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%\coverauthor{Valentin Skvortsov}
%\covertitle{A Martingale Closure Theorem for $A$-Integrable Martingale Sequences}
\received{April 20, 1998}
\MathReviews{26A39, 60G46}
\keywords {A-integral, Uniform A-integrability, Martingale sequence}
\firstpagenumber{1}
\markboth{Valentin~Skvortsov}{A Martingale Closure Theorem}
\author{Valentin Skvortsov,
Department of Mathematics, Moscow State University, Moscow, 119899,
Russia, e-mail: {\tt vaskvor@nw.math.msu.su}
or: Institute of Mathematics, WSP, Plac Weyssenhoffa 11, Bydgoszcz,
85-072, Poland, e-mail: {\tt skworcow@wsp.bydgoszcz.pl}}
\title{A MARTINGALE CLOSURE THEOREM FOR $A$-INTEGRABLE MARTINGALE SEQUENCES}
%%%%%%Put Author's Definitions Below Here %%%%%%%%%%%
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\def\F{{\cal F}}
\def\X{{X_\infty}}
\def\ep{{\varepsilon}}
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\begin{document}
\maketitle
\begin{abstract}
A generalized conditional expectation and the corresponding martingale is
defined in terms of the Kolmogorov $A$-integral. It is proved that the uniform
$A$-integrability of a martingale sequence is a sufficient condition for the
sequence to be closed on the right by the $A$-integrable last element.
\end{abstract}
\bigskip
A well known theorem in martingale theory states that a martingale
sequence
$\{X_n,\F _n$, $n=1,2,\ldots\}$ is closed on the right by the last element
$\X (\omega)=\lim\limits_{n\to\infty} X_n(\omega)$ iff
$\{X_n,\F _n\}$ is a
uniformly integrable sequence (see \cite[p.300]{ash},
\cite[p.239]{cho} or \cite[p.60]{loe}).
The conditional expectation in this theory is defined in terms of
the Lebesgue integral.
Meanwhile there are some other versions of a notion of the
mathematical expectation which involve integration more general
than the Lebesgue integration. One of such generalization was introduced
by Kolmogorov in \cite{kol} who defined generalized mathematical
expectation as a non-absolutely convergent integral which later became
known as the Kolmogorov $A$-integral (see \cite{bar}, \cite{lee}).
In this note we are extending this definition to the case of
the conditional expectation and applying this extension to
the investigation of $A$-integrable martingales.
We recall some definitions.
\definition{
A random variable (r.~v.) $X$ defined on a probability space
$(\Omega, {\cal{B}}, P)$ is said to be {\it $A$-integrable} over
a set $B\in {\cal{B}}$ if
\begin{equation}
P\bigl\{\omega\in B\colon |X(\omega)|>C\bigr\}=\bar{o}(1/C)
\mbox{ as } C\to\infty
\end{equation}
and if there exists a finite limit
$$
\lim\limits_{C\to\infty}\!\!\!\! \int\limits_{\{\omega\in B\colon\:
|X(\omega)|\le C\}}
\!\!\!\!\!\!\!\!\!\!\!\!X\, dP \,= \,I.
$$
Then $I$ is called {\it the $A$-integral} of $X$ over $B$ and is
denoted by $(A)\int\limits_B X\,dP$.
Note that if a r.~v. $X$ is $A$-integrable over some set $B\in {\cal{B}}$
and $X(\omega)\ge 0$ a.~s. on $B$ then $X$ is also $L$-integrable and
$(A)\int\limits_B X\,dP=
(L)\int\limits_B X\,dP$.
This fact implies that if an $\F$-measurable r.~v. $X$ ($\F$ being a
sub-$\sigma$-field of ${\cal{B}}$) is $A$-integrable over any
$\F$-measurable subset of some $\F$-measurable set $B$ then $X$ is
$L$-integrable over $B$. Indeed, put
$B_{+}=\{\omega\in B\colon X(\omega)\ge
0\}$ and $B_{-}=\{\omega\in B\colon X(\omega)<0\}$.
