January 17th 1929; Popeye the Sailor Man, a cartoon character created by Elzie Segar, first appears in the Thimble Theatre comic strip. Popeye the Sailor Man is a fictional hero notable for appearing in comic strips and animated films as well as numerous television shows. He was created by Elzie Crisler Segar, and first appeared in the daily King Features comic strip Thimble Theatre on January 17, 1929. Popeye has now become the strip's title as well.
Although Segar's Thimble Theatre strip, first published on December 19, 1919, was in its tenth year when Popeye made his debut, the sailor quickly became the main focus of the strip and Thimble Theatre became one of King Features' most popular properties during the 1930s. Thimble Theatre was carried on after Segar's death in 1938 by several writers and artists, most notably Segar's assistant Bud Sagendorf. The strip, now titled Popeye, continues to appear in first-run installments in its Sunday edition, written and drawn by Hy Eisman. The daily strips are reprints of old Sagendorf stories.
In 1933, Max and Dave Fleischer's Fleischer Studios adapted the Thimble Theatre characters into a series of Popeye the Sailor theatrical cartoon shorts for Paramount Pictures. These cartoons proved to be among the most popular of the 1930s, and the Fleischers—and later Paramount's own Famous Studios—continued production through 1957. The cartoons are now owned by Turner Entertainment, a subsidiary of Time Warner, and distributed by sister company Warner Bros. Entertainment.
Over the years, Popeye has also appeared in comic books, television cartoons, arcade and video games, hundreds of advertisements and peripheral products, and a 1980 live-action film directed by Robert Altman starring comedian Robin Williams as Popeye.
In most appearances (except during the World War II era), Popeye is a middle-aged sailor with a unique way of speaking, disproportionately muscular forearms with two anchor tattoos, thinning hair, and an ever-present corncob pipe (which he toots like a steamship's whistle at times). Popeye is generally depicted as having only one eye, his left. In at least one Fleischer cartoon, Bluto refers to Popeye as a "one-eyed runt." Mostly quiet as to how he lost his right eye, the sailor claims it was in "the mos' arful battle" of his life. Later versions of the character had both eyes, with one of them merely being squinty, or "squinky" as he put it. According to the official site, Popeye is 34 years old and was born in a typhoon off Santa Monica, California (however, in Popeye, the Ace of Space his original age is given as 40 by an alien aging machine).
In addition to a gravelly voice and a casual attitude towards grammar, Popeye is known for having an apparent speech impediment (a common character-distinguishing device in early cartoons), which either comes naturally or is caused by the ever-present pipe in his mouth. Among other things, he has problems enunciating a trailing "t"; thus, "fist" becomes "fisk" (as sung in his theme song, which makes it conveniently rhyme with "risk") and "infant" becomes "infink." This speech impediment even found its way into some of the titles of the cartoons.
Popeye is depicted as having superhuman strength, though the nature of his strength changes depending on which medium he is represented in. Originally, the comic-strip Popeye gained his strength and invulnerability in 1929 by rubbing the head of the rare Whiffle Hen. From early 1932 onward, especially the cartoons, Popeye was depicted as eating spinach to become stronger. The animated shorts depicted Popeye as ridiculously strong, but liable to be pummeled by the much larger Bluto before his eating of the spinach.
When fed up with this treatment or exhausted, he would eat spinach, which would instantly restore and amplify his strength to an even greater level. In the comic strips, spinach is presented as a panacea, infusing Popeye not only with his extraordinary strength, but also making him invulnerable to all sorts of threats (including bullets, a basilisk's petrifying gaze, or aliens' weapons) and even capable of feats like flight or extraordinarily fast swimming (usually with the aid of his pipe as a propeller). In the animated shorts, Popeye's ingestion of spinach – which is almost invariably canned – is equally fanciful and often involves squeezing the can until the top opens, or sucking the spinach through his pipe, and on rare occasions, even ingesting the can as well. Occasionally, spinach has a similar invigorating power on other characters.
Other differences in Popeye's story and characterization show up depending upon which medium he is presented in. While Swee'Pea is definitively the adopted child of Popeye in the comic strips, he is often depicted as being related to Olive Oyl in cartoons. The cartoons also occasionally feature family members of Popeye that have never appeared in the strip, notably his look-alike nephews Peepeye, Pupeye, Pipeye, and Poopeye.
Even though there is no absolute sense of continuity in the stories, certain plot and presentation elements remain mostly constant, including purposeful contradictions in Popeye's capabilities. Though at times he seems bereft of manners or uneducated, Popeye is often depicted as capable of coming up with solutions to problems that (to the police, or, most importantly, the scientific community) seem insurmountable. Popeye has, alternatively, displayed Sherlock Holmes-like investigative prowess, determining for instance that his beloved Olive was abducted by estimating the depth of the villains' footprints in the sand, scientific ingenuity (as his construction, within a few hours, of a "spinach-drive" spaceship), or oversimplified (yet successful) diplomatic argumentation, by presenting to diplomatic conferences his own existence (and superhuman strength) as the only true guarantee of world peace.
Popeye's vastly versatile exploits are deemed even more amusing by a few standard plot elements. One is the love triangle between Popeye, Olive and Bluto, and the latter's endless machinations to claim Olive at Popeye's expense. Another is his (near-saintly) perseverance to overcome any obstacle to please Olive – who, quite often, treats him like dirt, and ends up being the only character capable of beating him up. Finally, in terms of the endless array of villain plots, Popeye mostly comes to the truth by "accidentally" sneaking on the villains, the moment they are bragging about their schemes' ingenuity, thus revealing everything to an enraged Popeye, who uses his "fisks" in the name of justice.
Thimble Theatre's first main characters/actors were the thin Olive Oyl and her boyfriend, Harold Hamgravy. After the strip moved away from its initial focus, it settled into a comedy-adventure style featuring Olive, Ham Gravy, and Olive's enterprising brother, Castor Oyl. Olive's parents, Cole and Nana Oyl, also made frequent appearances.
Popeye first appeared in the strip on January 17, 1929 as a minor character. He was initially hired by Castor Oyl and Ham to crew a ship for a voyage to Dice Island, the location of a casino owned by the crooked gambler Fadewell. Castor intended to break the bank at the casino using the unbeatable good luck conferred by stroking the hairs on the head of Bernice the Whiffle Hen. Weeks later, on the trip back, Popeye was shot many times by Jack Snork, a stooge of Fadewell's but survived by rubbing Bernice's head. After the adventure, Popeye left the strip—but due to reader reaction he was quickly brought back.
The Popeye character became so popular that he was given a larger role, and the strip was expanded into many more newspapers as a result. Though initial strips presented Olive Oyl as being less than impressed with Popeye, she eventually left Ham Gravy to become Popeye's girlfriend-and Ham Gravy left the strip as a regular. Over the years, however, she has often displayed a fickle attitude towards the sailor. Castor Oyl continued to come up with get-rich-quick schemes and enlisted Popeye in his misadventures. Eventually he settled down as a detective and later on bought a ranch out West. Castor has seldom appeared in recent years.
In 1933, Popeye received a foundling baby in the mail, whom he adopted and named "Swee'Pea." Other regular characters in the strip were J. Wellington Wimpy, a hamburger-loving moocher who would "gladly pay you Tuesday for a hamburger today" (he was also soft-spoken and cowardly, which may have led to the coinage of the word wimpy); George W. Geezil, a local cobbler who spoke in a heavily affected accent and habitually attempted to murder or wish death upon Wimpy; and Eugene the Jeep, a yellow, vaguely dog-like animal from Africa with magical powers. In addition, the strip featured the Sea Hag, a terrible pirate, as well as the last witch on earth (her even more terrible sister excepted); Alice the Goon, a monstrous creature who entered the strip as the Sea Hag's henchman and continued as Swee'Pea's babysitter; and Toar, a caveman.
Segar's strip was quite different from the cartoons that followed. The stories were more complex, with many characters who never appeared in the cartoons (King Blozo, for example). Spinach usage was rare and Bluto made only one appearance. Segar would sign some of his early Popeye comic strips with a cigar, due to his last name being a homophone of "cigar" (pronounced SEE-gar).
Thimble Theatre soon became one of King Features' most popular strips during the 1930s and, following an eventual name change to Popeye in the 1970s, remains one of the longest running strips in syndication today. The strip carried on after Segar's death in 1938, at which point he was replaced by a series of artists. In the 1950s, a spinoff strip was established, called Popeye the Sailorman. Acknowledging Popeye's growing popularity, the Thimble Theatre strip was re-named Thimble Theatre Starring Popeye during the 1960s and 1970s, and was eventually retitled, simply, Popeye, the name under which the strip continues to run.


text:
\section{Introduction}
In the last decade, systems generalising $M/G/1/\infty,$ or
simply $M/G/1$ (cf. \cite{GK}) -- one of the oldest and most
popular queueing systems -- attracted much attention, see
\cite{Asmussen} -- \cite{fakinos1987},
 \cite{Thor83}
In this paper a single--server system similar to \cite{Ve2013_ait}
is considered,
in which {\em intensities} of new arrivals as well as of their
serving could depend on the ``whole state'' of the system and the
whole state includes the number of customers in the system --
waiting and on service -- {\em and} the elapsed time of the last
serving. Batch arrivals are not allowed. Emphasize that the {\em
elapsed} serving time is assumed known at any moment, but {\em
not} remaining service times for each customer. For definiteness,
the discipline of serving may be regarded as FIFO, although, this
is of no importance for the questions addressed in this paper: all
results are valid for any serving discipline without time
sharing. Recall that despite a non-markovian character of the
model, this model -- as any other process -- may be made Markov by
extending the state space, although not in all cases it may be
helpful; in our case it will be crucial. {\em Dynkin's formula},
or {\em Dynkin's identity} \cite[Ch. 1, \S 3]{Dynkin},
\begin{equation}\label{dynkin1}
E_Xf(X_t) - f(X) = E_X\int\limits_0^t {\mathcal G}f(X_s)\,ds
\end{equation}
for $f$ from some appropriate class of functions and its
non-homogeneous counterpart
\begin{equation}\label{dynkin_t}
E_X\varphi(t,X_t) - \varphi(0,X) = E_X\int\limits_0^t
\left(\frac{\partial}{\partial s}\varphi(s, X_s) + {\mathcal
G}\varphi(s,X_s)\right)\,ds,
\end{equation}
for appropriate functions of two variables $(\varphi(t,X))$ play
a very important role in the analysis of Markov models. Here
${\mathcal G}$ is understood as a {\em generator} or {\em
infinitesimal operator} of the process, and $X$ is a (non-random)
initial value of the process. Note that the formula
(\ref{dynkin_t}) may be considered as a version of (\ref{dynkin1})
written for the couple $(t,X_t)$. In this paper it is shown that
both formulae (\ref{dynkin1})--(\ref{dynkin_t}) hold true for a
large class of functions $f$, $\varphi$ with an {\em
extended generator} ${\mathcal G}$ (see below) and under the
minimal assumptions on regularity of intensities (i.e.,
without any regularity). For short, in the title it was called
Dynkin's identity, but, in fact, Dynkin's formula with
generalised generators will be established. As an application, convergence rate bounds to
stationary distribution are established under appropriate
recurrence conditions similar to those in \cite{Ve2013_ait}, but
without regularity assumed therein.
Minimal or no regularity may
be important, in particular, in queueing systems with control,
where {\em optimal control} is usually discontinuous.

This model may be considered as a partial case of piecewise deterministic Markov processes (PDMP) considered in \cite{Davis}, \cite{GK}, \cite[Ch. 7]{Jacobsen}, \cite{Kalash}, et al. Notice that the method in \cite{Davis},\cite{Jacobsen} is based on martingale theory and requires some further references. In this paper we aim to show that the ``naive'' approach based on a properly designed version of a complete probability formula works well, at least, in our particular model and may be presented in a self--contained way. Also notice that our model allows a presentation as a picewise linear Markov process (PLMP) as in \cite{GK}, \cite{Kalash}, while \cite{Davis} and \cite{Jacobsen} study a more general deterministic behaviour between ``Markov'' jumps.

The standard proof of the formulae
(\ref{dynkin1})--(\ref{dynkin_t}) given in \cite{Dynkin} in ``good cases'' is practically trivial; however,
this triviality is based on the notion of {\em generator} (not extended one), which does
assume some regularity. For example, in the theory of diffusion
processes a similar notion requires continuous coefficients of
drift and diffusion; at the same time, it is well-known that
Dynkin's formula itself still holds true for diffusion processes
with a weaker notion of extended
generator. This is possible due to PDE theory results, namely, {\em Alexandrov'x inequality} and {\em Krylov's estimate}.
In queueing theory there are no similar resources and, as a
consequence, in some classical works a strict justification of
Dynkin's formulae is just dropped.
In the authors' view, this issue must be addressed and the goal of
this paper is to show that such a rigorous justification is well possible practically without -- or with a minimum of -- martingale theory. As a first application, convegence rate bounds to the
stationary regime is shown for a class of models generalising the
system $M/G/1$ similar to \cite{Ve2013_ait} but for a wider class
of intensities. This may be useful in some other areas such as
stochastic averaging and Poisson equations, see, for example,
\cite{kulik2012}. The bounds in \cite{Ve2013_ait} were proposed as
a complement to the results from \cite{Seva}, \cite{Ve77}, et al.
about convergence to stationarity in non-Markov queueing models,
in particulare, of Erlang--Sevastyanov's type. The authors hope to
apply this approach to such systems as well.

