m+2) > 1/4,
\]
and this occurs infinitely often for some fixed $m$.
\end{proof}

\begin{claim}
If $x \notin L$, then $f(x) \notin N_1$.
\end{claim}

\begin{proof}[Proof of claim]
As in the second half of the proof of the previous claim, we observe that there is some $m$ which acts infinitely often for the $j_p$ and conclude that $f(x) \notin N_1$.
\end{proof}

This completes the proof.

\section{A generalization}

We now indicate how to generalize the preceeding argument to an arbitrary fixed base $b$.  In the first place, our definition of $d_\sigma (\alpha)$ may be extended to strings $\sigma , \alpha \in \{0,1, \ldots , b-1\}^{< \omega}$.  Namely, $d_\sigma (\alpha)$ is the number of times $\sigma$ occurs as a substring of $\alpha$, divided by $|\alpha|$.  Also, if $x \in \{0,1, \ldots , b-1\}^\NN$, we let
\[
(a \, . \, x)_b = a + \sum_{n=0}^\infty \frac{x(n)}{b^{n+1}}.
\]

\begin{definition}
For integers $b, r$, with $b \geq 2$ and $r \geq 1$, and $x \in \{0 , 1 , \ldots , b-1\}^\NN$, we say that a real number $(a \, . \, x)_b$ is {\em order-$r$ normal} in base $b$ iff, for each $\sigma \in \{0 , 1 , \ldots , b-1\}^r$,
\[
\lim_{k \rightarrow \infty} d_\sigma (x \upto k) = b^{-r}.
\]
We let $N^b_r$ denote the set of real numbers which are order-$r$ normal in base $b$.
\end{definition}

It is important to note that $N^b_s \subseteq N^b_r$, if $r < s$.  This follows from the fact that, for any $\sigma \in \{ 0, 1 , \ldots , b-1\}^r$, there are $b^{r-s}$ many $\tau \in \{ 0, 1 , \ldots , b-1\}^s$, having $\sigma$ as an initial segment.  Hence, if $(0\, . \, x)_b \in N^b_s$ and $\sigma \in \{ 0, 1 , \ldots , b-1\}^r$,
\[
\lim_{k \rightarrow \infty} d_\sigma (x \upto k) = b^s \cdot b^{r-s} = b^r.
\]
We now state and sketch the proof of a generalization of Theorem~\ref{T1}.

\begin{theorem}
For each base $b \geq 2$ and $s > r \geq 1$, the set $N^b_r \setminus N^b_s$ is $\mathcal D_2 (\mathbf \Pi^0_3)$-complete.
\end{theorem}

\begin{proof}[Sketch of proof]
I.~J.~Good \cite{good.normal.recurring} showed that, for each $b,r$ as in the definition above, there exists a finite string $\theta \in \{0,1, \ldots , b-1\}^{b^r}$ such that the real number $(0 \, . \, \theta ^\infty)_b \in N^b_r$.  If $s > r$, there are $b^s$ possible strings of length $s$ using digits $\{ 0, 1, \ldots , b-1\}$.  Thus, if $\theta \in \{0,1, \ldots , b-1\}^{b^r}$, then $(0 \, . \, \theta^\infty)_b$ cannot be order-$s$ normal in base $b$, since there are at most $b^r$ substrings of $\theta^\infty$ of any fixed length.  It follows that, if $\theta$ is as in Good's result with $(0\, . \, \theta^\infty)_b \in N^b_r$, then $(0\, . \, \theta^\infty)_b$ is not order-$s$ normal, for any $s > r$.

Now fix a base $b$ and $r<s$.  Let $\theta , \mu \in \{ 0 ,1, \ldots , b-1\}^{< \omega}$ be such that
\begin{itemize}
\item $|\theta| = b^s$,
\item $|\mu | = b^r$,
\item $(0\, . \, \theta^\infty)_b \in N^b_s$ and
\item $(0\, . \, \mu^\infty)_b \in N^b_r$.
\end{itemize}
Following the notation of the proof of Theorem~\ref{T1}, let $\alpha_n = \theta^n \concat \mu$ and $\beta_n = \theta^n \concat 0$.  Note that, for all $n$, $0 \, . \, \alpha_n^\infty \in N^b_r \setminus N^b_s$, whereas, for all $n$, $0\, . \, \beta_n^\infty \notin N^b_r$.  Observe that, if
\[
y = (0\, . \, \alpha_{i_0}^{a_0} \concat \beta_{j_0}^{b_0} \concat\alpha_{i_1}^{a_1} \concat \beta_{j_1}^{b_1} \concat \ldots)_b,
\]
then $y \in N^b_r$ if $j_p \rightarrow \infty$, as $p \rightarrow \infty$.  Similarly, $y \in N^b_s$, if $i_p \rightarrow \infty$ and $j_p \rightarrow \infty$.

Following the proof of Theorem~\ref{T1}, with these new $\alpha_n$ and $\beta_n$ and certain other minor modifications yields a proof of the theorem above.
\end{proof}

\section{The proof of Theorem~\ref{T2}}

Fix a descending sequence $F_1 \supseteq F_2 \supseteq \ldots$ of $\mathbf \Pi^0_3$ sets.  For each $k$, let $F_{k,m,n}$ be closed sets with
\[
F_k = \bigcap_m \bigcup_n F_{k,m,n}.
\]
We may assume that, for each pair $k,m$, we have $F_{k,m,0} \subseteq F_{k,m,1} \subseteq \ldots$.  Our objective is to show that $\bigcup_k F_{2k+1} \setminus F_{2k+2}$ is a continuous preimage of $\bigcup_k N_{2k+1} \setminus N_{2k+2}$, where $N_k$ denotes the set of real numbers in $[0,1]$ which are order-$k$ normal.  To this end, we will define a continuous function $f: \{0,1\}^\NN \rightarrow \{0,1\}^\NN$ such that, for each $x \in \{0,1\}^\NN$,
\[
x \in F_k \iff 0 \, . \, f(x) \in N_k.
\]
Given $x \in \{0,1\}^\NN$, we will define finite binary strings, $\sigma_t$, with $\sigma_0 \preceq \sigma_1 \preceq \ldots$ and let $f(x) = \bigcup \sigma_t$.

Before proceeding, we introduce some notation for the sake of the construction.  For each $i \in \NN$, $i > 0$, let $\eta_i \in \{0,1\}^i$ be, as in I.~J.~Good \cite{good.normal.recurring}, such that $0\, . \, (\eta_i)^\infty$ is order-$k$ normal.  Note that each $\alpha \in \{0,1\}^i$ must occur exactly once in each period of the repeating decimal $0 \, . \, (\eta_i)^\infty$.  Also, since $|\eta_i| = 2^i$, the real number $0 \, . \, (\eta_i)^\infty$ is not order-$(i+1)$ normal, as there are at most $2^i$ distinct substrings of $(\eta_i)^\infty$ of any fixed length.  For each $i$, we therefore fix an $\alpha_i \in \{0,1\}^i$ which is not a substring of $(\eta_{i-1})^\infty$.

