Show martingality property for r.v. $S_{\infty}$, given some assumptions. Could you please detail your answer to my two points?

Question

Given a probability space $(\Omega$, $\mathcal{F}$, $\mathbb{P})$, let $(S_n)_{n\geq1}$ be a martingale and a sequence of uniformly integrable random variables. Also assume $S_n\rightarrow S_{\infty}$ in $\mathcal{L}^1$, that is $\lim\limits_{n\rightarrow\infty}\mathbb{E}\{|S_n-S_{\infty}|\}=0$. Finally, $S_{\infty}$ is in $\mathcal{L}^1$.

I would like to show $\mathbb{E}\{S_{\infty}\mid\mathcal{F}_n\}=S_n$.

First of all, let $n\geq m$ and $\Lambda\in\mathcal{F}_m$. Then, by martingale property: $$ \mathbb{E}\{S_n1_{\Lambda}\}=\mathbb{E}\{S_m1_{\Lambda}\} $$

At this point, from JACOD-PROTTER I read:

However, \begin{align*} |\mathbb{E}\{S_n1_{\Lambda}\}-\mathbb{E}\{S_{\infty}1_{\Lambda}\}| &\leq \mathbb{E}\{|S_n-S_{\infty}|1_{\Lambda}\}\\&\leq\mathbb{E}\{|S_n-S_{\infty}|\} \end{align*} which tends to $0$ as $n$ tends to $\infty$.

A SHORT COMMENT OF MINE: Notice that first inequality follows from Jensen's inequality (since both $M_n$ and $M_{\infty}$ are integrable, i.e. in $\mathcal{L}^1$, and absolute value is a convex function), while the fact that $\mathbb{E}\{|S_n-S_{\infty}|\}$ "tends to $0$ as $n$ tends to $\infty$" follows from $S_n\rightarrow S_{\infty}$ in $\mathcal{L}^1$, i.e. $\lim\limits_{n\rightarrow\infty}\mathbb{E}\{|S_n-S_{\infty}|\}=0$.

$\color{red}{Thus}$ $\mathbb{E}\{S_m1_{\Lambda}\}=\mathbb{E}\{S_{\infty}1_{\Lambda}\}$ and $\color{red}{hence}$ $\mathbb{E}\{S_{\infty}\mid\mathcal{F}_n\}=S_n$ a.s.

I cannot understand the two logical implications in red.

1) "$\color{red}{thus}$" part: as far as I understood, in the first quoted block one gets to the fact that $\lim\limits_{n\rightarrow\infty}|\mathbb{E}\{S_n1_{\Lambda}\}-\mathbb{E}\{S_{\infty}1_{\Lambda}\}|=0$. So, the first question is: why does $\lim\limits_{n\rightarrow\infty}|\mathbb{E}\{S_n1_{\Lambda}\}-\mathbb{E}\{S_{\infty}1_{\Lambda}\}|=0$ imply that $\mathbb{E}\{S_m1_{\Lambda}\}=\mathbb{E}\{S_{\infty}1_{\Lambda}\}$?
2) "$\color{red}{hence}$" part (second question): why does $\mathbb{E}\{S_m1_{\Lambda}\}=\mathbb{E}\{S_{\infty}1_{\Lambda}\}$ imply that $\mathbb{E}\{S_{\infty}\mid\mathcal{F}_n\}=S_n$ a.s.?

Answer 1

$\def\F{\mathscr{F}}\def\G{\mathscr{G}}\def\d{\mathrm{d}}$1) Note that$$ \lim_{n → ∞} |E(S_n I_A) - E(S_∞ I_A)| = 0 \Longrightarrow \lim_{n → ∞} E(S_n I_A) = E(S_∞ I_A). $$ Since it is already shown by the martingale property that$$ E(S_m I_A) = E(S_n I_A), \quad \forall n \geqslant m $$ then$$ E(S_m I_A) = \lim_{n → ∞} E(S_n I_A) = E(S_∞ I_A). $$

2) To rephrase what has been proved up to the “hence”:

For any $n \geqslant 1$, if $A \in \F_n$ then $E(S_n I_A) = E(S_∞ I_A)$. $\qquad(*)$

Now recall the definition of conditional expectations:

For an integrable random variable $X$ on a probability space $({\mit Ω}, P, \F)$ and a sub-σ-algebra $\G \subseteq \F$, a random variable $Y$ in this space is said to be the conditional expectation of $X$ given $\G$ if:

1. $Y$ is $\G$-measurable, and
2. $\displaystyle\int_A X\,\d P = \int_A Y \,\d P$ for any $A \in \G$.

So ($*$) proves by the definition that $E(S_∞ \mid \F_n) = S_n$ since $S_n$ is $\F_n$-measurable.