The paper consists of the Section 1 -- Introduction, of the setting
and main results in the Section 2, of a set of auxiliary results in the
Section 3, of the proof of Dynkin's formulae in the Section 4,
of the proof of some extension in the Section 5 and of a brief
reminder of the techniques leading to convergence
rate bounds in the Section~6.

\section{The setting and main results}

\subsection{Dynkin's formulae and generalised generator}
Recall \cite{Dynkin} that the generator of a Markov process $(X_t,
\, t\ge 0)$ is an operator \( {\mathcal G}, \) such that for a
sufficiently large class of functions $f$
\begin{equation}\label{dynkin2}
 \sup_X \lim_{t\to 0} \left\|\frac{E_Xf(X_t) - f(X)}{t} -
{\mathcal G}f(X)\right\| = 0
\end{equation}
in the norm of the state space of the process. The notion of
generator does depend on this norm. However, in this paper the
classical generator is not used, so we skip this point here.
Recall further that the class of functions for which
(\ref{dynkin2}) holds true is called {\em domain} of a generator
(of course, in general, it depends on the norm) and in most cases its full
description is a hard task. Fortunately, often it suffices to have
a wide enough class of such functions. On the other hand, an
operator ${\mathcal G}$ is called {\em extended generator} if (\ref{dynkin2}) is replaced by its corollary
(\ref{dynkin1}), also for a wide enough class of functions $f$. It
turns out that in many situations this notion suffices.

Let us present the class of models under investigation in this
paper. The state space is a union of subspaces,
\[
{\mathcal X} = \{(0,0)\} \bigcup \{(n,x): \; n = 1,2,\ldots, N, \;
x\ge 0\}
\]
($1\le N\le \infty$) with topology arising from the metric
($X=(n,x)$, $X'=(n',x')$)
\[
\mbox{dist}(X, X')= |n-n'| + |x-x'|
\]
Functions of
class $C^1({\mathcal X})$ are understood as functions with
classical continuous derivatives with respect to the variable $x$.
Functions with compact support on ${\mathcal X}$ are understood
as functions vanishing outside some domain bounded in this metric:
for example, $C^1_0({\mathcal X})$ stands for the class of
functions with compact support and one continuous derivative.
There is a generalised Poisson arrival flow with intensity $
\lambda(X), $ where \( X = (n,x) \; \mbox{for $n>0$}, \;
\mbox{and} \; X=(0,0) \; \mbox{for $n=0$}\); group arrival are not allowed.
 If $n>0$, then the
server is serving one customer while all others are waiting in a
queue; if $n=0$ then the server remains idle until the next
arrival and the intensity of such arrival at the state $(0,0)$ is
constant. Here $n$ denotes the total number of customers in the
system and $x$ stands for the elapsed time of the current serving.
Denote $n_t=n(X_t)$ -- the number of customers corresponding to
the state $X_t$ and $x_t=x(X_t)$, the second component of the
process $(X_t)$, i.e. the elapsed time of the current serving. If
$n=0$, then there is no serving. Below, we will investigate formally only the
case with infinite number of waiting places, but the situation where this number
is bounded may be included by assuming $\lambda(n,x)\equiv 0$ for any $n$ large enough.
For
any $X=(n,x)$, intensity of serving $h(X) \equiv h(n,x)$ is
defined; it is also convenient to assume $h(X)=0$ for $n(X)=0$.
Both intensities $\lambda$ and $h$ are understood in the following
way, which is a {\em definition:} on any nonrandom interval of
time $[t,t+\Delta)$, conditional probability given $X_t$ that the
current serving will {\em not} be finished and there will be no
new arrivals reads,
\begin{equation}\label{intu1}
\exp\left(-\int\limits_0^{\Delta} (\lambda+h)
(n_{t},x_{t}+s)\,ds\right).
\end{equation}
This is similar to the approach developed in \cite{Davis}; the main difference is that in this paper the basis for the proof of Dynkin's identity is a complete probability (or complete expectation) formula rather than martingale problem.
In the sequel, $\lambda$ and $h$ are assumed to be {\em bounded}.
In this case, for $\Delta>0$ small enough, the expression in
(\ref{intu1}) may be rewritten as
\begin{equation}\label{intu2}
1-\int\limits_0^{\Delta} (\lambda+h)(n_{t},x_{t}+s)\,ds + O(\Delta^2),
\qquad \Delta\to 0,
\end{equation}
and this what is ``usually'' replaced by
\[
1 - (\lambda(X_t)+h(X_t))\Delta + O(\Delta^2).
\]
In our situation, the latter replacement may be incorrect because
of discontinuities of the functions $\lambda$ and $h$. Emphasize
that from time $t$ until the next jump, the evolution of the
process $X$ is {\em deterministic}. The (conditional given $X_t$)
density of the moment of a new arrival {\em or} of the end of the
current serving after $t$ at $x_t+ z$, $z\ge 0$ equals,
\begin{equation}\label{intu33}
 (\lambda(n_t, x_t+z)+h(n_t, x_t+z))\exp\left(-\int\limits_0^{z}
(\lambda+h)(n_t, x_{t}+s))\,ds\right).
\end{equation}
Further, given $X_t$, the moments of the next ``candidates'' for
jumps up and down are conditionally independent and have the
(conditional -- given $X_t$) density, respectively,
\begin{equation}\label{intu5}
\begin{array}{c}
 \lambda(n_t, x_t+z)\exp\left(-\int\limits_0^{z}
\lambda(n_t, x_t+s)\,ds\right) \; \\
 \\
 \mbox{and} \; \\
 \\
 h(n_t, x_t+z)\exp\left(-\int\limits_0^{z} h(n_t, x_t+s)\,ds\right), \; z\ge 0.\\

\end{array}
\end{equation}
Notice that (\ref{intu33}) does correspond to conditionally
independent densities given in (\ref{intu5}).

For modelling the evolution, both candidate moments should be
realised and the minimal of the two chosen, which will determine
whether the next jump will be, indeed, up or down. The component
$n_t$ is, generally speaking, not Markov, however, the couple
$(X_t = (n_t,x_t))$ in the state space ${\mathcal X}$ {\em is}
Markov by construction and we stress out that continuity of
$\lambda$ or $h$ was not used. The distribution of the process in
the space of trajectories is uniquely determined, that is, if any
other construction is suggested, yet the distributions of all
moments of next jumps up and down ought to be the same.

A natural candidate to the role of the extended generator
is (see a short comment in the beginning of the proof of the
Theorem \ref{thm1})
\begin{eqnarray*}
{\mathcal G}f(X) := \frac{\partial}{\partial x}f(X) + \lambda(X)
(f(X^+) - f(X))
+ h(X) (f(X^-) - f(X)),
\end{eqnarray*}
(recall that $h(0,0) = 0$), where for any $X=(n,x)$,
\[
X^+ := (n+1,x), \quad X^-:= ((n-1)\vee 0,0)
\]
(here $a\vee b = \max(a,b)$).

\begin{thm}\label{thm1}
If the functions $\lambda$ and $h$ are Borel measurable and
bounded, then the formulae (\ref{dynkin1}) and (\ref{dynkin_t})
hold true for any $t>0$ for every $f\in C^1_b({\mathcal X})$ and
$\varphi\in C^1_b([0,\infty)\times {\mathcal X})$, respectively.
Moreover, the process $(X_t, \, t\ge 0)$ is strong Markov with
respect to the filtration \(({\mathcal F}^X_t, \, t\ge 0)\).
\end{thm}

~

Let
$$
L_m(X) = (n+1+x)^m, \quad L_{k,m}(t,X) = (1+t)^k L_{m}(X).
$$
The following extensions of Dynkin's formulae for unbounded
functions hold true: it will be needed in the proof of the Theorem
\ref{thm2}, even though the main body of this proof will be
hidden.
\begin{cor}\label{cor1}
Under the assumptions of the Theorem \ref{thm1},
\begin{eqnarray}\label{M2}
L_{m}(X_t) - L_{m}(X) = \int\limits_0^t \lambda(X_s)
\left[ \left(L_{m}(X^{+}_s) -
L_{m}(X_s)\right) \phantom{\frac{\partial}{\partial} }\right.
 \nonumber \\ \\ \nonumber
\left. + h(X_s) \left(L_{m}(X^{-}_s) - L_{m} (X_s)\right) +
\frac{\partial}{\partial x}L_{m}(X_s)\right]\,ds +M_t,\hspace{0.5cm}
 \end{eqnarray}
with some martingale $M_t$, and also
\begin{eqnarray}\label{M2t}
L_{k,m}(t,X_t) - L_{k,m}(0,X) = \int\limits_0^t \left[\lambda(X_s)
\left(L_{k,m}(s,X^{+}_s) - L_{k,m}(s,X_s)\right) \phantom{\frac{1}{1}}\right.
 \nonumber \\ \\ \nonumber \left.
+ h(X_s) \left(L_{k,m}(s,X^{-}_s) - L_{k,m}(s,X_s)\right) +
\left(\frac{\partial}{\partial x} + \frac{\partial}{\partial
s}\right) L_{k,m}(s,X_s)\right]\,ds +\tilde M_t,\hspace{0.5cm}
 \end{eqnarray}
with some martingale
$\tilde M_t$.
\end{cor}
About martingale language in queueing models see, for example,
\cite{Jacobsen} or \cite{LSh}.

\subsection{Stability and convergence rate}
Denote
\[
\Lambda:= \sup_{n,x: \,n>0}\lambda(n,x).
\]
Recall that the process has no explosion with probability one due
to the boundedness of both intensities, i.e., the trajectory may
have only finitely many jumps on any finite interval of time.
For establishing convergence rate to the stationary regime, we
assume similarly to \cite{Ve2013_ait},
\begin{equation}\label{eq2}
\inf_{n>0} h(n,x) \ge \frac{C_0}{1+x}, \quad x\ge 0.
\end{equation}
\begin{thm}\label{thm2}
Let the functions $\lambda$ and $h$ be Borel measurable and
bounded, the assumption (\ref{eq2}) holds and $C_0$ satisfies the
condition
\begin{equation}\label{condi}
C_0 > 4 (1 + 2 \Lambda),
\end{equation}
and let let the value $k>1$ satisfy
\begin{equation}\label{condi2}
C_0 > 2^{k+1}(1 + \Lambda 2^{k}).
\end{equation}
Then there exists a unique stationary measure $\mu$ and there
exist $m>k$, $C>0$ such that for any $t\ge 0$,
\begin{equation}\label{est}
 \|\mu^{n,x}_t - \mu \|_{TV} \le
 C \,\frac{(1+n+x)^m}{(1+t)^{k+1}},
\end{equation}
where $\mu^{n,x}_t$ is a marginal distribution of the process
$(X_t, \, t\ge 0)$ with the initial data $X=(n,x)\in {\mathcal
X}$.
\end{thm}

\begin{rem
Under the assumption (\ref{condi}), the inequality (\ref{condi2})
holds true, at least, for all values of $k>1$ close enough to one.
\end{rem}

\begin{rem
Existence and uniqueness of invariant measure may be established
under weaker assumptions. However, we do not discuss this issue in
this paper, as our goal -- beside Dynkin's formula -- is only convergence rate, for
which we need the condition (\ref{condi}) anyway.
\end{rem}

\section{Auxiliary results}
\noindent The first Lemma is an $L_1$--version of the Theorem
2.8.2 from \cite{Krylov-eng} from integration theory, originally
formulated in $L_2$ as a tool in a Doob's construction related
to {\em stochastic} integrals (the latter are not used here). In
this paper, the Lemma \ref{lem3} will serve as a bridge from the
standard complete probability formula to its integral analogue.
\begin{lem}\label{lem3}
For any $g\in L_1[0,T]$,
\[
\int\limits_0^1 \int\limits_0^T |g_{ \kappa^a_{m'}(s)} - g_s|\,ds\,da \to 0,
\]
and for almost all $a\in [0,1]$,
\[
 \int\limits_0^T |g_{ \kappa^a_{m'}(s)} - g_s|\,ds \to 0,
\]
over some subsequence $m'\to\infty$, where it was denoted
\[
 \kappa^a_m(s):= [2^m(s+a)]2^{-m}-a,
\]
where, in turn, $[a]$ is the integer value of $a\in R$ (i.e., the
nearest integer approximation of $a$ from the left). It is
accepted that outside $[0,T]$ the function $g$ equals identically
zero.

\end{lem}
{\em The proof} is an exact repetition of the calculus in the
proof of \cite[Theorem 2.8.2]{Krylov-eng} with a tiny difference
of $L_2$ replaced by $L_1$, which does admit this small change.
Hence, the details are dropped. For convenience of the reader we
only recall that it follows from approximation of $g\in L_1$ by
continuous functions, which are dense in $L_1$ and establishing
the statement for continuous functions, which is evident. In the
proof of the Theorem \ref{thm1} below a similar trick will be used
for establishing a similar auxiliary assertion (\ref{la4}), in
which a one-dimensional nature of time seems to be important.

~

\noindent The second Lemma provides a rigorous proof of the
well-known properties that probability of ``one event'' on a small
nonrandom interval of length $\Delta$ is of the order $O(\Delta)$
and probability of ``two or more events'' on the same interval is
of the order $O(\Delta^2)$; although this is a ``common
knowledge'' in queueing theory, the authors believe that for
discontinuous intensities it must be justified.