For each triple $k,m,n$, we now let
\[
\tau_{k,m,n} = (\eta_{k,m})^i \concat (\eta_{k-1})^j,
\]
where $i,j \in \NN$ are chosen such that the following hold.
\begin{itemize}
\item For each triple $k,m,n$ and each $\alpha \in \{0,1\}^{\leq k+m}$,
\[
\left| \left(\lim_{s\rightarrow \infty} d_\alpha \big( (\tau_{k,m,n})^\infty \upto s\big)\right) - 2^{-|\alpha|} \right| < 2^{-(k+m)}.
\]
\item For each pair $k,m$, there exists $r_{k,m} < 2^{-k}$ such that, for all $n$,
\[
\lim_{s \rightarrow \infty} d_{\alpha_k} \big( (\tau_{k,m,n} )^\infty \upto s \big) < r_{k,m}.
\]
\item For each triple $k,m,n$ and each $\alpha \in \{0,1\}^{\leq k-1}$,
\[
\left| \left( \lim_{s \rightarrow \infty} d_\alpha \big( (\tau_{k,m,n})^\infty \upto s\big) \right)  - 2^{-|\alpha|} \right| < 2^{-\langle k,m,n \rangle}.
\]
\end{itemize}

\medskip

{\scshape The construction.}  At this point, fix $x \in \{0,1\}^\NN$.  As indicated above, we will define binary strings $\sigma_t$, determined by $x$. For each $t = \langle k,m,n \rangle$, we distinguish between two distinct cases.  We say that $t = \langle k,m,n \rangle$ is in {\em case 1} if
\begin{itemize}
\item $[x \upto \langle m,n\rangle ] \cap F_{k,m,n} \neq \emptyset$ and,
\item for each $n' < n$, if $[x \upto \langle m,n-1\rangle] \cap F_{k,m,n'} \neq \emptyset$, then $[x \upto \langle m,n\rangle ] \cap F_{k,m,n'} \neq \emptyset$.
\end{itemize}
Likewise, we say that $t = \langle k,m,n \rangle$ is in {\em case 2} if
\begin{itemize}
\item $[x \upto \langle m,n\rangle ] \cap F_{k,m,n} = \emptyset$ or
\item there exists $n'< n$ such that $[x \upto \langle m,n-1\rangle ] \cap F_{k,m,n'} \neq \emptyset$, but $[x \upto \langle m,n\rangle ] \cap F_{k,m,n'} = \emptyset$.
\end{itemize}

In the process of defining the binary strings $\sigma_t$, we also define binary sequences $y_t \in \{0,1\}^\NN$ such that $\sigma_t \prec y_t$ and functions $\mu_t : \{0,1\}^{< \omega} \times \NN \rightarrow \NN$ such that, for each $\alpha \in \{0,1\}^{<\omega}$ and $p \in \NN$, $\mu_t (\alpha, p)$ is the least $q \in \NN$ with
\[
\left| d_\alpha (y_t \upto q') - \left(\lim_{s \rightarrow \infty}d_\alpha (y_t \upto s)\right) \right| < 2^{-p},
\]
for all $q' \geq q$.  Note that the limit in the expression above is guaranteed to exist because $y_t$ is eventually periodic.  We call the map $\mu_t$ the {\em modulus of distribution} for $y_t$.

Suppose that $\sigma_{t-1}$ is given, we show how to define $\sigma_t$.  (In the case of $t = 0$, we let $\sigma_{-1}$ be the empty string, for notational purposes.)

First suppose that $t = \langle k,m,n \rangle$ is in case 1.  Let $y_t = \sigma_{t-1} \concat (\eta_t)^\infty$ and $\sigma_t = \sigma_{t-1}\concat (\eta_t)^i$, where $i$ is large enough that the following hold.
\begin{enumerate}
\item For all $\alpha \in \{0,1\}^{\leq t}$,
\[
\left| d_\alpha \big( \sigma_t \big) - 2^{-|\alpha|} \right| < 2^{-t}.
\]
\item[(2a)] If $t+1$ is in case 1 and $\mu : \{0,1\}^{< \omega} \times \NN \rightarrow \NN$ is the modulus of distribution for $\sigma_t \concat (\eta_t)^i \concat (\eta_{t+1})^\infty$, then
\[
\mu \upto \{0,1\}^{\leq t} \times \{ 0 , \ldots , t\} = \mu_t \upto \{0,1\}^{\leq t} \times \{ 0 , \ldots , t\}.
\]
\item[(2b)]  If $t+1 = \langle k',m',n' \rangle$ is in case 2, $\mu : \{0,1\}^{< \omega} \times \NN \rightarrow \NN$ is the modulus of distribution for $\sigma_t \concat (\eta_t)^i \concat (\tau_{k',m',n'})^\infty$ and $p = \min \{ t , k' + m' \}$, then
\[
\mu \upto \{0,1\}^{\leq p} \times \{ 0 , \ldots , p\} = \mu_t \upto \{0,1\}^{\leq p} \times \{ 0 , \ldots , p\}
\]
and, if $k^* = \min \{ t , k' - 1\}$,
\[
\mu \upto \{0,1\}^{\leq k^*} \times \{ 0 , \ldots , t\} = \mu_t \upto \{0,1\}^{\leq k^*} \times \{ 0 , \ldots , t\}.
\]
\end{enumerate}

Now suppose that $t = \langle k,m,n \rangle$ is in case 2.  Let $y_t = \sigma_{t-1} \concat (\tau_{k,m,n})^\infty$ and $\sigma_t = \sigma_{t-1} \concat (\tau_{k,m,n})^i$, where $i$ is large enough that the following hold.
\begin{enumerate}\setcounter{enumi}{2}
\item For each $\alpha \in \{0,1\}^{\leq k+m}$,
\[
\left| d_\alpha (\sigma_t) - 2^{-|\alpha|} \right| < 2^{-(k+m)}.
\]
\item For each $\alpha \in \{0,1\}^{\leq k-1}$,
\[
\left| d_\alpha (\sigma_t) - 2^{-|\alpha|} \right| < 2^{-t}.
\]
\item $d_{\alpha_k} ( \sigma_t ) < r_{k,m}$, where $\alpha_k$ and $r_{k,m}$ are as above.
\item[(6a)]  If $t+1$ is in case 1, $\mu : \{0,1\}^{< \omega} \times \NN \rightarrow \NN$ is the modulus of distribution for $\sigma_t \concat (\tau_{k,m,n})^i \concat (\eta_{t+1})^\infty$ and $p = \min \{ k+m , t+1 \}$, then
\[
\mu \upto \{0,1\}^{\leq p} \times \{ 0 , \ldots , p\} = \mu_t \upto \{0,1\}^{\leq p} \times \{ 0 , \ldots , p\}
\]
and, if $k^* = \min \{ k-1 , t+1 \}$,
\[
\mu \upto \{0,1\}^{\leq k^*} \times \{ 0 , \ldots , t\} = \mu_t \upto \{0,1\}^{\leq k^*} \times \{ 0 , \ldots , t\}.
\]
\item[(6b)]  If $t+1 = \langle k',m',n' \rangle$ is in case 2, $\mu : \{0,1\}^{< \omega} \times \NN \rightarrow \NN$ is the modulus of distribution for $\sigma_t \concat (\tau_{k,m,n})^i \concat (\tau_{k',m',n'})^\infty$ and $p = \min \{ k+m , k'+m' \}$, then
\[
\mu \upto \{0,1\}^{\leq p} \times \{ 0 , \ldots , p\} = \mu_t \upto \{0,1\}^{\leq p} \times \{ 0 , \ldots , p\}
\]
and, if $k^* = \min \{ k-1 , k'-1\}$,
\[
\mu \upto \{0,1\}^{\leq k^*} \times \{ 0 , \ldots , t\} = \mu_t \upto \{0,1\}^{\leq k^*} \times \{ 0 , \ldots , t\}.
\]
\end{enumerate}

We now let $f(x) = \bigcup_t \sigma_t$.  This completes the definition of $f$.