Then, being the r.~v. $X$ $A$-integrable over $B_{+}$ and over $B_{-}$,
it must be $L$-integrable over each of these sets and consequently over $B$.
We use this observation in the following definition.
\definition{
Let a r.~v. $X$ be $A$-integrable over any set $B\in\F$ where
$\F$ is a sub-$\sigma$-field of ${\cal B}$. {\it The
conditional $A$-expectation of $X$ with respect to $\F$} is
defined as an $\F$-measurable r.~v. $AE(X|F)$ such that
for every $B\in\F$ we have
\begin{equation}
\label{exp}
\int\limits_B AE(X|\F)\, dP=(A)\int\limits_B X\, dP.
\end{equation}
}
\definition{
An $A$-integrable r.~v. $\X$ is said to be {\it the last element} of a
martingale sequence $\{X_n,\F_n, n=1,2,\ldots\}$ if $X_n=AE(\X|\F_n)$ for
each $n=1,2,\ldots$. We also say in this case that $\X$ {\it closes
the martingale sequence from the right}.
}
We have omitted the "$A$"-sign in front of the left hand side of
(\ref{exp}) because of the above observation, meaning that the Lebesgue
integral can be used here.
It follows from the same observation
that the use
of the $A$-integral in the definition of a martingale sequence
gives an essentially more general notion only for the last element
and only in the case where each $\sigma$-field $\F _n$
is a proper subset of the $\sigma$-field $\F
_\infty$generated by $\cup_n\F_n$.
Here we are going to give a sufficient condition for a martingale sequence
to be closed on the right by the $A$-integrable
last element. This condition
is formulated in terms of uniform $A$-integrability
which is a non-absolute analogue of the uniform Lebesgue integrability.
\definition{
A family of r. v. $\{X_\gamma\}_{\gamma\in\Gamma}$, defined
on $(\Omega, {\cal{B}}, P)$ ($\Gamma$ is some index set)
is said to be {\it uniformly $A$-integrable on $B\in {\cal{B}}$} iff
$$
\sup\limits_{\gamma\in\Gamma}\; \Biggl|\:
(A)\int\limits_{D_{\gamma}(C)} X_\gamma\,
dP\:\Biggr|\longrightarrow 0\quad \mbox{ as } C\to\infty,$$
where $D_{\gamma}(C)=\{\omega\in B\colon\,|X_\gamma(\omega)|>C\}$.
}
Now let $\{\F _n\}$ be an increasing sequence of
sub-$\sigma$-fields of ${\cal B}$ and let
$\{X_n, \F _n$, $n=1,2,\ldots\}$ be a martingale.
For any $B\in\cup _n\F _n$ we define a set function
$\Phi$ by putting
\begin{equation}
\label{phi}
\Phi(B)=\int\limits_B X_n\, dP \quad \mbox{ if } B\in\F _n.
\end{equation}
We call $\Phi$ {\it the associated set function} for $\{X_n,\F _n\}$.
Note that $\Phi$ is well defined. Indeed, if $m\ge n$ then
for the same $B$ by (\ref{phi}) with $n$ substituted by $m$ we get
\begin{equation}
\label{phim}
\Phi(B)=\int\limits_B X_m\, dP \quad \mbox{ if } B\in\F _n\subseteq\F _m.
\end{equation}
Now by the definition of the martingale and by the definition of
the conditional expectation we get
$$
\int\limits_B X_m\, dP=\int\limits_B E(X_m|\F _n)\, dP=\int\limits_B
X_n\, dP
$$
and this proves that the values of $\Phi$ on $B$ given by
(\ref{phi}) and (\ref{phim}) coincide.