In the next Lemma we accept the following convention: all trajectories of the process are right continuous with left limits; respectively, if the value of the process $X_t$ at $t$ is given, the phrase ``no jumps on $(t,t+s]$ relates to no jumps after the moment $t$, but not at $t$, where we are not aware whether or not $X_{t-}=X_t$.  At the same time, ``no jumps on $(0,s]$'' includes no jump at zero, as negative values of $t$ are not allowed, so that automatically $X_0=X_{0+}$.

\begin{lem}\label{lem1}
Under the assumptions of the Theorem \ref{thm1}, for any $t\ge 0$,
\begin{eqnarray}
P_{X_{t}}(\mbox{no jumps on $(t, t+\Delta]$}) =
\exp\left(-\int\limits_0^\Delta (\lambda+h)(X_t+s)\,ds\right) = 1 + O(\Delta),\hspace{0.6cm}\label{z0}\\
P_{X_{t}}(\mbox{at least one jump on $(t, t+\Delta]$}) = O(\Delta),\hspace{6cm}\label{z1}\\
P_{X_{t}}(\mbox{exactly one jump up \& no down on $(t,
t+\Delta]$}) = \int\limits_0^\Delta \lambda(X_t+s)\,ds + O(\Delta^2),\label{z1up}\\
P_{X_{t}}(\mbox{exactly one jump down \& no up on $(t,
t+\Delta]$}) = \int\limits_0^\Delta h(X_t+s)\,ds + O(\Delta^2),\hspace{0.1cm}\label{z1down}
\end{eqnarray}
and
\begin{eqnarray}
P_{X_{t}}(\mbox{at least two jumps on $(t, t+\Delta]$}) =
O(\Delta^2).\hspace{5.7cm}\label{z2}
\end{eqnarray}
In all cases above, $O(\Delta)$ and $O(\Delta^2)$ are uniform
with respect to $X_{t}$ and only depend on the norm
$\sup_{X}(\lambda(X)+h(X))$, that is, there exist $C>0, \,
\Delta_0>0$ such that for any $X$ and any $\Delta<\Delta_0$,
\begin{eqnarray}\label{z_uni}
 \limsup_{\Delta\to 0} \left\{\Delta^{-1}P_{X}(\mbox{at least one
jump on $(0,\Delta]$})\phantom{\int\limits_1^1}\right.\hspace{2cm}
 \nonumber \\ \nonumber \\ \nonumber
+ \Delta^{-2} P_{X}(\mbox{at least two jumps on $(0, \Delta]$})\hspace{2.5cm}
  \nonumber \\  \nonumber \\\nonumber
+ \Delta^{-2}\left[P_{X_{t}}(\mbox{one jump up \& no down on $(t,
t+\Delta]$}) - \int\limits_0^\Delta \lambda(X_t+s)\,ds \right]
 \nonumber  \\ \nonumber \\
\left.+ \Delta^{-2}\left[P_{X_{t}}(\mbox{one jump down \& no up on
$(t, t+\Delta]$}) - \int\limits_0^\Delta h(X_t+s)\,ds \right]\right\} \nonumber \\ \nonumber\\
\le C< \infty.\hspace{5cm}
\end{eqnarray}
\end{lem}
{\bf Proof}. {\bf A.} The assertions (\ref{z0}), as well as
(\ref{z1}), follows from the construction (see (\ref{intu1}) and
(\ref{intu2}), since $\lambda$ and $h$ are bounded. (It also
follows from the next steps of the proof.)

~

\noindent {\bf B.} Now it is convenient to proceed with
(\ref{z2}); it will be used in the proof of (\ref{z1up}) and
(\ref{z1down}). It suffices to consider $t=0$ and any fixed
initial value $X$. Denote
\[
s^N_j = \frac{j\,\Delta}{2^N}, \qquad 1\le j\le 2^N, \;\; N\ge 1,
\]
and let
\begin{eqnarray*}
B :=\{\mbox{two or more jumps -- up or down -- on $(0,\Delta]$}\},\hspace{2cm}\\
\\
B^N:=\bigcup_{j} \left\{\mbox{no jumps on $(0,s^N_{j-1}]$,}\right.
 \mbox {at
least one jump on $(s^N_{j-1},s^N_j]$}
 \\ \\
\left. \mbox{and
 at least one jump on $(s^N_j, \Delta]$}\right\},\hspace{3.2cm}
\end{eqnarray*}
and
\[
 p:= P_{X}(B), \qquad p_N := P_{X}(B^N).\hspace{5cm}
\]
Then,
\begin{eqnarray*}
p_N= \sum_{j\le 2^N} P_X(\mbox{no jumps on $(0,s^N_{j-1}]$,} \mbox{at
least one jump on $(s^N_{j-1},s^N_j]$}
 \\ \\
\mbox{and at least one jump on $(s^N_j, \Delta]$}).\hspace{3.2cm}
\end{eqnarray*}
Notice that for any nonrandom $t$, $P(X_t\not= X_{t-}) =0$. We
have,
\[
B^N \subset B,
\]
and, moreover,
\[
B \setminus B^N \downarrow \varnothing, \quad N\uparrow \infty.
\]
The latter holds because for any two jumps there exists $N$ such
that they would be covered by two different intervals, say,
$(s^N_{j-1},s^N_j]$ and $(s^N_{k-1},s^N_k]$ with $k\not=j$. In
other words, for each $\omega\in B$ there exists $N$ such that
$\omega \in B^N$. Hence, by continuity of a probability measure,
\[
 p_N\to p, \quad N\to\infty.
\]
From the Markov property and (\ref{z0}--\ref{z1}), if $\Delta$ is
small enough then we have,
\begin{eqnarray*}
 p_N
= \sum_{j \le 2^N} E_X E \left( 1(\mbox{no jumps
on $(0,s^N_{j-1}]$}) \right. \hspace{6.6cm}
\\\\
\times1(\mbox{at least one jump on
$(s^N_{j-1},s^N_j]$} \mid {\mathcal F}^{X}_{s^N_j})
\times \left.1 (\mbox{at least one jump on $(s^N_j,
\Delta]$})\right
 \\\\
= \sum_{j \le 2^N} E_X \left(1(\mbox{no jumps on $(0,s^N_{j-1}]$})\right.  \hspace{7.5cm}
\\\\
\times \left.1(\mbox{at least one jump on $(s^N_{j-1},s^N_j]$})\right)
\times E\left(1 (\mbox{at least one jump on $(s^N_j,
\Delta]$}\mid X_{s^N_j}\right
 \\\\
= \sum_{j \le 2^N} E_X \left(1(\mbox{no jumps on $(0,s^N_{j-1}]$}) \right.  \hspace{7.5cm}
\\\\
\times\left. 1(\mbox{at least one jump on $(s^N_{j-1},s^N_j]$}) \right) \times
\left(1-\exp\left(-\int\limits_{s^N_j}^{\Delta}(\lambda+h)
(X_{s^N_j}+s)\,ds\right)\right
 \\\\
\le C \Delta \, \sum_{j \le 2^N} E_X 1\left( \mbox{no jumps on
$(0,s^N_{j-1}]$}, \mbox{at least one jump on
$(s^N_{j-1},s^N_j]$}\right
 \\\\
= C \Delta \,P_X (\mbox{at least one jump up on
$(0,s^N_N]$})\hspace{6.5cm}
 \\\\
= C \Delta \,\left(1-\exp\left(-\int\limits_0^{\Delta}(\lambda+h)
(X_{}+s)\,ds\right)\right)

\le C^2 \Delta^2.  \hspace{4.3cm}
\end{eqnarray*}
Important is that here the last value of $C^2$ is the same for all
values of $N$ and $X$. Hence, by monotone convergence, we also
have
\[
 p \le C^2\Delta^2,
\]
as required. Notice that (\ref{z1up}) and (\ref{z1down}) were not
used so far.

~

\noindent {\bf C.} Let us show ({\ref{z1up}). Informally,
\begin{eqnarray*}
P_{X_{t}}(\mbox{exactly one jump up \& no down on $(t,
t+\Delta]$}) \hspace{5cm}
 \\\\
= \int\limits_0^\Delta \lambda(X_t+s)\, \exp\left(-\int\limits_0^s
\lambda(X_t+r)\,dr\right) \exp\left(-\int\limits_s^\Delta \lambda(X_s^+
+r)\,dr\right)\hspace{0.5cm}
 \\\\
\times \exp\left(-\int\limits_0^s h(X_t+r)\,dr\right)
\exp\left(-\int\limits_s^\Delta h(X_s^+ +r)\,dr\right)\,ds
\hspace{0.5cm}
 \\\\
= \int\limits_0^\Delta \lambda(X_t+s)\,ds + O(\Delta^2),
\hspace{7.9cm}
\end{eqnarray*}
as required and with a uniform $O(\Delta^2)$. Some drawback of this
explanation is that it uses an integral identity as if it were
complete probability formula, while the latter formula is stated for a split of $\Omega$ into finitely or countably many disjoint events only. Usually there is no problem with such integration, at least for Riemann integrable functions, since integration is a limit of Darboux sums. However, in our case we assume the integrands, in general, only Lebesque integrable. So, let us do this estimate more
rigorously. We will use the already established formula (\ref{z2}) and
instead of (\ref{z1up}) we will estimate a slightly different probability
of {\em at least one jump up earlier than down} on $(0,\Delta]$
(we mean that jump down may or may not occur on this interval).
According to our construction, in particular, (\ref{intu33}), the
value of this probability may be written precisely by using the
(conditionally independent given $X_t$) densities (\ref{intu5}):
\[
f^u(z):=\lambda(X_t+z)\exp\left(-\int\limits_0^{z}
\lambda(X_{t}+s)\,ds\right)
\]
and
\[
f^d(z):= h(X_t+z)\exp\left(-\int\limits_0^{z}
h(X_{t}+s)\,ds\right).
\]
We have,
\begin{eqnarray*}
 P_X(\mbox{at least one jump up earlier than down on $(0,\Delta]$})\hspace{3cm}
 \\\\
= \int\limits_0^\infty \int\limits_0^\infty 1(z^1<z^2)1(z^1\le \Delta)
f^u(z^1)f^d(z^2)\,dz^1dz^2\hspace{3cm}
 \\\\
\le \int\limits_0^\infty \int\limits_0^\infty 1(0<z^2) 1(z^1\le \Delta)
f^u(z^1)f^d(z^2)\,dz^1dz^2\hspace{3.2cm}
 \\\\
= \int\limits_0^\Delta f^u(z^1) \,dz^1 = \int\limits_0^\Delta
\lambda(X+z)\exp\left(-\int\limits_0^{z} \lambda(X+s)\,ds\right) \,dz\hspace{1.0cm}
 \\\\
\le \int\limits_0^\Delta \lambda(X+z) \,dz.\hspace{8.2cm}
\end{eqnarray*}
On the other hand,
\begin{eqnarray*}
 P_X(\mbox{at least one jump up earlier than down on $(0,\Delta]$})\hspace{4cm}
 \\\\
= \int\limits_0^\infty \int\limits_0^\infty 1(z^1<z^2)1(z^1\le \Delta)
f^u(z^1)f^d(z^2)\,dz^1dz^2\hspace{4.1cm}
 \\\\
\ge \int\limits_0^\infty \int\limits_0^\infty 1(\Delta<z^2) 1(z^1\le \Delta)
f^u(z^1)f^d(z^2)\,dz^1dz^2\hspace{4.2cm}
 \\\\
= \int\limits_0^\Delta f^u(z^1) \,dz^1 \times \int\limits_\Delta^\infty f^d(z^2)
\,dz^2\hspace{7.2cm}
 \\\\
\ge \int\limits_0^\Delta \lambda(X+z)\exp\left(-\int\limits_0^{\Delta}
\lambda(X+s)\,ds\right) \,dz \times\exp\left(-\int\limits_0^{\Delta}
h(X+s)\,ds\right)
 \\\\
= \int\limits_0^\Delta \lambda(X+z) \,dz\times
\exp\left(-\int\limits_0^{\Delta} (\lambda + h)(X+s)\,ds\right)\hspace{3.5cm}
 \\\\
= (1+O(\Delta)) \int\limits_0^\Delta \lambda(X+z) \,dz = \int\limits_0^\Delta
\lambda(X+z) \,dz + O(\Delta^2),\hspace{2.4cm}
\end{eqnarray*}
with uniform $O(\Delta)$ and $O(\Delta^2)$, as required. Since we
already know that probability of two or more jumps is of the order
$\Delta^2$ -- see (\ref{z2}) -- this justifies (\ref{z1up}).

~

\noindent {\bf D.} The statement ({\ref{z1down}) is established similarly.
Note that in the previous steps, expressions like
$\displaystyle 1-\exp\left(-\int\limits_0^\Delta \psi(t)\,dt\right)$ with some non-negative bounded $\psi(t)$ are evaluated, say, $\psi(t)<M$, $\forall t$. In this case, this exponent may be expanded via  Taylor's series of the variable $\displaystyle \int\limits_0^\Delta \psi(t)\,dt$. For $\Delta <2/M$ this series satisfies the well-known condition for the alternating series test, which implies strict bounds $\displaystyle 1-\exp\left(-\int\limits_0^\Delta \psi(t)\,dt\right)<M\Delta$ and $\displaystyle 0<\left(1-\exp\left(-\int\limits_0^\Delta \psi(t)\,dt\right)\right)-\int\limits_0^\Delta \psi(t)\,dt< M^2\Delta^2/ 2$. This shows that the estimates in the proof above and in
({\ref{z_uni}}) are uniform, as required.
The Lemma \ref{lem1} is proved.