\medskip

{\scshape Verification.}  The claims below will complete the proof of Theorem~\ref{T2}.

\begin{claim}
The map $f:\{0,1\}^\NN \rightarrow \{0,1\}^\NN$ in continuous.
\end{claim}

\begin{proof}[Proof of claim]
This follows from the fact that, given $x \in \{0,1\}^\NN$, each bit of $f(x)$ is determined by finitely many bits of $x$.
\end{proof}

In what follows, let $x \in \{0,1\}^\NN$ be fixed and let $\sigma_t$, $y_t$, $\mu_t$, etc.~be defined as above for $x$.

\begin{claim}
If $x \in F_{k_0}$, then $\lim_{t\rightarrow \infty} \mu_t (\alpha , p)$ exists, for each $\alpha \in \{0,1\}^{\leq k_0}$ and $p \in \NN$.
\end{claim}

\begin{proof}[Proof of claim]
Assume $x \in F_{k_0}$.  Fix $p \in \NN$ and let $t_0 \geq \max\{k_0,p\}$ be large enough that, for all $t = \langle k,m,n \rangle \geq t_0$, whenever $k\leq k_0$ and $t$ is in case 2, we have $k+m \geq \max \{ k_0 , p \}$.  To see that there is such a $t_0$, observe that, given a fixed pair $k,m$, with $k \leq k_0$, we have $x \in F_{k,m,n}$, for all but finitely many $n$, say $n_0$ is the least such $n$.  Hence, we have that $\langle k,m,n \rangle$ is in case 1 for all $n \geq n_0$ large enough that
\[
n' < n_0 \implies [x \upto \langle m,n \rangle] \cap F_{k,m,n'} = \emptyset.
\]
Hence, given any pair $k,m$, with $k \leq k_0$, there are only finitely many $n$ such that $\langle k,m,n \rangle$ is in case 2.  Thus, there are only finitely many $\langle k,m,n \rangle$ in case 2, with $k \leq k_0$ and $k+m <  \max \{ k_0 , p \}$.

We check that $\mu_{t+1} (\alpha , p) = \mu_t (\alpha , p)$, for all $t \geq t_0$ and $\alpha \in \{0,1\}^{\leq k_0}$.  We then conclude, by induction, that $\mu_t (\alpha , p) = \mu_{t_0} (\alpha , p)$, for all $t \geq t_0$ and $\alpha \in \{0,1\}^{\leq k_0}$.

Suppose that $t$ is in case 1.  In the first place, if $t+1$ is also in case 1, then, by condition (2a),
\begin{equation*}\tag{$*$}
\mu_{t+1} \upto \{0,1\}^{\leq k_0} \times \{ 0 , \ldots , p\} = \mu_t \upto \{0,1\}^{\leq k_0} \times \{ 0 , \ldots , p\},
\end{equation*}
since $t \geq t_0 \geq \max \{ k_0 , p\}$.  On the other hand, if $t+1 = \langle k',m',n' \rangle$ is in case 2 and $k_0 < k'$, we have that ($*$) again holds by condition (2b), since $k_0 \leq \min \{ t , k'-1\}$.  Finally, if $t+1 = \langle k',m',n' \rangle$ is in case 2 and $k' \leq k_0$, then ($*$) still holds by (2b), since
\[
\max\{ k_0 , p \} \leq \min \{ t , k'+m' \}.
\]

If $t = \langle k,m,n \rangle$ is in case 2, the arguments are analogous, using (6a) and (6b) above.  For instance, if $k \leq k_0$ and $t+1$ is in case 1, then, by assumption, $k+m \geq \max \{ k_0 , p\}$ and hence condition ($*$) holds by (6a), using the fact that $k_0 \leq \min\{ k+m , t+1 \}$.
\end{proof}

\begin{claim}
If $x \in F_{k_0}$, then
\[
\lim_{s\rightarrow \infty} d_\alpha \big( f(x) \upto s \big) = 2^{-|\alpha|},
\]
for each $\alpha \in \{0,1\}^{\leq k_0}$.
\end{claim}

\begin{proof}
Observe that $y_t \rightarrow f(x)$, as $t \rightarrow \infty$.  The functions $\mu_t \upto \{0,1\}^{\leq k_0} \times \NN$ also form a (pointwise) convergent sequence, by the previous claim.  Fixing $\alpha \in \{0,1\}^{\leq k_0}$, it follows that the sequence $\big( d_\alpha \big( f(x) \upto s \big) \big)_{s \in \NN}$ is Cauchy and therefore convergent.  By conditions (1), (3) and (4) above, for each $\varepsilon > 0$, there are infinitely many $s \in \NN$ such that
\[
\left| d_\alpha \big( f(x) \upto s \big) - 2^{-|\alpha|} \right| < \varepsilon.
\]
It follows that $\lim_{s \rightarrow \infty} d_\alpha \big( f(x) \upto s \big) = 2^{-|\alpha|}$.
\end{proof}

From the last two claims, we conclude that, if $x \in F_{k_0}$, we have $0 \, . \, f(x) \in N_{k_0}$.  The next claim asserts the converse.

\begin{claim}
If $x \notin F_{k_0}$, then $0 \, . \, f(x) \notin N_{k_0}$.
\end{claim}

\begin{proof}
Assume $x \notin F_{k_0}$ and $m_0 \in \NN$ is such that $x \notin F_{k_0 , m_0 , n}$, for all $n \in \NN$.  For each $n$, let $s_n \in \NN$ be least such that $[ x\upto s_n ] \cap F_{k_0 , m_0 , n} = \emptyset$.  We consider two distinct cases.

First, suppose that there are infinitely many $n$ such that $s_n > \langle m_0 , n \rangle$.  In this case, there exist $n_0 < n_1 < \ldots$ and $p_0 < p_1 < \ldots$ such that, for each $j$,
\begin{itemize}
\item $p_j > n_j$ and
\item $\langle m_0 , p_j - 1\rangle < s_{n_j} \leq \langle m_0 , p_j \rangle$.
\end{itemize}
Thus, for each $j$,
\[
[x \upto \langle m_0 , p_j - 1\rangle ] \cap F_{k_0, m_0 , n_j} \neq \emptyset \qquad \& \qquad [x \upto \langle m_0 , p_j \rangle] \cap F_{k_0 , m_0 , n_j} = \emptyset.
\]
It follows that each $t_j = \langle k_0 , m_0 , p_j \rangle$ is in case 2 and, hence, for each $j$,
\[
d_{\alpha_{k_0}} \big( f(x) \upto |\sigma_{t_j}| \big) < r_{k_0 , m_0} < 2^{-k_0},
\]
by condition (5) above.  Thus, $0 \, . \, f(x) \notin N_{k_0}$.

On the other hand, if $s_n \leq \langle m_0 , n\rangle$, for all but finitely many $n$, we have
\[
[x \upto \langle m_0 , n \rangle] \cap F_{k_0 , m_0 , n} = \emptyset,
\]
for cofinitely many $n$.  Thus, $\langle k_0 , m_0 , n\rangle$ is in case 2 for cofinitely many $n$ and again $0 \, . \, f(x) \notin N_{k_0}$.
\end{proof}

We conclude that, for each $k \geq 1$ and $x \in \{0,1\}^\NN$, we have $ x \in F_k$ iff $0\, . \, f(x) \in N_k$.  It follows that
\[
x \in \bigcup_k F_{2k+1} \setminus F_{2k+2} \iff 0\, . \, f(x) \in  \bigcup_k N_{2k+1} \setminus N_{2k+2},
\]
for each $x \in \{0,1\}^\NN$.  This completes the proof of Theorem~\ref{T2}.

redpajama_set_name: RedPajamaArXiv


While law school may teach you about the law, your local county law librarian can teach you how to find and use legal resources. In fact, your local county law librarians are probably your best resource for teaching you how to do your own legal research because that's what we do all day - study law and legal things.
Tax season, it sneaks up on you. For several years our Main Library has been providing print tax forms and instruction booklets courtesy of the IRS and the State of California. Guaranteed every January we start receiving calls asking for tax forms and we have learned we're one of the only places for miles that still carry them.
If you know what you should do or know that you don't know what to do but think you can wing it because you think no one will notice that you didn't do what you should have done the way it should have been done in the first place, odds are everyone will notice you did it wrong and you'll be the person everyone points to when they say, "whatever you do, don't do it like that person!"