$\Phi$ is of course additive on $\cup _n\F _n$ but we do not assume
that $\Phi$ can be extended to the $\sigma$-field $\F _\infty$
generated by $\cup _n\F _n$.
\begin{lemma}
A r.~v. $\X$ is the last element of a martingale
$\{X_n, \F _n, n=1,2,\ldots,\infty\}$ in the sense of the $A$-integral
iff for the associated set function $\Phi$ of the martingale sequence
$\{X_n,\F _n, n=1,2,\ldots\}$ we have $(A)\int _B \X\, dP=\Phi(B)$
for any $B\in\cup _n\F _n$.
\end{lemma}
\begin{proof}[{\sc Proof.}]
This follows directly from the definition
of $\Phi$ and from the definition of the last element.\end{proof}
Note that this Lemma is true for any other
integral which can be used in the above equality.
\begin{theorem}
Let $\{X_n,\F _n, n=1,2,\ldots\}$ be a martingale
sequence convergent a.~s. to a r.~v. $\X$. Let
\begin{equation}
P\bigl\{\omega\in\Omega\colon |X_n(\omega)|>C\bigr\}=\bar{o}(1/C)
\mbox{ uniformly in $n$ as } C\to\infty
\end{equation}
and $\{X_n\}$ be uniformly $A$-integrable on any $B\in\cup _n\F _n$
in the sense of Definition 3. Then $\X$ is $A$-integrable on
each $B\in\cup _n\F_n$ and closes the martingale sequence $\{X_n,\F _n\}$
on the right, i.~e. $\{X_n,\F _n, n=1,2,\ldots,\infty\}$ is
an $A$-integrable martingale with
$$
X_n=AE(\X|\F _n)
$$
$\X$ being its last element.
\end{theorem}
\begin{proof}[{\sc Proof.}]
We show first that (5) implies (1) with $X=\X$.
It is enough to prove (1) with $B=\Omega$.
\ Denote \ $D(C)=\bigl\{\omega\in\Omega\colon |\X (\omega)|>C\bigr\},\quad
D_n(C)=$ $\bigl\{\omega\in\Omega\colon |X_n(\omega)|>C\bigr\},\quad
G_n(C)=\bigcap\limits_{m=n}^\infty D_m(C)$.
Then obviously\
$P(D(C))\le P(\bigcup\limits_{n=1}^\infty G_n(C)),\quad
G_n(C)\subseteq G_{n+1}(C),$
\begin{equation}
P\bigl(D(C)\bigr)\le \lim\limits_{n\to\infty} P\bigl(G_n(C)\bigr),
\end{equation}
\begin{equation}
G_n(C)\subseteq D_n(C).
\end{equation}
Fix any $\ep >0$. By (5) there exists $C_{\ep}>0$ such that
$P\bigl(D_n(C)\bigr)\le \ep/C$ for all $n=1,2\ldots$ and for any $C\ge
C_{\ep}$.
Fix such $C$. Then by (6) and (7) $P\bigl(D(C)\bigr)\le \ep/C$. As $\ep >0$
is arbitrary and $C$ is such that $C\ge C_{\ep}$, then (1) with $X=\X$
is proved for $B=\Omega$ and therefore for any $B\in {\cal{B}}$.
For any r. v. $X$ and $C>0$ define
$$
X^C(\omega)=
\left\{
\begin{array}{rcl}
X(\omega),& & \mbox{ if }\: |X(\omega)|\le C,\\
C \operatorname{sign} X(\omega),& & \mbox{otherwise}.\\
\end{array}
\right.
$$
Notice that
\begin{equation}
\lim\limits_{n\to\infty} X^C_n(\omega)=X_\infty^C(\omega)\quad
\mbox{ a.~s. on } \Omega.
\end{equation}
Now fix $B\in\cup _n\F _n$. Then $B\in \F _n$ for some $n$
and hence for the associated function $\Phi$ the equality (4) holds for any
$m\ge n$.
Fix $\ep >0$.