~

\begin{lem}\label{Le2}
Under the assumptions of the Theorem \ref{thm1}, for any $f\in C_b({\mathcal X})$ the function
\linebreak $T_tf(X) = E_Xf(X_t)$ is continuous in $t$.
\end{lem}
{\bf Proof}. It suffices for any $f\in C_b({\mathcal X})$ and for
small $\Delta>0$ to show that for any $t\ge 0$,
\begin{equation}\label{Feller}
 |E_X (f(X_{t+\Delta}) - f(X_{t}))| \to 0, \quad
\Delta \to 0.
\end{equation}
and for any $t>0$,
\begin{equation}\label{Feller2}
|E_X (f(X_{t-\Delta}) - f(X_{t})|\to 0, \quad \Delta \to 0.
\end{equation}
The convergence (\ref{Feller}) follows straightforward from the
first statement of the Lemma \ref{lem1} by virtue of Lebesgue's
dominated convergence theorem, due to the cadlag property of the
process. Similarly, we have,
\[
|E_X (f(X_{t-\Delta}) - f(X_{t-})|\to 0, \quad \Delta \to 0.
\]
However, by construction (see (\ref{intu1})), jump at any nonrandom
time $t$ has probability zero, that is,
\[
P(X_t=X_{t-}) = 1.
\]
This implies (\ref{Feller2}). The Lemma \ref{Le2} is proved.

~

\noindent Denote $X'=(n',x')\uparrow X=(n,x)$ iff $x'\uparrow x$
and $n' = n$ when $x'$ is close enough to $x$. Similarly,
$X'=(n',x')\downarrow X=(n,x)$ iff $x'\downarrow x$ and $n' = n$
for $x'$ close enough to $x$.
\begin{lem}\label{lem4}
Under the assumptions of the Theorem \ref{thm1} the process
$(X_t,\,t\ge 0)$ is Feller, that is, $T_tf(\cdot)\in C_b({\mathcal
X})$ for any $f\in C_b({\mathcal X})$.
\end{lem}
{\em Proof.} \noindent {\bf A.} Denote $\phi(t,X):=E_Xf(X_t)$. Let
$(n', x')=X'\uparrow X=(n,x)$. Without loss of generality we may
assume $n'\equiv n$ and $x'\uparrow x$. Let $s:=x-x'>0$. Note that
for the process with initial data $X'=(n',x')$,
\[
x_s \
1(\mbox{no jumps on $(0,s]$})
= x
\,1(\mbox{no jumps
on $(0,s]$}).
\]
So, for $x'$ close enough to $x$, by the Lemma \ref{lem1},
\begin{eqnarray*}
\phi(t,X') = E_{X'}f(X_t) \hspace{8cm}\\\\
=E_{X'}f(X_t)1(\mbox{no jumps on
$(0,s]$})\hspace{3.5cm}
 \\\\
+ E_{X'}f(X_t)1(\mbox{at least one jump on $(0,s]$})\hspace{1cm}
 \\\\
= E_{X'}1(\mbox{no jumps on $(0,s]$})E_{X_s}f(X_{t-s}) + O(s)\hspace{1.1cm}
 \\\\
=E_{X'}1(\mbox{no jumps on $(0,s]$})E_{X}f(X_{t-s}) + O(s)\hspace{1.3cm}
 \\\\
= E_{X'}1(\mbox{no jumps on $(0,s]$})\phi(t-s,X) + O(s)\hspace{1.3cm}
 \\\\
= (1-O(s))\phi(t-s,X) + O(s).\hspace{3.7cm}
\end{eqnarray*}
By virtue of the Lemma \ref{Le2} and due to our notation $s=x-x'$,
\[
\lim_{X'\uparrow X} \phi(t,X') = \lim_{s\downarrow 0} \phi(t-s,X)
= \phi(t,X).
\]

~

\noindent {\bf B.} Now let $X'\downarrow X$. In this case denote
$s:=x'-x$ ($s>0$). Then similarly to the above,
\[
E_{X}f(X_t) = O(s) + (1 - O(s))E_{X'}f(X_{t-s})
\]
Hence, we have,
\[
\phi(t,X) = \phi(t-s, X') + O(s).
\]
Since by the Lemma \ref{Le2},
$$
\phi(t,X') = \phi(t-s, X') + o(1),
$$
we obtain,
\[
\phi(t,X') = \phi(t-s, X') + o(1) = \phi(t, X) + O(s) + o(1),
\]
or, equivalently,
\[
\lim_{X'\downarrow X}\phi(t,X') = \phi(t, X).
\]
This completes the proof of the Lemma \ref{lem4}.

\begin{rem
Note that in \cite{Davis} Feller's property is proved for a more general model, however, under the additional condition that intensities are continuous. In our case continuity is not necessary -- which is in line with the very idea of construction of a process with discontinuous intensities -- and in the sequel this Feller's property will be used for establishing strong Markov property by using classical tools. On the other hand, in \cite{Davis} strong Markov property is established independently from Feller's one.
\end{rem}

\section{Proof of Theorem \ref{thm1}}
As was already mentioned above, both formulae (\ref{dynkin1}) and
(\ref{dynkin_t}) are, actually, versions of complete probability
formula (it would be better to say, complete expectation). If it
were possible to replace the integrals by series, they both would
have been, indeed, complete probability (expectation) formulae as
suggested. However, in our case, integrals and series are not the
same things. Notice that for {\em continuous} and bounded
$\lambda$ and $h$, the formula (\ref{dynkin1}) with any $f\in
C^1_b({\mathcal X})$ is, indeed, a simple corollary of a standard
analysis of probabilities of transitions over a small time, as in
the derivation of the Kolmogorov forward equations. This is why
${\mathcal G}$ may be regarded as a natural candidate to a
generalised generator for discontinuous intensities, too.

~

\noindent {\bf Ia.} In the first part of the proof we assume $f\in
C^1_0({\mathcal X})$, rather than $f\in C^1_b({\mathcal X})$ in
the formula (\ref{dynkin1}). In some lines it is useful to know
that $f$ and its derivative are uniformly continuous. In the end
of the proof, the assertion (\ref{dynkin1}) will be extended from
$C^1_0({\mathcal X})$ to the whole $C^1_b({\mathcal X})$. Let $\Delta:=2^{-m}t$,
\[
t_0=0, \;\; t_i = \kappa^a_m\left(\frac{it}{2^m}\right) \vee 0\equiv ([it+a 2^m]2^{-m}-a)\vee 0, \; 0\le i\le 2^m, \quad
t_{2^m+1}=t.
\]
Let $\Delta_i:= t_{i+1}-t_i$. Notice that $\Delta_i=\Delta$, $1\le
i\le 2^m$ and $\Delta_i\le \Delta$ for any $i$. \\
Consider the difference
\begin{eqnarray*}
 E_X f(X_t) - f(X) = E_X \sum_{i=0}^{2^m} E_{X_{t_i}}
(f(X_{t_{i+1}})-f(X_{t_{i}})).
\end{eqnarray*}
Consider one term from this sum. Emphasize that they are all
treated similarly. This term may be split into four parts:
\begin{eqnarray*}
 E_{X_{t_i}}(f(X_{t_{i+1}})-f(X_{t_{i}})) = E_{X_{t_i}}
(f(X_{t_{i+1}})-f(X_{t_{i}}))1(\mbox{no jumps on
$(t_{i},t_{i+1}]$})
 \\\\
 + E_{X_{t_i}}(f(X_{t_{i+1}})-f(X_{t_{i}}))1(\mbox{one
jump up \& no down on $(t_{i},t_{i+1}]$})
 \\\\
 + E_{X_{t_i}}(f(X_{t_{i+1}})-f(X))1(\mbox{one jump
down \& no up on $(t_{i},t_{i+1}]$})\hspace{0.2cm}
 \\\\
 + E_{X_{t_i}}(f(X_{t_{i+1}})-f(X_{t_{i}}))1(\mbox{at
least, two jumps on $(t_{i},t_{i+1}]$})\hspace{1cm}
 \\\\
 \equiv I^i_1 + \ldots + I^i_4. \hspace{5cm}
\end{eqnarray*}

\noindent {\bf Ib.} We have, with notation $X+s:= (n,x+s)$ for any $X=(n,x)$,
\begin{eqnarray}\label{i1}
 I^i_1 = \exp\left(-\int\limits_0^{\Delta_i}
(\lambda(X_{t_i}+s)+h(X_{t_i}+s))\,ds\right)
(f(X_{t_i}+\Delta_i)-f(X_{t_i}))
 \nonumber \\ \\ \nonumber
 = (1+O(\Delta))\int\limits_0^{\Delta_i}\frac{\partial}{\partial
x}f(X_{t_i}+s)\,ds. \hspace{3cm}
\end{eqnarray}
Here it is likely that we could use the Lemma \ref{lem3}, but we
prefer easier methods where possible. Notice that
\begin{eqnarray}\label{i11}
 \int\limits_0^{\Delta_i} \frac{\partial}{\partial x}f(X_{t_i}+s)\,ds
 = \int\limits_0^{\Delta_i} \frac{\partial}{\partial
x}f(X_{t_i+s})\,ds
 \nonumber \\ \\ \nonumber
+ \int\limits_0^{\Delta_i} \left(\frac{\partial}{\partial x}f(X_{t_i}+s)
- \frac{\partial}{\partial x}f(X_{t_i+s})\right)\,ds.
\end{eqnarray}
After summation and taking expectation, this term gives us
\begin{eqnarray*}
E_X\sum_{i=0}^{2^m} I^i_1 = E_X \sum_{i=0}^{2^m}
\int\limits_0^{\Delta_i} \frac{\partial}{\partial x}f(X_{t_i}+s)\,ds
 = E_X \sum_{i=0}^{2^m} \int\limits_0^{\Delta_i} \frac{\partial}{\partial
x}f(X_{t_i+s})\,ds
 \nonumber \\ \\ \nonumber
+ E_X \sum_{i=0}^{2^m} \int\limits_0^{\Delta_i}
\left(\frac{\partial}{\partial x}f(X_{t_i}+s) -
\frac{\partial}{\partial x}f(X_{t_i+s})\right)\,ds.
 \hspace{1cm}
\end{eqnarray*}
By virtue of the cadlag property of $(X_s, \, s\ge 0)$ and $f\in
C^1_b$, we have by Lebesgue's dominated convergence Theorem,
\begin{eqnarray*}
E_X \sum_{i=0}^{2^m} \int\limits_0^{\Delta_i}
\left(\frac{\partial}{\partial x}f(X_{t_i}+s) -
\frac{\partial}{\partial x}f(X_{t_i+s})\right)\,ds\hspace{6cm}
 \\\\
= E_X \int\limits_0^t \left(\frac{\partial}{\partial
x}f(X_{\kappa^a_m(s)}+s-\kappa^a_m(s) - \frac{\partial}{\partial
x}f(X_{s})\right)\,ds\hspace{2.8cm}
 \\\\
= E_X \int\limits_0^t \left(\frac{\partial}{\partial
x}f(X_{\kappa^a_m(s)}+s-\kappa^a_m(s) - \frac{\partial}{\partial
x}f(X_{s-})\right)\,ds \to 0, \;\; \Delta \to 0,
\end{eqnarray*}
the latter equality a.s., because
\[
P_X(X_{\kappa^a_m(s)}+s-\kappa^a_m(s)-X_{s-} \to 0, \quad
m\to\infty)=1.
\]
Hence -- and due to $\int\limits_0^t |f(X_s)-f(X_{s-})|\,ds = 0$ a.s. -- we get,
\begin{eqnarray*}
E_X \sum_{i=0}^{2^m} I^i_1 = E_X \int\limits_0^t
\frac{\partial}{\partial x}f(X_{s})\,ds + o(1), \;\; \Delta \to 0.
\end{eqnarray*}

~

\noindent {\bf Ic.} Further,
\begin{eqnarray}\label{Ic}
 I^i_2
= E_{X_{t_i}}(f(X_{t_{i+1}})-f(X_{t_{i}}))1(\mbox{one jump up \&
no down on $(t_{i},t_{i+1}]$})\hspace{1cm}
 \nonumber \\ \nonumber \\ \nonumber
\equiv E_{X_{t_i}}(f(X^+_{t_{i}})-f(X_{t_{i}}))1(\mbox{one jump up \&
no down on $(t_{i},t_{i+1}]$})\hspace{0.5cm}
 \\ \nonumber \\ \nonumber
+ E_{X_{t_i}}(f(X_{t_{i+1}})-f(X^+_{t_{i}}))1(\mbox{one jump up \&
no down on $(t_{i},t_{i+1}]$})
 \\ \nonumber \\ \nonumber
= (f(X^+_{t_i}) - f(X_{t_i})) \left(\int\limits_0^{\Delta_i}
\lambda(X_{t_i}+s) \,ds + O(\Delta^2)\right)\hspace{3cm}
 \nonumber \\\\ \nonumber
+ E_{X_{t_i}}(f(X_{t_{i+1}})-f(X^+_{t_{i}}))1(\mbox{one jump up \&
no down on $(t_{i},t_{i+1}]$}),
\end{eqnarray}
by the Lemma \ref{lem1}. After summation and taking expectation,
the first term here gives
\begin{eqnarray*}
E_X \int\limits_0^t (f(X^+_{\kappa^a_m(s)}) - f(X_{\kappa^a_m(s)}))
\lambda(X_{\kappa^a_m(s)}+s - \kappa^a_m(s)) \,ds + O(\Delta)),
\end{eqnarray*}
and we will explain below -- via the Lemma 1 -- why it approaches
the desired
\begin{eqnarray*}
E_X \int\limits_0^t (f(X^+_{s-}) - f(X_{s-})) \lambda(X_{s}) \,ds = E_X
\int\limits_0^t (f(X^+_{s}) - f(X_{s})) \lambda(X_{s}) \,ds, \;\;
\mbox{a.s.}
\end{eqnarray*}
The absolute value of the second term in (\ref{Ic}) does not
exceed
\[
\rho_f(\Delta) E_{X_{t_i}}1(\mbox{one jump up \& no down on
$(t_{i},t_{i+1}]$}),
\]
where $\rho_f$ is the modulus of continuity of the function $f$.
By virtue of the Lemma \ref{lem1} (see (\ref{z1up})),
after summation and expectation this gives us
\[
\rho_f(\Delta) E_{X}\int\limits_0^t \lambda(X_{\kappa^a_m(s)}+s -
\kappa^a_m(s))\,ds + O(\Delta) = o(1), \quad \Delta\to 0.
\]