FlipHTML5 provides an easy-to-use interface for creating and sharing flipping books online.
Hong Kong, China -- (ReleaseWire) -- 11/18/2018 -- FlipHTML5 is a software program that makes it easy to transform images or PDF files into smart flipping books that support most browsers and devices. With a few clicks, users can customize the flipping book by adding hyperlinks, audio, and videos to enhance effects. The user-friendly interface enables both fledging and amateur users to navigate and create interactive flipping books in multiple languages. By utilizing the free templates and themes provided by FlipHTML5, every flipping book created becomes more captivating to the readers.
"Our FlipHTML5 software has provided a new dimension to digital reading," said Anna Lee, Designer of FlipHTML5. "We have empowered our users with more enhancements to help them offer readers a more engaging reading experience. As part of increasing their productivity, we have added more exceptional page-flipping effects and animations that will make every reader want to turn the page."
FlipHTML5 supports the accessibility of the flipping book on all browsers and devices including Google Chrome, Internet Explorer, iPhones, PCs, and iPads, among others. FlipHTML5 also provides free online hosting and self-hosting services for users to publish their flipping books online without the need for a website domain. This allows for maximum marketing exposure of the flipping book to the world. Readers can download flipping books from their devices at any time and enjoy a pleasant reading experience.
"We also offer a fully customized design panel that allows users to maximize their potential and creativity," continued Anna Lee. "Users can modify their flipping books and add more content including logos, headers, colors, backgrounds, animations and more, to make them more exciting to read. They can also personalize the flipping book and make it easy to interact, share and track readers' activities."
Enterprises can use FlipHTML5 to create flipping books about the business and share with online readers in efforts to promote the brand online. The HTML5 and jQuery technologies enable quick conversion of a PDF into an amazing and interactive flipping book. FlipHTML5 has benefited many users who can attest to the efficiency and productivity achieved after creating mind-blowing flipping books online. Signing up for a free FlipHTML5 account is free. By following a few simple steps, users often create classic flipping books in a few minutes. Take a peek at all the highest-selling FlipHTML5 flipping books at http://fliphtml5.com/.
The company provides free FlipHTML5 software that makes creating a flipping book and sharing it online a walk in the park. Sign up for a free FlipHTML5 account and receive 20GB of space to start creating the most wonderful flipping book readers can correlate with.


text:
Paterson woman indicted for allegedly setting fire to Madison Street apartment complex
September 3, 2021 by Jonathan Greene
A city woman has been indicted for allegedly setting fire to an apartment complex on Madison Street, according to the Passaic County Prosecutor's Office.
Ahkearah Muhammad, 24-year-old, has been indicted on two counts of terroristic threats, two counts of aggravated arson, and risking widespread injury or damage.
Muhammad remains a fugitive. She is considered dangerous, authorities said.
On March 20 at around 10:35 p.m., firefighters and police arrived at 55 Madison Street, a housing complex with 240 individual apartments, on report of a fire. They found a fire had started in front of an apartment leased by Muhammad's former acquaintance, authorities said.
Muhammad was identified as the person who set the fire using lighter fluid, authorities said. She was observed on surveillance video starting the fire. Investigators recovered a bottle of lighter fluid at the scene.
Authorities did not say why Muhammad set the fire. Her former acquaintance's name is being withheld because of the Prevention of Domestic Violence Act, authorities said.
The fire was extinguished by another resident of the apartment complex. No one was injured and no one was displaced.
Muhammad faces decades in state prison if convicted.
Anyone with information about Muhammad's whereabouts should contact the Passaic County Prosecutor's Office at 1-877-370-PCPO or call the Paterson Police Department at (973) 321-1111.
Deport her to Afghanistan

redpajama_set_name: RedPajamaCommonCrawl


Mountain Bike Breckenridge! The "Fat-Tire" destination. Breckenridge was named one of the top 10 great places to take the kids on a bike ride by USA Today. The Town of Breckenridge is always looking for ways to improve and expand their recreation territory. With the help of friends such as the Forest Service, Friends of the Dillon Range District & Friends of Breckenridge Trails more and more designated and maintained trails are opening up to bikers and creating connections from one side of Summit County to the other. Biking in Breckenridge has never been better or easier.


