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I am kind of confused in reading the next proof of a theorem in "Stochastic Analysis: Itô and Malliavin Calculus" in Tandem by Matsumoto & Taniguchi;

Matsumoto & Taniguchi

Here, a closable submartingale $\{ Z_n \}_n$ is defined as a submartingale that is bounded by an integrable random variable $Z$ in $$ Z_n \leq \textrm{E}[Z|\mathcal{F}_n]. $$

Where I am stuck: The equation in the middle, \begin{equation} \mathrm{E}[Y|\mathcal{F}_n] = \lim_{m \rightarrow \infty} \mathrm{E}[X_m^+ | \mathcal{F}_n]. \tag{1} \label{eq:question} \end{equation}

I don't see why we can get the limit out of the expectation. The text says "since $X_n^+$ converges also in $L^1$", but I don't think when $X_n \rightarrow X$ in $L^1$, it always holds for a sub-$\sigma$-algebra $\mathcal{G}$, $$ \textrm{E}[X_n | \mathcal{G}] \rightarrow \textrm{E}[X | \mathcal{G}] \quad \textrm{a.s.} $$ For we have a counterexample: we take $\mathcal{G} = \mathcal{F}$ (where $\mathcal{F}$ is the $\sigma$-algebra of the whole probability space) and $\{X_n\}$ such that $X_n \rightarrow X$ in $L^1$ but not $\textrm{a.s.}$, then $$ \textrm{E}[X_n | \mathcal{F}] = X_n \not\rightarrow X = \textrm{E}[X | \mathcal{F}] \quad \textrm{a.s.} $$ So how can we validate Eq.\eqref{eq:question}?

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1 Answer 1

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$E(Y|\mathcal F_n) \geq X_n^{+}$ $P-$ as. is correct but the first equality may not hold. Use the fact that $L^{1}$ convergence implies a.s. convergence for a subsequece Since $E(X_m^{+}|\mathcal F_n)\to E(Y|\mathcal F_n)$ in $L^{1}$ as $m \to \infty$we can go to a subsequence to finish.

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  • $\begingroup$ Thank you for your answer. Let me clarify some points. The subsequence $\{ \textrm{E} [X^+_m | \mathcal{F}_n]\}_{m = n, n+1, \cdots}$ is a submartingale (which can be proved easily by Tower property), and since $\textrm{E} [X^+_m | \mathcal{F}_n] \rightarrow \textrm{E} [Y | \mathcal{F}_n]$ in $L^1$, it also converges a.s.ly. That is why Eq.(1) stands. Am I correct? $\endgroup$
    – mathmrk
    Aug 16, 2021 at 0:47
  • $\begingroup$ Suppose $U_m \geq U$ a.s. and $U_m \to V$ in $L^{1}$. Then $V \geq U$ a.s.. Ths is because $u_{m_i} \to U$ a.s for some $(m_i)$. The inequality $U_m \geq U$ a.s. automatically holds for the subsequence also: $U_{m_i} \geq U$ a.s. Hence $V \geq U$. @mathmrk $\endgroup$ Aug 16, 2021 at 5:14

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