~

\noindent {\bf Id.} Similarly
with some $\theta\in (0,\Delta_i)$,
\begin{eqnarray*}
I^i_3 = (1 - \exp\left(-\int\limits_0^{\Delta_i} h(X_{t_i}+s) \,ds\right)
+ O(\Delta^2))
 (f(X^-_{t_i}+\theta) - f(X_{t_i}))
 \\\\
 = \left(\int\limits_0^{\Delta_i} h(X_{t_i}+s) \,ds + O(\Delta^2)\right)
 (f(X^-_{t_i}) - f(X_{t_i}) + \rho_f(\Delta)).
\end{eqnarray*}
Since $\rho_f(\Delta) = o(1)$, we get,
\begin{eqnarray*}
 E_X \sum_{i=0}^{2^m} I^i_3=\hspace{12cm}
 \\\\
= E_X \int\limits_0^t h(X_{\kappa^a_m(s)}+s - \kappa^a_m(s)) \,ds
 (f(X^-_{\kappa^a_m(s)}) - f(X_{\kappa^a_m(s)}) + o(1)).
\end{eqnarray*}
It will be shown below that the main term here approaches
\begin{eqnarray*}
E_X \int\limits_0^t h(X_{s}) (f(X^-_{s-}) - f(X_{s-}))\,ds
 \stackrel{\mbox{\small a.s.}}{=} E_X \int\limits_0^t h(X_{s}) (f(X^-_{s}) - f(X_{s})\,ds.
\end{eqnarray*}

~

\noindent {\bf Ie.} Finally (see the Lemma \ref{lem1}), with
uniform $O(\Delta^2)$,
\[
 I^i_4 = O(\Delta^2) \Longrightarrow E_X \sum_{i=0}^{2^m} I^i_4 = O(\Delta).
\]

~

\noindent {\bf If.} Now in the expression arising from the terms
$E_X I^i_2$, as well as from $E_X I^i_3$, we would like to replace
$f(X^+_{\kappa^a_m(s)})
 - f(X_{\kappa^a_m(s)})$ by
$f(X^+_{s})
 - f(X_{s})$ and
$f(X^-_{\kappa^a_m(s)})
 - f(X_{\kappa^a_m(s)})$ by
$f(X^-_{s})
 - f(X_{s})$,
respectively; also,
$h(X_{\kappa^a_{m'}(s)})$ and $\lambda(X_{\kappa^a_{m'}(s)})$ will be
replaced by
$h(X_{s})$ and $\lambda(X_{s})$, respectively. By virtue of continuity of $f$ and
cadlag property of the process, we estimate
for almost every $a\in [0,1]$,
\begin{eqnarray*}
 \left(E_X \int\limits_0^t \lambda(X_{\kappa^a_m(s)})
(f(X^+_{\kappa^a_m(s)})
 - f(X_{\kappa^a_m(s)}))\,ds\right.
 \\\\
\left. - E_X \int\limits_0^t \lambda(X_{s})(f(X^+_{s})
 - f(X_{s}))\,ds\right)
 \to 0, \quad m \to\infty,
\end{eqnarray*}
and
\begin{eqnarray*}
 \left(E_X \int\limits_0^t h(X_{\kappa^a_{m'}(s)})(f(X^-_{\kappa^a_{m'}(s)})
 - f(X_{\kappa^a_{m'}(s)}))\,ds\right.
 - \\\\
 \left.- E_X \int\limits_0^t h(X_{s})(f(X^-_{s+})
 - f(X_{s}))\,ds\right)
 \to 0, \quad m' \to\infty,
\end{eqnarray*}
where $m'\to \infty$ is some subsequence guaranteed by the Lemma \ref{lem3}.
Indeed, say, for $h$, by the Lemma \ref{lem3} for a.e. $a$,
\begin{eqnarray*}
 E_X \int\limits_0^t |(h(X_{\kappa^a_{m'}(s)}) - h(X_s))
(f(X^+_{\kappa^a_{m'}(s)})
 - f(X_{\kappa^a_{m'}(s)}))|\,ds
 \\\\
 \le C_f \, E_X \int\limits_0^t |(h(X_{\kappa^a_{m'}(s)}) - h(X_s))|\,ds
 \to 0, \quad m' \to\infty.
\end{eqnarray*}
Further, by continuity of $f$ and cadlag property of $X$,
\begin{eqnarray*}
\left( E_X \int\limits_0^t h(X_{s})(f(X^+_{\kappa^a_{m'}(s)})
 - f(X_{\kappa^a_{m'}(s)}))\,ds -\right.\hspace{1cm}
\\\\
\left. - E_X \int\limits_0^t h(X_{s})(f(X^+_{s-})
 - f(X_{s-}))\,ds\right)
 \to 0, \quad m' \to\infty,
\end{eqnarray*}
by virtue of the Lebesgue dominated (bounded) convergence theorem, as
$f(X_{\cdot})$ is cadlag, both $h$ and $f$ are bounded and
\[
 \kappa^a_{m'}(s) \to s-, \quad m'\to\infty.
\]
Finelly, since \(P(X_s\not=X_{s-})=0, \, \forall s\),
we get,
\begin{eqnarray*}
 E_X \left|\int\limits_0^t h(X_{s})(f(X^+_{s-})
 - f(X_{s-}))\,ds - \int\limits_0^t h(X_{s})(f(X^+_{s})
 - f(X_{s}))\,ds
\right| = 0.
\end{eqnarray*}

~

\noindent {\bf Ig.} Further, we would like to have $ \lambda(X_s)$
under the integral instead of $\lambda(X_{t_i}+(s-t_i))$ in each term (\ref{Ic})
\[
(f(X^+_{t_i}) - f(X_{t_i})) \left(\int\limits_0^{\Delta_i}
\lambda(X_{t_i}+s) \,ds + O(\Delta^2)\right).
\]
So, let us estimate the difference
\begin{eqnarray*}
 E_X\sum_{i=0}^{2^m} \left(\int\limits_{t_i}^{t_{i+1}}
\lambda(X_{t_i}+(s-t_i)) \,ds - \int\limits_{t_i}^{t_{i+1}}
 \lambda(X_s)\,ds\right)
 (f(X^+_{t_i}) - f(X_{t_i})).
\end{eqnarray*}
Recall that $t_i = \kappa^a_m\left(\frac{it}{2^m}\right)$. Let us show that
for almost any $a\in [0,1]$,
\begin{eqnarray}\label{main_dif1}
 E_X \sum_{i=0}^{2^m} \left|\int\limits_{t_{i}}^{t_{i+1}} \lambda(X_{t_i}+(s-
t_i)) \,ds
 - \int\limits_{t_{i}}^{t_{i+1}} \lambda(X_s)\,ds\right|\,
 |f(X^+_{t_{i}}) - f(X_{t_{i}})|= \hspace{2.1cm}
 \nonumber \\\\\nonumber
E_X\sum_{i=0}^{2^m} \left|\int\limits_{t_{i}}^{t_{i+1}}
(\lambda(X_{\kappa^a_m(s)} {t_i}+(s- \kappa^a_m(s)))
 - \lambda(X_s))\,ds\right|\,
 |f(X^+_{t_{i}}) - f(X_{t_{i}})|
 \to 0,
\end{eqnarray}
as $\; \Delta\to 0.$

~

Firstly, according to the Lemma \ref{lem3}, there exists a
subsequence $(m'\to\infty)$ such that for almost every $a\in
[0,1]$
\[
\int\limits_0^t |\lambda(X_s) - \lambda(X_{\kappa^a_{m'}(s)})|\,ds \to
0,
\]
and for the whole sequence $(m\to\infty)$,
\begin{equation}\label{la1}
\int\limits_0^1\int\limits_0^t |\lambda(X_s) - \lambda(X_{\kappa^a_m(s)})|\,ds
\,da\to 0.
\end{equation}
Now let us show that
\begin{equation}\label{dif2}
\int\limits_0^1\int\limits_0^t |\lambda(X_{\kappa^a_m(s)} + (s-\kappa^a_m(s))) -
\lambda(X_{\kappa^a_m(s)})| \,ds\,da \to 0, \; m\to\infty.
\end{equation}
The idea is very similar to that of the Lemma \ref{lem3}, however, here a direct reference to this Lemma is questionable and, hence, we need to provide an independent proof. Recall that the trajectory $(X_s)$ may have only finitely many
jumps on $(0,t]$. Also, except for the points of jumps, the
evolution of $(X)$ is deterministic and linear with a constant
positive speed. In other words, integration of the composite
function $\lambda(X_{\cdot})$ turns out to be equivalent to the
integration of $\lambda(\cdot)$ over some interval between two
consequent jumps of the process $(X)$. This allows us to use the
same trick from the Lemma \ref{lem3}, but this time we
approximate the function $\lambda(\cdot)$ rather than the
composite function $\lambda(X_{\cdot})$. Emphasize that this is
possible exactly because of the piecewise linear law of evolution
of $(X_s)$ between the consequent moments of jumps and because
there are only finitely many jumps on each trajectory.

~

\noindent So, let us approximate the function $\lambda(\cdot)$ by
bounded uniformly continuous functions -- say, $\lambda^\epsilon$
-- in the topology of convergence in $L_1[0,T]$ for every $T>0$.
 Note that it is not enough to
approximate $\lambda(\cdot)$ on $(0,t]$. Denote by $(\tau_j, \,
j=0,1,\ldots)$ the sequence of moments of jumps of the trajectory
of $X$; recall that for each $\omega$ there are only finitely many
of them on $(0,t]$.

We have, on each $(\tau_j,\tau_{j+1}]$ and for every $m$,
\begin{eqnarray*}
 \int\limits_0^1\int\limits_{\tau_j}^{\tau_{j+1}} |\lambda^\epsilon
(X_{\kappa^a_m(s)}) - \lambda(X_{\kappa^a_m(s)})|
 \,ds\,da

= \int\limits_{\tau_j}^{\tau_{j+1}} \int\limits_0^1 | \lambda^\epsilon
(X_{\kappa^a_m(s)}) - \lambda(X_{\kappa^a_m(s)})|
 \,da\,ds
 \\\\
\le \int\limits_{\tau_j}^{\tau_{j+1}} 2^m\,\|\lambda^\epsilon - \lambda
\|_{L_1([\tau_j - 2^{-m},\tau_{j+1}])}\,ds

= \|\lambda^\epsilon - \lambda \|_{L_1([\tau_j - 2^{-
m},\tau_{j+1}])} \to 0,\; \epsilon \to 0.
\end{eqnarray*}
Since the number of moments $\tau_j$ on $(0,t]$ is finite for
(almost) all $\omega$, it also follows for each $m$ that
\begin{eqnarray}\label{la2}
 \int\limits_0^1\int\limits_{0}^{t} |\lambda^\epsilon
(X_{\kappa^a_m(s)}) - \lambda(X_{\kappa^a_m(s)})| \,ds\,da \hspace{3.8cm}
 \nonumber \\\\ \nonumber
= \int\limits_0^1\sum_j \int\limits_{\tau_j}^{\tau_{j+1}} | \lambda^\epsilon
(X_{\kappa^a_m(s)}) - \lambda(X_{\kappa^a_m(s)})|\,ds\,da \to 0,\;
 \epsilon \to 0.
\end{eqnarray}

 ~

Now consider the difference
\begin{eqnarray*}
\int\limits_0^1\int\limits_0^t | \lambda^\epsilon(X_{\kappa^a_m(s)} +
(s-\kappa^a_m(s))) - \lambda(X_{\kappa^a_m(s)} +
(s-\kappa^a_m(s)))|\,ds\,da.
\end{eqnarray*}
On each $(\tau_j,\tau_{j+1}]$ and for every $m$, due to
integration over $s$,
\begin{eqnarray*}
\int\limits_0^1\int\limits_{\tau_j}^{\tau_{j+1}} |
\lambda^\epsilon(X_{\kappa^a_m(s)} + (s-\kappa^a_m(s))) -
\lambda(X_{\kappa^a_m(s)} + (s-\kappa^a_m(s)))| \,ds\,da
 \\\\
\le \int\limits_0^1 \|\lambda^\epsilon - \lambda\|_{L_1([\tau_j -
2^{-m},\tau_{j+1}])} \,da = \|\lambda^\epsilon -
\lambda\|_{L_1([\tau_j - 2^{-m}, \tau_{j+1}])} \to 0, \;
\epsilon\to 0.
\end{eqnarray*}
Therefore, also
\begin{eqnarray}\label{la3}
 \left(\int\limits_0^1\int\limits_{0}^{t} | \lambda^\epsilon(X_{\kappa^a_m(s)} +
(s-\kappa^a_m(s))) \right. \hspace{1cm}
\nonumber \\\\ \nonumber
\left.\phantom{\int\limits_0^1}
- \lambda(X_{\kappa^a_m(s)} +
(s-\kappa^a_m(s)))|\,ds\,da \right)\to 0,
\quad  \epsilon\to 0.
\end{eqnarray}
Notice that by virtue of continuity of $\lambda^\epsilon$,
\begin{eqnarray}\label{la4}
\int\limits_0^1\int\limits_0^t | \lambda^\epsilon(X_{\kappa^a_m(s)} +
(s-\kappa^a_m(s))) - \lambda^\epsilon(X_{\kappa^a_m(s)})|\,ds\,da
\to 0, \; m \to \infty,
\end{eqnarray}
for each $\omega$ and $\epsilon$. So, due to (\ref{la2}),
(\ref{la3}) and (\ref{la4}), the left hand side in (\ref{dif2})
goes to zero as $m\to\infty$, as it does not depend on $\epsilon$.
From here it easily follows that there exists a subsequence $(m')$
such that for almost every $a$,
\begin{equation}\label{dif22}
\int\limits_0^t |\lambda(X_{\kappa^a_{m'}(s)} + (s-\kappa^a_{m'}(s))) -
\lambda(X_{\kappa^a_{m'}(s)})| \,ds \to 0, \; m'\to\infty.
\end{equation}
Moreover, this subsequence may be chosen from the earlier fixed
subsequence for which (\ref{la1}) is valid. Hence, we may assume
that for $(m')$ both (\ref{la1}) and (\ref{dif22}) are valid
simultaneously.