text:
Russian diplomat found dead in Berlin, embassy refused autopsy Accused Cleo Smith kidnapper, Terence Kelly in maximum security prison Thirteen-year old Palestinian said shot dead by IDF in clashes near Nablus COP26 Climate Summit: What Happened During the First Days Greta Thunberg says COP26 climate summit is a failure and a PR event
What to expect from Biden's voting rights speech in Atlanta : NPR
President Biden and Vice President Harris are renewing the administration's push for federal action to protect voting rights during a trip to Georgia on Tuesday.
Greg Nash/Paul/Getty Images
Caption switch
President Biden plans to renew his push for action in Congress to protect voting rights in a speech in Georgia on Tuesday, where proposed federal legislation for the Democratic Party has stalled and the president faces mounting pressure to intervene.
Biden is expected to throw his support behind the blocking change to make it easier for voting rights bills to pass, and to ensure they "can be restored and defend that fundamental right," according to a White House official who provided a preview of Biden's remarks on the condition of anonymity.
"The next few days, when these bills are voted on, will be a turning point in this nation," Biden said, according to prepared statements shared by the White House. "Do we choose democracy over tyranny, light over shadow, justice over injustice? I know my place."
But some activists, frustrated with the lack of action on the issue, plan to skip the statements, arguing that Biden's time would be better spent brokering a plan to overcome Republican opposition in the Senate than to deliver another speech.
Biden will be joined by Vice President Harris, who has been the administration's key person on voting rights, to deliver a speech at the University of Atlanta Center. The speech will be delivered in the district formerly represented by the late civil rights icon Representative John Lewis, in a state that has been a focal point in the struggle over voting rights.
"He's not going to Georgia tomorrow – a place where there is a tremendous history of civil rights leaders, and his friend John Lewis – defending voting rights, if he's not willing and willing to make this case and continue to do so," White House Press Secretary Jen Psaki said on Monday.
Biden has been pressured to combat threats to voting rights more vigorously, which has been one of the key promises of his presidency. In a speech in July, he called the fight against restrictive Republican-led voting laws in states including Georgia "the most important test of our democracy since the Civil War."
"I think we all want to see the White House do everything it can. Drag every point, hit every lever to get that done," said Reverend Leah Dattrey, a veteran Democratic strategist, who will be attending the address in Atlanta. "Because without all of us being able to count our votes and honor our votes, there will be nothing else in our democracy. And what we saw on January 6, 2021 is just the tip of the iceberg."
In the Senate, Democrats pushed forward two voting rights bills that were blocked by Republicans using stalling: the Freedom of Voting Act that would set new minimum standards for early voting and mail, among other provisions. The John Lewis Voting Rights Advancement Act seeks to restore key elements of the historic Voting Rights Act that have been weakened by Supreme Court rulings.
Senators will vote to stall within days
Senate Majority Leader Chuck Schumer of New York State said the Senate will vote no later than January 17 on changing Senate rules if Republicans continue to block voting rights legislation.
"We hope that our fellow Republicans will change course and work with us," Schumer said in a letter to his campaign rally. "But if they don't, the Senate will debate and consider changes to Senate rules on or before January 17, Martin Luther King Jr. Day, to protect the foundations of our democracy: free and fair elections."
But in order to change the Senate's rules, all Democrats must be at the top of the list. Senator Joe Manchin of West Virginia and Senator Kirsten Senema of Arizona have repeatedly defended legislative stalling and may not be open to amending it, despite support for the voting legislation itself.
"As she has done throughout her tenure in the US House and Senate, Senator Senema also continues to support the Senate's 60-vote threshold, to protect the country from frequent radical setbacks in federal policy that would entrench uncertainty, deepen divisions, and further erode Americans," Hannah said. Hurley, a spokeswoman for Cinemas, said in a statement: "It's confidence in our government."
Biden and Democrats have faced increasing pressure from advocates for voting bills to either change Senate rules to eliminate obstruction, or to obtain an exemption for some bills, including measures focused on voting rights.
Psaki declined to say whether Biden had been involved in talks with senators in recent weeks about voting rights legislation or potential rule changes.
"Don't stop in Georgia"
While Republicans viewed the Democrats' goals as a planned federal election takeover, Democrats, including Biden and Harris, were under increasing pressure to address the issue of voting rights more forcefully.
Derek Johnson, president of the NAACP, said some policymakers have approached the issue "in a very lackluster way."
"What we witnessed on January 6th was an attack on our democracy. But what we saw after January 6th, with legislatures across the country passing legislation for voter suppression, was also an attack on our democracy," Johnson said, attending Biden's speech in Atlanta on Tuesday. "We are not asking for anything extraordinary. The constitution guarantees the right to vote and we must ensure that this right is protected."
Melanie Campbell, executive director of the National Alliance for Black Civic Engagement, said she hopes Biden and Harris will deliver similar speeches in other states where laws to restrict voting have been passed.
"Don't stop in Georgia. Go to Arizona. Go to these places where these laws have been passed in 19 states… for people to really see what's going on. They have the power of a bully pulpit to help elevate these laws," she said in an interview.
But Campbell, who also planned to attend Biden's remarks on Tuesday, said speeches are not a substitute for legislative action.
Some defenders will not attend
Prior to the visit of the president and vice president, some voting rights groups in Georgia expressed opposition to Biden's planned approach.
Please stay in Washington tomorrow because we don't need you here in Georgia.
Faye Nguyen, Asian American, Atlanta Advance Justice
A number of organizations — including Black Voters Matter, Asian American Advocacy Fund, and New Georgia Project Action Fund — have called on Biden and Harris to cancel their trip if they don't plan to announce "a final plan for voting rights that will be approved by both houses." Procrastination doesn't stop him, he gets signed off to become law.
Leaders of those groups said Monday that they would not attend Biden's speech.
Latosha Brown, co-founder of Black Voters Matter, said she and other activists did not go beyond the statements to be "hostile" or "combatant," but said she wanted to convey the urgency of the moment.
"We've been screaming and screaming and talking about this for over a year," Brown said.
"Please stay in Washington tomorrow because we don't need you here in Georgia. We've got it here in Georgia, and we always do," said Faye Nguyen, executive director of Asian American Adventure Justice – Atlanta. Monday.
The address will be attended by family members of civil rights leader Martin Luther King, Jr., including King's oldest surviving son, Martin Luther King III. King and his wife, Arendrea Waters King, will meet with President Biden in Atlanta.
King said in a statement that he supports Georgia activists who refused to attend Biden's speech, noting that they are "frustrated after a year of inaction, and so are we."
King said he plans to meet with Biden "to say that his visit can't be a mere formality."
"We've seen what is possible when President Biden uses the full weight of his office to provide bridges, and now we need to see him do the same for voting rights," King said.
NPR's Deirdre Walsh contributed reporting.
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text:
Get help now: 312-243-5900
Statement of Karen Bloom & John Huber in Response to the Rittenhouse Verdict
Kenosha, Wisconsin – We are heartbroken and angry that Kyle Rittenhouse was acquitted in his criminal trial for the murder of our son Anthony Huber. There was no justice today for Anthony, or for Mr. Rittenhouse's other victims, Joseph Rosenbaum and Gaige Grosskreutz.
We did not attend the trial because we could not bear to sit in a courtroom and repeatedly watch videos of our son's murder, and because we have been subjected to many hurtful and nasty comments in the past year. But we watched the trial closely, hoping it would bring us closure.
That did not happen. Today's verdict means there is no accountability for the person who murdered our son. It sends the unacceptable message that armed civilians can show up in any town, incite violence, and then use the danger they have created to justify shooting people in the street. We hope that decent people will join us in forcefully rejecting that message and demanding more of our laws, our officials, and our justice system.
Make no mistake: our fight to hold those responsible for Anthony's death accountable continues in full force. Neither Mr. Rittenhouse nor the Kenosha police who authorized his bloody rampage will escape justice. Anthony will have his day in court.
No reasonable person viewing all of the evidence could conclude that Mr. Rittenhouse acted in self-defense. In response to racist and violent calls to action from militia members, Mr. Rittenhouse travelled to Kenosha illegally armed with an assault rifle. He menaced fellow citizens in the street. Though he was in open violation of a curfew order, Kenosha police encouraged him to act violently. Kenosha police told militia members that they would push peaceful protestors toward the militia so that the militia could "deal with them." Soon after, Mr. Rittenhouse killed Joseph Rosenbaum. The police did nothing. Concerned citizens, confronted with a person shooting indiscriminately on the street, stepped in to stop the violence. Anthony was shot in the chest trying to disarm Mr. Rittenhouse and stop his shooting spree. Still, the police did nothing. Mr. Rittenhouse continued to shoot, maiming Gaige Grosskreutz. The police let Mr. Rittenhouse leave the scene freely. Mr. Rittenhouse came to Kenosha armed to kill. Kenosha police encouraged him to act violently, and our son is dead as a result.
We are so proud of Anthony, and we love him so much. He is a hero who sacrificed his own life to protect other innocent civilians. We ask that you remember Anthony and keep him in your prayers.
Additional questions can be directed to our attorneys Steve Art (steve@loevy.com), Anand Swaminathan (anand@loevy.com), Dan Twetten (dan@loevy.com), and Quinn Rallins (quinn@loevy.com).
The statement reproduced below was released by John Huber & Karen Bloom on August 17, 2021, after they filed their civil rights lawsuit against the City and County of Kenosha. That lawsuit is pending in federal district court in Milwaukee, Wisconsin, under the caption Huber v. Beth, et al., No. 2:21-cv-00969. A copy of the suit can be found here. Questions about the pending lawsuit can be directed to John Huber & Karen Bloom's attorneys Steve Art (steve@loevy.com), Anand Swaminathan (anand@loevy.com), Dan Twetten (dan@loevy.com), and Quinn Rallins (quinn@loevy.com).
Milwaukee, August 17, 2021 – Almost one year ago, Anthony Huber was tragically shot and killed in downtown Kenosha by Kyle Rittenhouse during peaceful demonstrations against police violence that followed the police shooting of Jacob Blake. Today, the family of Anthony Huber has filed a constitutional civil rights lawsuit in federal court against the City of Kenosha Police Department and Kenosha County Sheriff's Department for their role in Rittenhouse's deadly rampage that left the Hubers' son Anthony dead.
Kenosha City and County Police took sides during the peaceful protest and deputized Kyle Rittenhouse and the other armed, pro-police militia members to patrol the streets and control demonstrators who were protesting police violence and racism. The lawsuit alleges that the police invited in Rittenhouse and other illegally armed individuals to patrol the streets with deadly weapons to mete out justice as they saw fit.
The lawsuit also lays out a detailed timeline of the events leading up to Rittenhouse's shooting spree and its aftermath. Armed militia members posted racist and violent messages on social media and gave the Kenosha Chief of Police advance notice they were coming. Kenosha Police praised and encouraged the armed militia members when they arrived, allowed them to patrol the street with dangerous weapons, and made no effort to stop or disarm illegally armed individuals. Making matters worse, militia members were captured on video saying that the police planned to push the demonstrators toward armed militia members so that they could "deal with them." Shortly afterward, Rittenhouse's shooting spree began. When it ended, Anthony Huber and Joseph Rosenbaum were dead, and Gaige Grosskreutz had been shot and injured. Instead of arresting Rittenhouse, Kenosha Police allowed him to walk right past them and away from the scene of the crimes.
"After he had killed and maimed multiple individuals, Kyle Rittenhouse walked up to a dozen Kenosha police officers, assault rifle in hand, with crowds yelling that he had just killed innocent people. What did the police do? They spoke to him and let him walk away," said Huber family attorney Anand Swaminathan. "Make no mistake: if Kyle Rittenhouse had been Black, or had been wearing a Black Lives Matter t-shirt, there is no way on God's green earth he would have been allowed to walk away. More likely he would have been shot dead in the street."
The lawsuit seeks to hold the City of Kenosha Police Department and the Kenosha County Sheriff's Department responsible for inviting armed militia members to patrol the streets, deputizing and conspiring with them to control peaceful civilians, and creating a dangerous and unsafe environment in which armed militia members took matters into their own hands, leading predictably to Rittenhouse's deadly rampage.
"The police are supposed to serve and protect," Anthony's father, John Huber, said in a statement, "but that's not what the Kenosha police did. They walked away from their duties and turned over the streets of Kenosha to Kyle Rittenhouse and other armed vigilantes. If they had done their job, my son would still be alive today."
The lawsuit also alleges that the decision of Kenosha Police to treat demonstrators and militia members differently was motivated by racial discrimination and was in retaliation for protests critical of police violence. While Kenosha Police allowed armed militia members who were white and supported the police to roam the streets illegally with weapons, they controlled and strictly enforced a curfew against peaceful protestors, who were a diverse group speaking out against police violence. That discriminatory and retaliatory conduct violated the U.S. Constitution's guarantees of a right to free speech and peaceful assembly and to equal protection of the law.
"There is an ongoing criminal process to hold Kyle Rittenhouse accountable for his actions. But so far there has been zero accountability for the Kenosha Police who let Rittenhouse fire an assault weapon at peaceful civilians. That ends today." said Steve Art, another one of the Huber family's attorneys. "In Kenosha, Wisconsin, Kyle Rittenhouse was allowed to shoot three people in the street with an assault weapon and then walk away, while Jacob Blake was shot by police seven times in the back while he walked unarmed to a car in which two children were waiting. Kenosha authorities should be ashamed of themselves."
Anthony Huber's mother, Karen Bloom, said the lawsuit is about justice for her son, and his legacy. "After Anthony's death, there will always be a hole in my heart. But his memory lives on, and I will never stop telling his story." The Huber family is represented by Jon Loevy, Anand Swaminathan, Steven Art, Dan Twetten and Quinn Rallins, of the civil rights law firm Loevy & Loevy Attorneys at Law. Loevy & Loevy is one of the nation's largest civil rights law firms.
By Karen Bloom & John Huber
Kimberly Long Sues Corona, CA Police Who Framed Her for Murder
By Andy Thayer
Parents of Cedric Lofton Demand Immediate Access to Video Showing Their Teenage Son's Death in Custody
One of the Nation's Longest Wrongfully Imprisoned People Sues Cops Who Allegedly Framed Him
John Huber & Karen Bloom Sue Kenosha Police for the Shooting Death of Their Son Anthony Huber
Ronald Watts' Victim Sues for City's Report
Parents of Cedric Lofton Demand Immediate Access to Video Showing Their Teenage Son's Death in Custody	Kimberly Long Sues Corona, CA Police Who Framed Her for Murder
#anthony-huber
#kenosha
#kenosha-police
#kyle-rittenhouse
#vigilante
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text:
Is This Really How the World of HR Will Look in 2022?
By E. James (Jim) Brennan December 24, 2014 July 21, 2015
Editor's Note: It's a TLNT annual tradition to count down the most popular posts of the year. This is No. 46. Our regular content will return on Jan. 5, 2015. Happy Holidays!
The world of workplace rewards in 2022 may feature scary aspects where constantly-monitored employees are overworked, paid only for performance and pitted against each other, only to be discarded if found wanting, according to a major survey.
An alternative prediction was that the employer of the future would be dedicated to enriching individual worker talents and fulfilling personal interests while minimizing environmental impact. The third possibility described a world featuring networks of independent contractors cooperating in virtual work relationships.
These predictions came from PwC, the consulting firm that polled 10,000 people in the U.S., UK, Germany, India and China and over 500 HR managers across the world for this report.
3 scenarios for the world of work in 2022
They say that "the recruitment, reward and employee engagement strategies likely to be most relevant" merely eight years from now will include sensors monitoring employee psychological behaviors, a Chief Performance Officer title combining HR and Finance functions, and remote work extending to include surgery and other professional services.
Wisely hedging their bets to cover all possibilities, the consultants offer three scenarios for the future world of work they see for 2022.
Large corporations become mini-states with social roles;
Specialization promotes the rise of modular collaborative networks;
Business strategy follows social and environmental agendas.
As skill gaps grow and the velocity of change accelerates, HR is predicted to go one of three different ways, becoming more proactive and vitally strategic, outsourced as a transactional function, or the driver of corporate social responsibility.
Possible roles for HR
Quite a range of options are expected in this three-cornered universe:
Corporate is King – HR becomes a major player at the big table, driving a relentless capitalistic pressure to perform, with intrusive effects on personal freedoms;
Small is Beautiful — HR is considered essentially irrelevant except as an outside vendor coordinating independent contractor services;
Companies Care — HR plays the "company conscience" role as ethical/environmental motives dominate employer strategies.
Few of these interesting predictions seem to include most of the traditional work relationships found in HR today. Most enterprises tend to take the capitalistic approach. Evidence is scarce to expect that many will refocus their workplace rewards systems to meet those new general standards predicted to be the reality only eight years from now.
I wonder if that means that the new models will swiftly effectively replace all the old ones. That hasn't happened in the past.
Will all this change happen in just 8 years?
Even now in the 21st Century, feudal work patterns and sweatshops similar to the beginning of the industrial revolution can still be found along with Fredrick Taylor's scientific management versions. It is hard to imagine all that vested tradition being suddenly swept away by new outbursts of capitalism on steroids, social responsibility objectives and/or snazzy new technology enabling more remote workers and networked independent contractor relationships.
If the seeds of the new HR approaches forecasted for the next decade are not found already deeply imbedded in current practices, it seems doubtful that the globe will experience changes as dramatic as those, particularly not as quickly as they are promised.
Read the report yourself. Do you see many reward systems consistent with those predictions existing in today's workplace?
This was originally published at the Compensation Café blog, where you can find a daily dose of caffeinated conversation on everything compensation.
E. James (Jim) Brennan
E. James (Jim) Brennan is Senior Associate of ERI Economic Research Institute, the premier publisher of interactive pay and living-cost surveys. Semi-retired after over 40 years in HR corporate and consulting roles throughout the U.S. and Canada, he's pretty much been there done that (articles, books, speeches, seminars, radio/TV, advisory posts, in-trial expert witness stuff, etc.), and will express his opinion on almost anything. Contact him at ej.brennan@erieri.com.
Change ManagementCultureFuture of HRHR InsightsHR ManagementHR TrendsPlanningEditor's Pick


