# If the positive part of submartingale is uniformly integrable, it is closable

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; 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}?

$$E(Y|\mathcal F_n) \geq X_n^{+}$$ $$P-$$ 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.
• 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? Aug 16, 2021 at 0:47
• 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 Aug 16, 2021 at 5:14