~

Overall, by virtue of Lebesgue's dominated convergence, we
conclude that by virtue of (\ref{la1}) and (\ref{dif22}),
convergence (\ref{main_dif1}) holds true with almost any $a\in
[0,1]$, that is, over some subsequence $(m'\to\infty)$,
\begin{eqnarray}\label{main_dif11}
E_X \sum_{i=0}^{2^{m'}} \left|\int\limits_{t_{i}}^{t_{i+1}}
(\lambda(X_{\kappa^a_{m'}(s)})
 - \lambda(X_s))\,ds\right|\,
 |f(X^+_{t_{i}}) - f(X_{t_{i}})|
\to 0.
\end{eqnarray}
Similarly, without loss of generality, we may  assume that over the same subsequence,
\begin{eqnarray}\label{main_dif111}
E_X \sum_{i=0}^{2^{m'}} \left|\int\limits_{t_{i}}^{t_{i+1}} (h(X_{\kappa^a_{m'}(s)}) - h(X_s))\,ds\right|\,
 |f(X^-_{t_{i}}) - f(X_{t_{i}})|
\to 0.
\end{eqnarray}
From
(\ref{main_dif11})--(\ref{main_dif111}), the statement
(\ref{dynkin1}) (Dynkin's formula) for $f\in C^1_0({\mathcal X})$
follows.

~

\noindent {\bf II.} Now assume $f\in C^1_b({\mathcal X})$. Let us
approximate this function by a uniformly bounded sequence $f^{N}
\in C^1_0({\mathcal X})$, so that
\[
 \|f^N-f\|_{C^1(K)} \to 0, \quad \mbox{for any compact set $K \in {\mathcal X}$}.
\]
Then, by the part I of the proof, we have
\[
E_Xf^N(X_t) - f^N(X) = E_X\int\limits_0^t {\mathcal G}f^N(X_s)\,ds,
\]
and we need to justify the passage to the limit as $N\to\infty$.
This follows from Lebesgue's dominated convergence Theorem.
Indeed, for each $s,t$ and $\omega$,
\[
f^N(X_t) \to f(X_t), \quad {\mathcal G}f^N(X_s) \to {\mathcal
G}f(X_s),
\]
and $f^N(X_t)$ and ${\mathcal G}f^N(X_s) $ are uniformly bounded.
Hence, the statement of the Theorem \ref{thm1} about Dynkin's
formula (\ref{dynkin1}) for any $f\in C^1_b({\mathcal X})$ is
proved.

~

\noindent {\bf III.} Now let us show the formula (\ref{dynkin_t}).
Here we have to consider the difference
\begin{eqnarray*}
 E_X \varphi(t,X_t) - \varphi(0,X) = E_X \sum_{i=0}^{2^m} E_{X_{t_i}}
(\varphi(t_{i+1}, X_{t_{i+1}})-\varphi(t_i,X_{t_{i}})).
\end{eqnarray*}
Consider one term from this sum. This term may be split into five
parts,
\begin{eqnarray*}
E_{X_{t_i}}(\varphi(t_{i+1}, X_{t_{i+1}})-\varphi(t_{i},X_{t_{i}})) \hspace{7cm}
 \\\\
= E_{X_{t_i}}
(\varphi(t_i, X_{t_{i+1}})-\varphi(t_i, X_{t_{i}}))1(\mbox{no
jumps on $(t_{i},t_{i+1}]$})\hspace{2.5cm}
 \\\\
 + E_{X_{t_i}}(\varphi(t_i, X_{t_{i+1}})-\varphi(t_i, X_{t_{i}}))1(\mbox{one
jump up on $(t_{i},t_{i+1}]$})\hspace{1.1cm}
 \\\\
 + E_{X_{t_i}}(\varphi(t_i, X_{t_{i+1}})-\varphi(t_i, X_{t_i}))1(\mbox{one jump
down on $(t_{i},t_{i+1}]$})\hspace{0.6cm}
 \\\\
 + E_{X_{t_i}}(\varphi(t_i, X_{t_{i+1}})-\varphi(t_i, X_{t_{i}}))1(\mbox{at
least, two jumps on $(t_{i},t_{i+1}]$})
 \\\\
+ E_{X_{t_i}}(\varphi(t_{i+1},X_{t_{i+1}})-\varphi(t_i,
X_{t_{i+1}})). \hspace{5.3cm}
\end{eqnarray*}
All terms but the last one are considered similarly to the case
$(f(X))$. The last term, clearly, gives us
\[
E_{X_{t_i}}(\varphi(t_{i+1},X_{t_{i+1}})-\varphi(t_i,
X_{t_{i+1}})) = E_{X_{t_i}}\int\limits_{t_i}^{t_{i+1}}
\frac{\partial}{\partial s} \varphi(s, X_{t_{i+1}})\,ds,
\]
which after summation and taking expectation converges as follows,
\[
E_X\sum_{i=0}^{2^m} E_{X_{t_i}}\int\limits_{t_i}^{t_{i+1}}
\frac{\partial}{\partial s} \varphi(s, X_{t_{i+1}}))\,ds \to E_{X}
\int\limits_{0}^{t} \frac{\partial}{\partial s} \varphi(s, X_{s})\,ds,
\quad m\to\infty,
\]
by Lebesgue's dominated convergence Theorem for any $\varphi\in
C^1_0([0,\infty)\times {\mathcal X})$, as required. Now we may
repeat the arguments about extending the formula from
$C^1_0([0,\infty)\times {\mathcal X})$ to $C^1_b([0,\infty)\times
{\mathcal X})$.

~

\noindent {\bf IV.} It remains to show strong Markov property.
This follows from continuity of the function $E_X f(X_t)$ in $X$
by virtue of the Lemma \ref{Le2} and due to the cadlag property of
the process, see \cite[Theorem 3.4]{Dynkin}. The Theorem
\ref{thm1} is proved.

\section{Proof of Corollary \ref{cor1}}
\noindent Notice that the issue is to extend both
Dynkin's formulae to polynomially growing functions. First of all,
recall that the formulae (\ref{dynkin1}) and (\ref{dynkin_t}) for
$f\in C^1_b( {\mathcal X})$ and $\varphi\in C^1_b([0,\infty)\times
{\mathcal X})$ in martingale language read as follows: the
processes
\begin{equation}\label{mart100}
M_{t}:=f(X_{t}) - f(X) - \int\limits_0^{t} {\mathcal G} f(X_s)\,ds,\; \;
\; t\ge 0,
\end{equation}
and
\begin{equation}\label{mart111}
\tilde M_{t}:=\varphi(t, X_{t}) - \varphi(0,X) - \int\limits_0^{t}
 \left(\frac{\partial}{\partial s}\, \varphi(s,X_s) + {\mathcal
G}\varphi(s,X_s)\right)\,ds,\; \; \; t\ge 0,
\end{equation}
are both martingales (cf. \cite{LSh}). For the reader's
convenience let us show this, say, for $\tilde M$. Indeed,
\[
 E|\tilde M_t| < \infty
\]
since all terms in the right hand side in (\ref{mart111}) are
bounded for $\varphi\in C^1_b$, and
\[
 E_X(\tilde M_{t} - \tilde M_{s}\mid {\mathcal F}^X_s)
 = E_{X_s}(\tilde M_{t-s} - \tilde M_{0}) = E_{X}(\tilde M_{t-s})|_{X=X_s},
\]
due to Markov property. But due to (\ref{dynkin_t}), for {\em
every} initial data $X$,
\[
E_{X}(\tilde M_{t-s})= 0
\]
for every $X$. Hence,
\[
E_{X}(\tilde M_{t-s})|_{X=X_s} = 0.
\]
Therefore, $\tilde M$ is, indeed, a (cadlag) martingale.

~

\noindent Now the task is to extend (\ref{mart111}) to the
function $L_{k,m}$, or, equivalently, to extend (\ref{dynkin_t})
to such function. It is easy to see that $L_{k,m}$ may be
approximated by functions $\varphi^N \in C^1_b([0,\infty)\times
{\mathcal X})$ so that
\[
 \|\varphi^N(\cdot,\cdot) - L_{k,m}(\cdot,\cdot)\|_{C^1([0,N]\times [0,N])} \to 0,
 \quad N\to\infty,
\]
for any $t\ge 0,\, X\in {\mathcal X}$, and
\begin{equation}\label{co1}
 \sup_N \left(|\varphi^N(t,X)| + \left|\frac{\partial}{\partial x} \varphi^N(t,X)\right|
 + \left|\frac{\partial}{\partial t} \varphi^N(t,X)\right| \right) \le C(1+L_{k,m}(t,X)).
\end{equation}
Then it is possible to pass to the limit in all terms of the
equation (\ref{dynkin_t}) written for $\varphi^N$, if the
following a priori bound is established,
\begin{equation}\label{apri}
 \sup_{t\le T} E_{(n_0,x_0)} (n_t+x_t)^m \le C(T,n_0,x_0,m) < \infty,
\end{equation}
for any $m>0$ with some function $C(T,n,x,m)$, for any $T>0$.
Indeed,
\[
\varphi^N(t,X_t) \to \varphi(t,X_t), \quad
\frac{\partial}{\partial s}\varphi^N(s,X_s) \to
\frac{\partial}{\partial s}\varphi(s,X_s), \quad N\to\infty,
\]
and
\[
{\mathcal G}\varphi^N(s,X_s) \to {\mathcal G}\varphi(s,X_s), \quad
N\to\infty.
\]
Since (\ref{apri}) will be established for any $m>0$, we obtain by
Lebesgue's convergence Theorem under the uniform integrability
condition,
\begin{eqnarray*}
\varphi^N(t, X_{t}) - \varphi^N(0,X) - \int\limits_0^{t}
 \left(\frac{\partial}{\partial s}\, \varphi^N(s,X_s) + {\mathcal G}\varphi^N(s,X_s)\right)\,ds
 \\\\
\to \varphi(t, X_{t}) - \varphi(0,X) - \int\limits_0^{t}
 \left(\frac{\partial}{\partial s}\, \varphi(s,X_s) + {\mathcal G}\varphi(s,X_s)\right)\,ds, \;\; N\to\infty.
\end{eqnarray*}
This would also imply that $E_X|\tilde M_t| + E_X|M_t| < \infty$,
for $f=L_{m}$ and $\varphi = L_{k,m}$.

~

\noindent The easiest explanation of (\ref{apri}) is, apparently,
to use the fact that the process $(X_t = (n_t, x_t), \,t\ge 0)$ is
dominated by a similar process, say, $(\bar X_t= (\bar n_t, \bar
x_t), \,t\ge 0)$ {\em without actually serving the customer
at the server} and with a constant arrival
rate $\bar\lambda:= \Lambda \vee \lambda_0$, that is, for each
$\omega$,
\[
\bar x_t\ge x_t, \quad \bar n_t\ge n_t, \quad t\ge 0.
\]
For the process $(\bar X)$ we have,
\[
 \bar x_t = x+t
\]
and
\[
 \bar n_t = n_0 + \xi_t,
\]
where $\xi_t$ has a Poisson distribution with parameter $\bar
\lambda t$. We may imagine a situation as if the current serving has
suddenly pended and the customer remained at the idle server
forever. We have,
\begin{eqnarray*}
 \psi(t,m):=
 E\xi_t^m = \sum_{j=0}^{\infty} j^m\, \frac{(\bar \lambda t)^j}{j!} e^{-\bar \lambda
 t}
 <\infty,
\end{eqnarray*}
so,
\[
 \sup_{t\le T} E_{(n_0,x_0)} (\bar n_t+\bar x_t)^m \le 3^{m-1}(x_0+T)^m +
 3^{m-1} n_0^m + 3^{m-1} \psi(T,m).
\]
Hence, (\ref{apri}) holds true and the statement of the Corollary
\ref{cor1} is proved.