We take data security very seriously so we use only trusted service providers with the highest security standards. On this page, you will find details on what data we store, where we store it and how we secure it. If you think something is missing or you have any security related questions please let us know at Spartez Support Portal or support@spartez.com.
Canned Responses application runs on Heroku environment. Besides the application itself, Heroku stores application access logs that contain user IP, user key, instance URL and user JWT token. These logs are purged after 7 days.
We use Heroku addon mLab MongoDB to store the templates. Each Jira instance uses a separate collection. Only users from particular Jira instance can access the templates from that instance.
The database is backed up every week and 8 last backups are stored.
We don't store issue summaries, descriptions, comments nor other sensitive information. We don't store full usernames nor e-mails but we use user keys and project keys provided by Jira instead.
We use Heroku addon Papertrail to store application logs. Logs are in the system for 7 days. Log archives are stored from the last 365 days. Papertrail allows us to detect suspicious patterns in the logs and notify us whenever they are detected which improves our reaction time. It also allows us to analyze application's behavior after an incident occurs.
For the better understanding of our clients, we collect anonymous statistics of the add-on usage. These statistics tell us how we should develop our plugin to make our customers happy.


text:
Q: Cakephp 2.6 : Role wise isAuthorized not working I have several method in UsersController, I am trying to give role wise access.
if

*

*user_types == 1 (user can access all methods)

*user_types == 2 (user can't access admin_list method.