Since the sequence $\{X_n,{\cal F}_n, n=1,2,\ldots\}$ is
uniformly $A$-integrable on B and (5) holds, we can
find $C_0$ such that for all $C\ge C_0$ and for all $m$
$$
\Biggl|\: \int\limits_B \bigl(X_m-X_m^C\bigr)\, dP \:\Biggr|=
\Biggl|\; \int\limits_{\{\omega\in
B\colon\,|X_m(\omega)|>C\}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!
{X_m\, dP}\,
$$
$$
- C P{\bigl\{\omega\in B\colon\, X_m(\omega)>C\bigr\}}\, +
C P{\bigl\{\omega\in B\colon\, X_m(\omega)<-C\bigr\}} \:\Biggr|\le
$$
\begin{equation} \le
\Biggl|\:\int\limits_{\{\omega\in
B\colon\,|X_m(\omega)|>C\}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
X_m \,dP\:\Biggr|+C P\bigl\{|X_m|>C\bigr\} < \ep/2
\end{equation}
Let $C$ be also fixed for the moment. Then (8) and the Lebesgue dominated
convergence theorem imply that for some $m=m_{\ep,C}\ge n$
\begin{equation}
\int\limits_B \bigl|X_m^C-X_\infty^C\bigr| \,dP < \ep/2
\end{equation}
Now combining (4), (9) and (10) we get for the chosen $m$
$$\Biggl|\:\Phi(B)-\int\limits_B X_\infty^C \,dP\:\Biggr|=
\Biggl|\: \int\limits_B X_m \, dP -\int\limits_B X_\infty^C \,dP \:\Biggr|\le
$$
$$
\le \Biggl|\: \int\limits_B \bigl(X_m-X_m^C\bigr) \,dP \:\Biggr| +
\int\limits_B \bigl|X_m^C-X_\infty^C\bigr| \,dP < \ep
$$
This together with (1) proved already for $X=\X$ implies that
$\X$ is $A$-integrable on $B$ to $\Phi(B)$ and by Lemma, the r.~v.
$\X$ is the last element, in the sense of the $A$-integral, of
the considered martingale sequence. This completes the proof.
\end{proof}
Note that unlike in the case of the uniform Lebesgue integrability,
the above condition in terms of the uniform $A$-integrability is not
necessary for existence of the last element. This can be shown by
constructing an example of a Haar series such that it is the $A$-Fourier
series of an $A$-integrable function and its partial sums are not
uniformly $A$-integrable. (See \cite{kos} for details.)
\begin{thebibliography}{99}
\bibitem{ash} {\sc R.~B.~Ash.}\quad Real Analysis and Probability,\quad
Academic Press, New York, 1972.
\bibitem{bar} {\sc N.~K.~Bary.}\quad A Treatise on Trigonometric Series,\quad
Macmillan, New York, 1964.
\bibitem{cho} {\sc Y.~S.~Chow, H.~Teicher.}\quad Probability Theory,\quad
2-nd edition, Springer-Verlag, 1988.
\bibitem{kol} {\sc A.~N.~Kolmogorov.}\quad Foundations of the Theory of
Probability,\quad
Chelsea Publishing Company, New York, 1950.
\bibitem{lee} {\sc L.~P.~Lee.}\quad Lanzhou Lectures on Henstock
Integration,\quad World Scientific, Singapore, New Jersey, London, 1989.
\bibitem{loe} {\sc M.~Loeve.}\quad Probability Theory II, \quad
4th edition, Springer-Verlag, New York, 1978.
\bibitem{kos} {V.~A.~Skvortsov, V.~V.~Kostin.}\quad Martingale Sequences
in the Theory of Orthogonal Series,\quad Vestnik of Moscow University,
\quad in press.
\end{thebibliography}
\end{document}
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{A Martingale Closure Theorem for $A$-Integrable Martingale Sequences}
{Valentin Skvortsov}
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Professor
{Valentin Skvortsov\newline
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