\section{Proof of Theorem \ref{thm2}}
The proof repeats the calculus in \cite{Ve2013_ait} based on
Lyapunov functions
\[
L_{m}(X):= (n+1 + x)^m \quad \mbox{and} \quad L_{m,k}(t,X):=
(1+t)^k(n+1 + x)^m
\]
and on Dynkin's formulae (\ref{dynkin1}) and (\ref{dynkin_t}) due
to the Corollary \ref{cor1}. The news is only a wider class of
intensities, which may be discontinuous, however, this does not
affect the calculus at all once (\ref{dynkin1}) and
(\ref{dynkin_t}) are established. Hence, we drop the details.

redpajama_set_name: RedPajamaArXiv


text:
Main focus was to be there for the team and score as many runs as possible: Shubman Gill
It's no secret that of late, Shubman Gill has been a class act while opening the batting in ODIs. But against New Zealand at the Rajiv Gandhi International Stadium in Hyderabad on Wednesday, Gill's star in ODIs ascended even further in front of 31,187 spectators when he smashed a sensational 208 off 149 balls, propelling India to 349/8, which proved to be a winning total.
Since being Player of the Tournament in India's victorious run at the 2018 U19 Men's World Cup in New Zealand, Gill had already been talked about as the next backbone and future star of the senior men's team batting line-up.
Despite justifying the talk through a consistent 2022, his place in the ODI side was being questioned because Ishan Kishan, who himself hit 210 while opening the batting against Bangladesh in Chattogram, was moved down the order to accommodate Gill.
But under the skies of Hyderabad, on a two-paced pitch, Gill dazzled in an opening masterclass where he was in complete control of his strokeplay and ran hard to leave fans mesmerised with his 208.
"It (the double-century against New Zealand) means a lot to me obviously. I think in the first ODI against Sri Lanka and in the third ODI, I was set, and I was looking to get a big score, but unfortunately, that didn't happen for me. Once I was set here, the main focus for me was to obviously be there for the team and score as many runs as possible. And it feels good when it pays off," said Gill in the post-match press conference.
Gill added that being consistent in ODIs is something he aspires to as a batter. "Consistency is the key to everything. It's one of the things that I strive for. Being consistent as a batsman is very important in any format. It feels nice when you are doing things over a period of time repeatedly."
One of the most thrilling moments of his knock came at the fag end of the Indian innings. Gill broke a run of four boundary-less overs by smashing sixes over leaping long-on and square leg off Blair Tickner in the 48th over.
Starting the 49th over at 182, Gill reached his double century by smacking a hat-trick of sixes over fine leg, long-off and straight down the ground in the 49th over off Lockie Ferguson. The loud roar had an element of joy and waving his bat in delight as the crowd chanted his name. He acknowledged the cheers by bowing gently after getting his hundred and on reaching the double century, the euphoria reached a crescendo in Uppal.
"First I decided to go hard in the last five overs. But with Washington (Sundar) getting out in the 44th over, then the message was to go hard in the last three overs. I think in the 46th or 47th over, I hit a couple of sixes and that's when I decided to go hard because I thought that bowler was my match-up."
"Not a conscious effort (to keep batting aggressively), I would say, but with an extra fielder inside the circle, we see other teams pushing in the middle overs. Even when wickets were falling, my main focus was to show some intent to the bowler because it becomes very easy for the bowler to be able to bowl dot balls if the batsman is not going to show any intent that he's going to hit any boundaries — even if we have lost a wicket. So that was my intent — even when wickets were falling to show the bowler that I will hit you if you're going to bowl bad balls."
Amongst the many wishes pouring in from social media, the tweet from former India all-rounder Yuvraj Singh saying it was a proud moment for him and Gill's father took the cake. "It feels very good. Yuvi paaji has been a mentor to me, like a big brother."
"During the lockdown and even after that, he talks a lot about my batting and how we can work more on it. My dad has been my primary coach since I started playing cricket. Making Yuvi paaji and my dad proud is definitely a special feeling for me," concluded Gill.
There is still a pretty long way to see if Gill becomes the next generational batting great from India. But for now, his 208 at Hyderabad showed that he is here to stay for a long time in ODIs.
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IPL 2022: Punjab Kings beat RCB by five wickets
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Dilip Vengsarkar feels Umran Malik is ready to play international cricket

redpajama_set_name: RedPajamaCommonCrawl


Fermentables hasserved Arkansas since 1995 with quality products and a knowledgeable staff. We stock everything needed to ferment wine or beer and have a large selection of cheesemaking products, coffee roasting equipment / green beans, kombucha and vegetable fermentation supplies as well as a complete line of hydroponic and indoor gardening products including nutrients, growing media, grow lights, organic soils and fertilizers. Our gardening center encompasses a large inventory and displays the latest technology in hydroponic systems, LED grow lights, indoor green houses and organic methods.


Much of the success of Rogue One: A Star Wars Story, the first Star Wars standalone movie, can be attributed to the exhilarating final act, which, as I suggested in my review, took Star Wars to the nth degree and into A New Hope.
During the SXSW Film Festival, director Gareth Edwards spoke to Fandango and shed light on Rogue One's extensive summer reshoot that included the addition of Darth Vader boarding the Rebel ship, with horrifying consequences, in the final moments.
"He arrives and obliterates the Calamari ship, and then the blockade runner gets out just in time and he pursues the blockade runner. And then Jabez was like, 'I think we need to get Darth on that ship,' and I thought, yeah, that's a brilliant idea and would love to do it, but there's no way they're going to let us do it. It's a big number and we had, what, like 3 or 4 months before release.
Fans finally witnessed the wrath of Darth Vader in a way the prequels failed to convey.
Rogue One: A Star Wars Story is released on Blu-ray this April.
Yoda to return in The Last Jedi?


text:
Q: `[innerHTML]` - html tags are missing In my variable  I declared like :
title:any="
Arif
";

In my template I tried both way :
<content [innerHTML]="title"></content>

Both of them has no h2 tag at all. what is the problem? also how to just replace the content tag using h2 element?
any one help me?

A: I changed the condition from :

to

Now the innerHTML works fine.

redpajama_set_name: RedPajamaStackExchange


Two master classes were held during the festivals Games Week Denmark and Internet Week Denmark to capitalise on the interest in these events.
The second masterclass 'VR today & tomorrow' was live transmitted.
The Summit was about privacy in digital communication and behaviour with diverse speakers on digital privacy. Well over 100 participants came from both entertainment and non-entertainment sectors. It addressed the hot spots and blind spots of the transformation of privacy in digital communications and behaviour and brought together researchers, activists and artists in the intersection of human rights in the digital age.
It was a cross-industry conference and participants spanned the fashion industry, universities and other educational institutions, public organizations, healthcare, the Royal Danish Academy of Music and IBM.
Speakers included Katrine K. Pedersen, Aral Balkan, Emma Holten, Bogi Eliasen and Dries Depoorter.
Here's what they talked about….
Katrine commutes to and from Silicon Valley, the epicenter of the digital evolution. She introduced the conference with a report from Silicon Valley on the trends that will dominate the future of online privacy.
The year is 2017. Donald Trump is president of the United States. 8 men hold half of the world's wealth. Between them, they preside over a digital panopticon that enslaves and farms the world's population. It is time to understand that human rights are cyborg rights and deserve constitutional protection. It is time to rise up, reclaim our rights, and start paving the path forward toward a more egalitarian and sustainable future, starting in Europe by reclaiming individual sovereignty and nurturing a healthy, decentralised, digital commons.
Worker's rights can seem like a dated concept in a time when robots are taking over and we hardly even need people to drive cars. But the reality is that worker's rights are more important than ever before. New technologies mean new ways of surveilling the way a person works, moves around, and thinks. Are we surrendering our right to keep our body to ourselves, in the name of productivity and efficiency?
Dries is a Belgian media artist who turns modern surveillance technologies into thought provoking installations. On the basis of his privacy projects he challenged the viewer's perception of today's surveillance culture.


text:
<?php

class NavigationClassWalker extends \Walker_Nav_Menu {
  private $cpt; // Boolean, is current post a custom post type
  private $archive; // Stores the archive page for current URL

  public function __construct() {
    add_filter('nav_menu_css_class', array($this, 'cssClasses'), 10, 2);
    add_filter('nav_menu_item_id', '__return_null');
    $cpt           = get_post_type();
    $this->cpt     = in_array($cpt, get_post_types(array('_builtin' => false)));
    $this->archive = get_post_type_archive_link($cpt);
  }

  public function cssClasses($classes, $item) {

    $classes[] = 'c-Navigation-listItem';

    $classes = array_unique($classes);
    $classes = array_map('trim', $classes);

    return array_filter($classes);
  }
}

redpajama_set_name: RedPajamaGithub


Want to be notified of new releases in GerritEichner/kader?
This example shows you how to estimate at x0 = 2 the value of the density function of the probability distribution underlying Old-Faithful's eruptions data using the (nonrobust) method of Srihera & Stute (2011). The initial grid (given to Sigma) on which the minimization of the estimated MSE as a function of a (kernel-adjusting) scale parameter σ is started is rather coarse here to save computing time.


text:
Q: Passing and assign data to state using Composition API in Vue I new in web programming and I try to make simple dynamic web using Vue.
I have a little problem with passing and assign data to state using composition API. Here the snippet code :
Main.vue :
export default defineComponent({
    setup(_, { root: { $api } }) {

    const state = reactive({
      isFetching: true,
      build: [],
      data:{},
      searches: {
         name: [],
         age: []
      },
    });

onMounted(async () => {
  // get data from API
  state.data = dataFromAPI;
});

const onSearch = (state2) => {
    ....
};

return {
  ...toRefs(state),
  tableHeaders,
  onSearch,
  pagination: {
    rowsPerPage: 10,
  },
};

Child.vue :
export default defineComponent({
    props: {
        searches: { type: Object, required: true },
    },
    setup(_, { emit }) {
        const state = reactive({
           name: ref(null), // I need assign this with value that I get from API in Main.vue
           age: ref(null),
        });

   const onSearch = () => {
       emit('onSearch', { ...state });
   };

   return {
      ...toRefs(state),
      onSearch,
   };
});

I need to passing data and assign that data which I get from API in Main.vue to state in Child.vue. Few days ago I didn't try with composition API and I success pass the data between parent to child or child to parent but I got confused with composition API. I read the composition API documentation but not find any answer.