*user_types == 3 (user can access only forget_password method)

In controller I have tried below code
public $components = array('Session','RequestHandler','Auth'=>array(

            'loginRedirect' => array('controller' => 'users','action' => 'dashboard'),
            'logoutRedirect' => array('controller' => 'users','action' => 'login'),
            'authError'=>'You can not access this page!!',
            'authenticate' => array(
            'Form' => array(
                'fields' => array(
                  'username' => 'email', //Default is 'username' in the userModel
                  'password' => 'password'  //Default is 'password' in the userModel
                )
              ),
            ),
            'authorize' => array('Controller')
        ));

        public function isAuthorized($user) {
            return true;
        }

In before filter I have allowed
    $this->Auth->allow('login','logout');

Now in UserController I have tried below code
public function isAuthorized($user) {
    // Admin can access every action
    if (isset($user['usertype_id']) && $user['usertype_id'] == 1) {
            return true;
    }
    else if(isset($user['usertype_id']) && $user['usertype_id'] == 2)
    {
         $this->Auth->deny('admin_list');
    }else
        $this->Auth->allow('change_password');

    return parent::isAuthorized($user);
    }

Problem is it's always returning true. If I login with user_type = 3 , I can access all methods.

A: Auth::allow() and Auth::deny() are meant to define which actions are allowed to be accessed by non logged-in users (authentication), and are not to be used for authorization.
For this purpose, you have to define isAuthorized() in your controller, as you have done. However, this method is expected to return true (logged in user/group is authorized to access action) or false (authorization is denied).
Your UsersController::isAuthorized() method should be rewritten as:
public function isAuthorized($user) {

    if (!isset($user['usertype_id'])) {
        return false;
    }

    if($user['usertype_id'] == 2 && $this->action=='admin_list') {
       return false;
    }

    if($user['usertype_id'] == 3) && $this->action!='forget_password'){
        return false;
    }

    return true;
}

See the Cookbook for more on ControllerAuthorize.


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Tim Oksanen – Red
December 5, 2017 By The Ark of Music Leave a Comment
Native to Brisbane, Queensland, Australia, singer/songwriter, Tim Oksanen, developed a love for making music long ago. The son of a pastor, it all began as a kid playing keyboard in his mother's church band.
Fast-forward many years, and to no surprise, music has become an important—and at times therapeutic—feature of his life. For, despite the fact that Tim has been immersed in his studies as a medical student for the past four years, he's still managed to carve out enough time to record his official debut, solo album, Red; a project he created with the hope and intention of bringing about healing, inspiration, encouragement, and empathy to others going through tough times.
The album's talented supporting cast includes: wife Naomi Oksanen, Janet O'Sullivan, and mother Christine Oksanen (vocals), The Little River Band's George McArdle (bass), Teegan Dowdell (violin), Steve Huf and Darran Muller (drums, percussion), Mika Oksanen, Sam See, and Paris Barclay (guitars), and Daniel Spirovski (saxophone).
Here's what we dug most…
The album opens with the soft and jazzy, Wonders (feat. Janet O'Sullivan); a song with a gentle sound and a divinely curious message…
"Eyes are now wandering
Into the sky and above
Searching what one cannot comprehend
Seeing wonders that sing the same song"
OUR FAVORITE TRACK:
With an impressive and inspiring suggestion for all of us to find the beauty and the wholesomeness in every situation, the album's title track, Red, is as strong and inspiring as a title track should be. A piano-based melody supports Oksanen's strong vocal and writing efforts to create one truly lovely song…
"But don't push the red away
No matter how hard it seems
For pushing the red away
Means the red rose falls from your gaze…"
Delicate finger-picking, keys, violin, vocal harmonies and more, all merge to create the beautiful ballad, Love's Dance (feat. Naomi Oksanen). The track takes its time, and is a clear example (for any music-head) of the tender-loving-care and focus which was taken to create this album.
The rich, fully-symphonic soundscape of the instrumental, The Joanne, creates an imagination-stirring sonic voyage which, at 2:50, left me wanting more.
Petrichor, the album's third instrumental, tastefully incorporates celestial synth tones, a rolling piano melody, and a sweet bit of violin. The track climaxes at the 2:00 mark, begging to be placed on the sound-score of the next great romance film.
I'm going to do something here which I've never done before. I'm going to admit that before I listened to this album—I made an assumption. I assumed it was going to be your average singer/songwriter collection. I assumed that Oksanen's talent would be ample and worthy of our review, but that the album as a whole would not leave a lasting mark on my musical memory. I assumed all of those things. I judged a book by its cover.
As a whole, Red is rich and full of well-executed sonic-contrast. Highly competent, vocally stirring pieces are rivaled—in equal measure—by their instrumental-based counterparts. Concerning the vocal presence of Oksanen, O'Sullivan and company (which is just damn lovely), what this means is that it leaves the listener to simultaneously love their presence—and their absence. And, in an age where most popular music is packed to the gills with empty words and meaningless monikers, it's wildly refreshing to discover a good instrumental piece. (This album offers three.) All in all, Oksanen's four-year labor of love has paid-off, as he has created a project worthy of music fans everywhere, particularly those inclined to lean more toward sound with substance, intention, and heart.
Let me also add that investing in the talents of George McArdle on bass was a wonderful decision, as his contribution creates a strong melodic foundation throughout the album.
WHOSE LOVECHILD…?
Magazine Gap meets Jason Mraz…
BELOW: Listen to music from Tim Oksanen's debut album, Red, and connect with his website and social media platforms. Please support Tim Oksanen by visiting him online, and playing, downloading, and/or purchasing his music. And, as always, thank you for supporting real music!
Listen to music from Tim Oksanen's debut album, Red:
Follow, stream, download & connect with Tim Oksanen online:
Share Tim Oksanen with your people:
*Album Reviews, Easy Listening, Pop, Singer / Songwriter