text:
Prince Harry And Meghan Markle Considered Which Dog Was Most In Need When They Adopted A Beagle, The Rescue Group's Founder Said
The couple adopted a 7-year-old beagle named Mamma Mia, who was rescued from a facility that breeds dogs for biomedical research.
Posted on August 26, 2022, 1:15 am
Michael M. Santiago / Getty Images
Prince Harry, Duke of Sussex, and Meghan, Duchess of Sussex, arrive at the UN Headquarters on July 18, 2022, in New York City.
Prince Harry and Meghan, the Duchess of Sussex (aka Meghan Markle), were eager to give a home to a dog who was rescued from a breeding facility that shut down over alleged animal abuse and immediately connected with the 7-year-old beagle when they met her, a Los Angeles–based rescue organization told BuzzFeed News.
"[Harry and Meghan] really impressed me," said Shannon Keith, the president and founder of the Beagle Freedom Project, in an interview Thursday. "Their energy was so loving and sweet." She identified the Sussexes' new canine companion as a 7-year-old beagle named Mamma Mia, who was rescued alongside her eight puppies. (A spokesperson for the duke and duchess confirmed the adoption when BuzzFeed News reached out for comment.)
Keith said that, earlier this month, she received an email asking if she had time for a call with a "VIP donor," and she agreed without knowing the person to whom she'd be talking. The duchess called her personally, but only identified herself as "Meghan" — and Keith said that she had to Google "Meghan actress Montecito" after the call to figure out who she'd been talking to. Despite not knowing exactly who was on the phone, Keith told BuzzFeed News that she immediately connected with Meghan during that first conversation.
"I was enthralled by her," she said. "The entire time I was thinking, Whoever this person is, she's amazing, she's our people."
Instagram: @beaglefreedom
The Sussexes' new pup is one of 4,000 beagles rescued from an Envigo RMS LLC facility in Cumberland, Virginia. The Indianapolis-based company breeds and sells animals for use in biochemical and pharmaceutical scientific research, and the US government filed suit against it in May, alleging egregious violations of the Animal Welfare Act. The lawsuit was filed after five US Department of Agriculture inspections of the facility during the period from July 2021 to March 2022.
In July, the Justice Department secured a court order forcing the company to permanently shut down and turn over all of the canines housed at the breeding facility to the Humane Society, which then launched an interstate campaign to find homes in coordination with shelters and rescue organizations around the country. The Beagle Freedom Project is one of those organizations.
Keith said that although Meghan set up the call with the Beagle Freedom Project to talk about animals already listed as being available for adoption on the foundation's website, she was very interested when Keith told her that they were in the process of adopting nine dogs from the Envigo facility.
"She knew about Envigo, and when she heard about [Mia and her puppies] she said, 'Oh my goodness, we would love to adopt the mother,'" Keith told BuzzFeed News. "Nobody ever wants to adopt the mother."
"[Meghan] is somebody who wants to rescue those who are in the most need of care," she said.
The inspection reports from the Envigo facility — particularly the two from the first inspection in July 2021 — reveal the horrors of Mia's early life.
The USDA reported that in a seven-month period, more than 300 puppies died of unknown causes — and the facility took no steps to determine why these young dogs died in order to prevent future deaths. Records revealed that animals with easily treated conditions were often euthanized instead of being given care. Insects and insect larvae of all types were found living in food that was left out for the animals. The enclosures where the dogs were kept were overcrowded and in such horrible condition, the report said, that inspectors observed animals whose paws and legs were trapped in the cracked and damaged flooring of their pens. The inspection report also notes that the temperature within buildings where the dogs were kept nearly reached 90 degrees during the five hours that government officials were on the premises, causing the animals to experience "discomfort, lethargy, and stress." (Envigo employees said the air conditioning was broken, the report noted.)
Courtesy Beagle Freedom Project
When Keith and Meghan spoke earlier this month, Keith said Meghan said her family felt "settled" in Montecito and wanted to bring another dog into their life. They talked about the family's other two rescue dogs (Guy, also a beagle, and Pula, a black Labrador retriever) and flock of chickens, which were rescued by Meghan from a slaughterhouse. In that first conversation, Keith said, she knew Meghan was "the real deal" and was "super stoked" at the possibility of her adopting from their organization.
"I've been doing this so long, I have a stringent process for fosters and adopters," Keith said. "I work off [my] gut as well. When I meet people, I just know. This is going to work out or not work out."
Her gut feeling proved right when the duke and duchess made a "low-key and casual" visit to LA on Aug. 11 to meet Mia, Keith said.
"I fell in love with them when they came here," she said.
There was an immediate connection between Mia and the Sussexes, she said, and, within seconds of walking in, the duchess passed one of Keith's potential adopter "tests" by getting down on the floor to greet the dog.
"Mia just ran to them, something she'd never done — she's pretty subdued, she's been traumatized — and Meghan immediately just got on the floor with her and was giving her love," Keith said.
When Harry sat down on a couch, she added, "Mia ran right over to him, put her paws on his knees, and looked into his eyes." Keith said that Harry had done his research and peppered her and fellow volunteer Malina Shirley, who drove Mia and her puppies from the East Coast to California, with questions.
"He asked about the lump on her chin, if it needed a biopsy, her loves, her dislikes," Keith said. "It was really heartwarming, it was everything you'd want [a potential adopter] to ask and then some."
Shirley told BuzzFeed News that she knew it was "something special" and "meant to be" as soon as Harry and Meghan walked into the house. "Mamma Mia was so super excited, just so happy."
The Sussexes spent some time in the yard with Mia on her own, Keith said, and when Meghan came back into the house, she was carrying the dog in her arms. "That was it," Keith said.
After finalizing the adoption, the couple were about to leave with Mia when Harry stopped because the dog was sniffing around the building in the backyard where the rescue dogs are housed.
"He said, 'We can't leave yet, there's something [in there] that she wants, does she have a favorite toy or something?'" Keith said. "He was so adamant that Mamma Mia [wanted] something in that back house," Shirley added.
They all entered the building and Mia immediately went to a container full of dog toys. "I tipped the toy box and she dug her little nose in," Shirley said. She rooted around until she found a fox toy that Shirley had given her on their cross-country trip to California. "She walked outside, and she was so proud of herself," she said.
Harry was delighted, Keith said. "He said, 'I knew it, I knew she wanted something in this room. Now we can go home because she has her toy.'"
"It was so sensitive and intuitive," she said. "At that point, I knew she's going to have the best home."
Mamma Mia with her fox toy
Meghan Markle Revealed How She And Harry Were Forced To Leave Archie Behind To Attend An Official Engagement After His Nursery Caught Fire Ellie Hall · Aug. 23, 2022
The Curious Timing Of The Confirmation That Prince Charles Met His Granddaughter Lilibet Ellie Hall · July 1, 2022
The Cambridge Children And The Fashion Of Growing Up Royal Ellie Hall · June 24, 2022


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THIS SEMINAR IS FULL. PLEASE ADD YOUR NAME TO THE WAIT LIST AND WE WILL CONTACT YOU IF ADDITIONAL SPACE OPENS UP OR IF WE HAVE A CANCELLATION. WE ARE ALSO OFFERING THIS SEMINAR IN FLOSSMOOR ON 5/8/19. REGISTRATION IS OPEN ON OUR WEBSITE.
TOPIC: This seminar will provide attendees with helpful insights and case studies related to lead service lines. The speakers will be discussing how communities can plan for water main replacement projects; funding lead service line replacements utilizing IEPA Low Interest Loans; how to perform a system lead service line inventory; describe lead service line replacement projects; and repair ordinance development, including issues related to private property; public notice requirements and policies related to planned water main replacement projects and system emergency repairs; lead testing protocols and procedures; private property access agreements and suggested public involvement in lead service line issues. The science and health effects of lead will also be discussed along with proper water treatment corrosion control practices to help prevent leaching of lead into drinking water. New, or proposed, IEPA/IDPH legislative issues related to lead service line repairs and replacements will be reviewed. Finally, as time allows, the panel of speakers will be taking questions from attendees to assist Section members with the nuances of lead service line repair and replacement, and best practices, as related to the lead service line regulations.


The Crossfire TRM-507 Tremelo Effects Pedal delivers a warm vintage tremolo effect. The fully adjustable rate and depth controls allow you to dial in the perfect speed and strength of effect. The wave adjusts the overall richness of the tremolo.


But all that is in the past Dwarf Fortress Walkthrough! It comes with a pre-made world and save and full game download for you to play along with. And a ton of screenshots. Yes, many, many screenshots.
Download the walkthrough pack here.
This tutorial uses a pre-configured install of DF, with the same save and all the extras you need - but no more. If that's not going to work for you, see the guide to Setting up your install.
After downloading the pack, you need to unzip the folder onto your hard drive (but not in Program Files). Then create a shortcut on your desktop leading to the launcher or directly to Dwarf Fortress.exe.
Now we're ready to get going! Click on your new shortcut, hit "Play" if you're using the launcher, and you should get the screen below.
So, you've got the game running and enjoyed the exciting ASCII intro movie. Well done! I can tell you'll go far! Next step is to hit Enter on Continue Playing, which should be highlighted. You'll then be presented with this screen.
This is the save game list, which gets cluttered very quickly as seasonal saves build up. But for now all you have to do is use the down-arrow key (not the number pad arrow key) to select region1-post-embark. Hit Enter and the game will load.
Once the game loads you will be presented with this screen. Although it's confusing on first glance, don't panic! It will all be made clear shortly. First thing though, hit Space to pause the game. Esc backs out of any menu you're currently in.
Lets talk about what you can see. First up, the screen is divided into three sections: The left shows the local view. The center shows the action menu, the right shows the area map. To make things a lot neater hit Tab, which cycles through various menu arrangements. Stop when your view matches the one below.
While you examine your surroundings you should be able to spot tree trunks and the grassy areas easy enough, same with (frozen) ponds and the stream and bushes. Other tiles won't make much sense, these tend to be the slightly-arcane zones like open air spaces (pale blue tiles, or tiles with dots) and the earth (brown tiles) and slope or ramp tiles (upward triangles). The screenshot below should help point those features out.
If you look on the bottom right of the window you'll see we're currently at level 108. To move between levels you need to hit < to go up, and > to go down. If you go up a level the map will look like image below. Try it yourself now.
You'll see that what was black dirt is now grass and trees. It's like we've gone up a floor in a lift and we're no longer looking at a slope, now we're looking across a grassy plain. Below us is open space and tree tops. Make sense? Now, try going back down to 108 and then down to 107, a level below "ground" level.
Your view should look something like this. We're underground now with mostly dirt around and earth between the pool, the tree roots, and the rest of the ground. You may have noticed the pool doesn't appear to have changed much, but don't be confused. The difference is that on this lower level we're on the same level as the body of the pool itself, at the level above, we're actually above the pool a little looking at the top of it.
Continue having a look around. When you're done, return to Z-level 108 and find the wagon with our dwarves. The helpful graphic below explains what you're looking at.
So now what do we do? In Dwarf Fortress we're usually trying to get a bustling community of dozens of dwarves up and running, while fending off goblins, the whims of nobles and assorted other evils. To do this we need to build ourselves a mighty fortress! Preferably underground. So lets find a cliff face to dig into and get this show on the road!
Hit the d key. You'll see the menu on the right changes and in the local view a yellow X has appeared. The menu on the right should show the Mine option highlighted. If it doesn't, hit d again, to select it.
The browned out area shows where your miners are going to come along and dig. But they won't act until you back out of the Designations menu as the game is paused. Hit Esc and you should see the game unpause and the menu reset to its master list. Oh, when you have the d menu up you can actually click on the map with your mouse and select areas to mine. Some people prefer to select areas this way.
With the game unpaused you should notice two dwarves race to the cliff face and start digging. With our entrance under way we should also think about laying out some rooms for our dwarves to live in, who wants to spend time out under that hot yellow disk when lovely rock and earth beckon!
See if you can match the layout below by marking out areas to dig, as you've learnt. If you make a mistake you can hit x from the Designations (d) menu and you'll note that the menu on the right has Remove Designation highlighted. Now when you hit enter and select an area, any area set for digging will be cleared of that designation. Anyway, enough detail, on with the room building!
Look at the little dwarves go! Aren't they industrious! They're quickly digging away and leaving a lot of dirt floor behind them. Good lads! Lets leave them to their work while we sort out some other important jobs.
Hit d. Hopefully this time you'll notice the game has automatically got Chop Down Trees selected, if it doesn't, hit t and it will be highlighted.
Hit d again and then p. One the right Gather Plants is now selected.
Now lets sit back and watch the dwarves work for a minute! It won't take long before our miners have cleared out our temporary living quarters and our woodcutter and plant gatherer have begun their work. Once the interior space is clear we have lots of work to do, so lets break it down into useful chunks.
Farming is how you will make most of your food in Dwarf Fortress, and it's important to get farming quite quickly. If food runs out your dwarves will starve and your game will end prematurely, and we don't want that, right? dwarves usually farm underground and handily, we have a space all prepared. So follow along, and lets get some tasty mushrooms growing for our stumpys to chow down on!
Hit b for the Building menu and then p for Farm Plot.
The screen instructions tell you how you can increase the size, we need a good 6 by 6 plot, so hit u and k a few times until you've got a big green grid, like the one below.
Move the grid around with the arrow keys until it's in about the same position as the one in the screenshot. If you get the size wrong, h and m will reduce the plot.
Hit Enter and the green area marker will change to a flashing brown field marker. When stuff flashes it indicates that dwarves are on their way to come build the construction, in this case, a field. And lookee-here! A farmer has come to build our field for us!
He will take a short while to build your field. When you see the dwarf run off, you know he's done. Now it's time to set the field to grow delicious plump helmets all year round.
Hit the q key. This is the key for Set Building Tasks and Preferences. Again you'll notice a cursor on the local screen and if it's anywhere near the field, the field will be flashing. You can move that cursor around with your arrow keys, this is useful later on when you need to select different workshops and buildings.
What we want to do is set Plump helmets as the food to be planted every season. You can see on the top right of the menu we have the various plants we could try and grow and in the middle area we have the different seasons. Right now Dimple cups are selected (but not highlighted, note) and the season is set on Spring, as it's Spring currently. We need to change these settings.
Now comes an important skill! Menu scrolling! Using = (not the down-arrow), scroll down through the list to Plump helmets and hit enter. You should now see Plump helmets highlighted. By the way, -/= is the way to scroll in menus; DF usually uses - and +, but having to use the shift key gets annoying.
Good work! But right now we've only set the spring planting. It's time to set the planting for Summer, Winter and Autumn too. To do this, with the field task menu up, hit b for Summer and then scroll to Plump helmets and hit enter again. Hit c for Autumn, set Plump helmets and then d for Winter, repeating your selection. You may have noticed that what you can plant changes with the seasons, but don't worry about that for now.
Hit Esc to back out of the menu and resume the game. Your dwarves will quickly begin planting in your new field, well done!
Oh, by now you may have noticed the announcements along the bottom of the screen. This is generally because the game wants to let you know that something important has happened. You can hit space to let the game resume, or hit a to see the alert if you miss it at the bottom of the screen). Hitting Esc will take you back out of the alerts menu and resume the game.
While you can just leave all of your dwarves' stuff strewn around the countryside, it's much more efficient to have it all inside near where it's needed. You may have noticed your farmer dwarves running in and out of the fortress to grab the seeds that they need (when they flash between their icon and a little red dot (the seed icon) you know they are carrying seeds).
This is where stockpiles come in. They make everything more organised. A stockpile for all our food and seeds right next door to the farm would probably be pretty handy, right? We don't want the dwarves running across the map to get a bite to eat or a seed to plant, so lets make a food stockpile inside!
Hit the p key (for "piles" of course), a cursor will appear.
Hit f to select Food, we want to make a food stockpile, after all.
Move the cursor to the bottom right of the room, hit enter, move it to the top center and hit enter again. You have now set that space as a food stockpile, well done!
Hit w to make a Wood stockpile, and place it on the other side of the room.
Before reading Chapter 2, how about setting some more plants to be gathered and more wood to be chopped. Then, head to the next installment and we'll learn all about workshops, bedrooms, dinning rooms and stairs! It will be ever so exciting, I promise!