text:
McPhee, Joe / John Edwards / Klaus Kugel
A Night In Alchemia
(Not Two - Poland)
The working trio of New York trumpeter & saxophonist Joe McPhee, London double bassist John Edwards, and German drummer Klaus Kugel, actively touring through Europe since 2017, are heard live at Alchemia in Krakow, Poland in 2018, for four monumental examples of free improvisation in superlative playing with intense rhythmic interaction and deeply expressive soloing.
Label: Not Two
Catalog ID: MW992-2
Recorded at Alchemia, in Krakow, Poland, on October 20th, 2018, by Ralf Drewiany.
Joe McPhee-trumpet, saxophones
John Edwards-double bass
Klaus Kugel-drums
• Show Bio for Joe McPhee
"Joe McPhee, born November 3,1939 in Miami, Florida, USA, is a multi-instrumentalist, composer, improviser, conceptualist and theoretician. He began playing the trumpet at age eight, taught by his father, himself a trumpet player. He continued on that instrument through his formative school years and later in a U.S. Army band stationed in Germany, at which time he was introduced to performing traditional jazz. Clifford Thornton's Freedom and Unity, released in 1969 on the Third World label, is the first recording on which he appears as a side man. In 1968, inspired by the music of Albert Ayler, he took up the saxophone and began an active involvement in both acoustic and electronic music.
His first recordings as leader appeared on the CJ Records label, founded in 1969 by painter Craig Johnson. These include Underground Railroad by the Joe McPhee Quartet (1969), Nation Time (1970), Trinity (1971) and Pieces of Light (1974). In 1975, Swiss entrepreneur Werner X. Uehlinger release Black Magic Man by McPhee, on what was to become Hat Hut Records.
In 1981, he met composer, accordionist, performer, and educator Pauline Oliveros, whose theories of "deep listening" strengthened his interests in extended instrumental and electronic techniques. he also discovered Edward de Bono's book Lateral Thinking: A Textbook of Creativity, which presents concepts for solving problems by "disrupting an apparent sequence and arriving at the solution from another angle." de Bono's theories inspired McPhee to apply this "sideways thinking" to his own work in creative improvisation, resulting in the concept of "Po Music." McPhee describes "Po Music" as a "process of provocation" (Po is a language indicator to show that provocation is being used) to "move from one fixed set of ideas in an attempt to discover new ones." He concludes, "It is a Positive, Possible, Poetic Hypothesis." The results of this application of Po principles to creative improvisation can be heard on several Hat Art recordings, including Topology, Linear B, and Oleo & a Future Retrospective.
In 1997, McPhee discovered two like-minded improvisers in bassist Dominic Duval and drummer Jay Rosen. The trio premiered at the Vision Jazz Festival in 1998 but the concert went unnoticed by the press. McPhee, Duval, and Rosen therefore decided that an apt title for the group would be Trio X. In 2004 he created Survival Unit III with Fred Lonberg-Holm and Michael Zerang to expand his musical horizons and with a career spanning nearly 50 years and over 100 recordings, he continues to tour internationally, forge new connections while reaching for music's outer limits."
-Joe McPhee Website (http://joemcphee.com/bio.html)
^ Hide Bio for Joe McPhee
• Show Bio for John Edwards
"After taking up the bass, around 1987, John Edwards co-formed The Pointy Birds who went on to win awards for their music for The Cholmondeleys and Featherstonehaughs dance troupes. The group appeared at festivals in Glasgow, Edinburgh, Moers, Leverkusen, Copenhagen. Around 1990, Edwards played his first gigs with London improvisers such as Roger Turner, Lol Coxhill, Maggie Nicols, Phil Minton.
Between 1990 and 1995 Edwards was a member of three touring groups simultaneously: B-Shops For The Poor, The Honkies and GOD. During this period he also became an increasingly regular player on the London improvised music scene and performed his first solo gigs; he composed and performed music theatre with the bass and cello duo The Great Explorers, street-busked a lot and appeared at many more festivals in Germany, Estonia, France, Italy, Czech, etc.
Since 1995 John Edwards has become a "mainstay" of the London scene, playing with just about everybody, an activity that has seen him clocking up between 150 and 200 gigs a year. He has become regular player with Evan Parker, in many groupings, and with Tony Bevan, Veryan Weston, and Elton Dean, often in collaboration with Mark Sanders on percussion. He has become a more frequent player on the European (and festival) scene, appearing at Taktlos, Ulrichsburg, Nickelsdorf, Budapest, New Zealand and in the USA. He continues to work on solo performances."
-EFI (http://www.efi.group.shef.ac.uk/musician/medwards.html)
^ Hide Bio for John Edwards
• Show Bio for Klaus Kugel
"Klaus Kugel studied at the School of Jazz in Munich. Since 1989 he attracted attention worldwide through projects with Petras Vysniauskas, Karl Berger, Tomasz Stanko, Charlie Mariano, Kent Carter, Michel Pilz, Theo Jorgensmann, Kenny Wheeler, Vyacheslav Ganelin, Bobo Stenson, Glen Moore, Steve Swell, Sabir Mateen, Robert Dick, Peter Evans, Perry Robinson, Hilliard Greene, John Lindberg, Gerd Dudek, William Parker, Bobby Few, Roy Campbell, Jemeel Moondoc, Waclaw Zimpel, Mars Williams, Charles Gayle, Conny Bauer, Ken Vandermark, Joe McPhee, John Edwards a.m.o..
Over the past 30 years, he gave numerous concerts and appeared at festivals throughout Europe, the Baltic States, Canada, USA, Syria, Japan, Mexico, Russia, China, Ukraine, Azerbaijan and Israel."
-Klaus Kugel Website (https://www.klauskugel.com/)
^ Hide Bio for Klaus Kugel
1. Burden Of Proof 10:50
2. Just To Wait 6:12
3. At The Waters Edge 23:17
4. To Watch A Mist Rise 7:01
NY Downtown & Metropolitan Jazz/Improv
Joe McPhee
Collective Free Improvsation
Not Two.
Mateen, Sabir / Christopher Dell / Christian Ramond / Klaus Kugel
Krakow Improvisers Orchestra
KIO at Cricoteka
Big Tent (Jerome Kitzke / Steve Rust / Harvey Sorgen)
I Am Waiting
Alcorn, Susan / Joe McPhee / Ken Vandermark
Invitation To A Dream
Imaginary Numbers (McPhee / Niggenkemper / Solberg)
Brotzmann / Edwards / Noble
Soulfood Available
Jorgensmann, Theo Quartet
Snijbloemen
Decoy (Alexander Hawkins / John Edwards / Steve Noble) With Joe McPhee
Sun Ra Arkestra
Heliocentric Worlds 1 and 2
Ayler, Albert Trio
1964 Prophecy Revisted
Croaks, The (Kuchen / Turner / Rogers)
One Of The Best Bears!
Watts, Trevor Quartet The (Veryan Weston / John Edwards / Marc Sanders)
The Real Intention - Cafe OTO concert
Morris, Joe
Instantiation: Paradoxical
(Glacial Erratic)
Eisenstadt, Harris (w/ Wooley / Hawkins / Niggenkemper)
Canada Day Quartet Live
Kid Millions / Sarah Bernstein
Broken Fall (VINYL)
Carter, Daniel / Patrick Holmes / Matthew Putman / Hilliard Greene / Federico Ughi
Electric Telepathy, Vol. 1 [VINYL]
Berman / Lytton / Roebke
Trio Discrepancies [VINYL]
Copland, Marc / Drew Gress / Joey Baron
(Illusions/Mirage)
Pursglove, Percy / Paul Dunmall / Mark Sanders / John Etheridge
Hawkins, Chester
Cops Need Serial Numbers [USB mp3 & wav + pdf]
(Intangible Arts)
von Orelli, Marco / Tommy Meier / Luca Sisera / Sheldon Suter
Lotus Crash
Shipp, Matthew
Invisible Light, Live Sao Paulo
SLD Trio (Shocron / Lamonega / Diaz)
El Contorno del Espacio
Rocher, Christophe / Joe Fonda / Harvey Sorgen
New Origin


Where do I start? Last month, I ended my Publisher's Letter by reminding people to register to vote if they weren't already registered. Now, the midterm election is here, so, please, vote. It's important. And if you don't believe me, check out the Lani Says column on page 10. She's not old enough to vote yet, but she gets it.
The paper this month hits on a lot of the issues. Laura Brady-Allen writes about education on page 1, with L.A. teachers ready to go on strike, as teachers have done in other cities. T.A. Hendrickson writes about immigration, also on page 1, with local leaders trying to take on the Trump Administration. Homelessness is the focus of stories by our columnist Lionel Shockness (page 11) and by Bill Hendrickson (also on page 11). Homelessness is also the theme in this month's "Get to Know" column by Jeffery Marino, which profiles the extraordinary Father Clarke Prescott of All Saints Episcopal Church in Highland Park.
Crime is also a perennial issue. Laura Brady-Allen has been talking to the cops about the shooting last month at Johnny's Bar in Highland Park (page 7) and the lockdown last month at Franklin High (page 6). She also had her own experience that led her to call the cops – a non-emergency, thank goodness – that she relates on page 7, with really interesting insights.
I haven't forgotten it's Thanksgiving. Enjoy your families. Give thanks. And give thanks to veterans too on Veterans Day this month. You can find local Veterans Day events on page 16.
If shopping is part of your Thanksgiving weekend tradition, Shop Small, and patronize the advertisers on page 9 who will be taking part in Shop Small Saturday in Eagle Rock on Nov. 24. And then, check out pages 12 and 13, where we review the local nonprofits, any and all of which would be worthy of a donation on Giving Tuesday, Nov. 27, the Tuesday after Thanksgiving.