If L1 Convergent, Then Martingale is Closable
Theorem
Given a probability space $( \Omega , \mathcal{F} , P)$ and a martingale $\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}$, if a stochastic process $\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}$ converges to a random variable $Y$ through $\mathcal{L}_{1}$, then $\left\{ ( X_{n} , \mathcal{F}_{n} ): n = 1 , \cdots , \infty \right\}$ is a closable martingale.
Description
Even if $X_{n}$ converges to $Y$ through $\mathcal{L}_{1}$ and almost surely converges to $X_{\infty}$, it cannot be guaranteed that $Y$ and $X_{\infty}$ have some relation. $$ X_{n} \overset{\mathcal{L}_{1}}{\to} Y \land X_{n} \overset{\text{a.s.}}{\to} X_{\infty} \nRightarrow Y = X_{\infty} $$ In formula terms, it can be restated as above, and during the proof, it is verified that in the case of martingales, $Y \overset{\text{a.s.}}{=} X_{\infty}$ holds.
Proof
Part 1. $E |X_{n}| \to E |Y|$
It is said that $\left\{ X_{n} \right\}$ converges to some random variable $Y$ through $\mathcal{L}_{1}$, as follows. $$ \lim_{n \to \infty} \int_{\Omega} | X_{n} - Y | dP = \lim_{n \to \infty} E | X_{n} - Y | = 0 $$ Without loss of generality, since $\left| |a| - |b| \right| \le | a - b|$, $$ \begin{align*} \left| E | X_{n} | - E | Y | \right| \le & \left| E \left( | X_{n} | - | Y | \right| \right) \\ \le & E \left| |X_{n}| - |Y| \right| \\ \le & E \left| X_{n} - Y \right| \end{align*} $$ That is, when $n \to \infty$, $E |X_{n}| \to E |Y|$ holds.
Part 2. $Y \overset{\text{a.s.}} = X_{\infty} $
Submartingale Convergence Theorem: Given a probability space $( \Omega , \mathcal{F} , P)$ and a submartingale $\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}$.
If it is assumed that $\displaystyle \sup_{n \in \mathbb{N}} E X_{n}^{+} < \infty$, then $X_{n}$ almost surely converges to some random variable $X_{\infty}: \Omega \to \mathbb{R}$, and $$E X_{\infty} < E X_{\infty}^{+} < \infty$$
Since it is $E |X_{n}| \to E |Y|$, it becomes $\displaystyle \sup_{n \in \mathbb{N}} E | X_{n} | < \infty$. Of course, $E X_{n}^{+} \le E | X_{n} | < \infty$ and since a martingale is a submartingale, according to the submartingale convergence theorem, the stochastic process $\left\{ X_{n} \right\}$ almost surely converges to some random variable $X_{\infty}$. Hence, $Y \overset{\text{a.s.}}{=} X_{\infty}$ holds.
Part 3. $E \left( X_{\infty} | \mathcal{F}_{n} \right)= X_{n}$
Without loss of generality, since $Z \ge 0 \land E Z = 0 \implies Z \le 0$, it is sufficient to show that $E \left| E \left( X_{\infty} | \mathcal{F}_{n} \right) - X_{n} \right| = 0$ holds. Considering $m > n$, $$ \begin{align*} E \left| E \left( X_{\infty} | \mathcal{F}_{n} \right) - X_{n} \right| =& E \left| E \left( X_{\infty} | \mathcal{F}_{n} \right) - E \left( X_{n} | \mathcal{F}_{n} \right) \right| \\ =& E \left| E \left( X_{\infty} | \mathcal{F}_{n} \right) - E \left( X_{m} | \mathcal{F}_{n} \right) \right| \\ =& E \left| E \left( X_{\infty} - X_{m} | \mathcal{F}_{n} \right) \right| \\ \le & E \left[ E \left( \left| X_{\infty} - X_{m} \right| | \mathcal{F}_{n} \right) \right] \\ =& E | X_{\infty} - X_{m} | \end{align*} $$ Applying $\displaystyle \lim_{m \to \infty}$ to both sides and according to Parts 1 and 2, $$ \begin{align*} E \left| E \left( X_{\infty} | \mathcal{F}_{n} \right) - X_{n} \right| &= \lim_{m \to \infty} E \left| E \left( X_{\infty} | \mathcal{F}_{n} \right) - X_{n} \right| \\ \le & \lim_{m \to \infty} E | X_{\infty} - X_{m} | \\ \le & 0 \end{align*} $$ Therefore, it is confirmed that $\left\{ ( X_{n} , \mathcal{F}_{n} ): n = 1 , \cdots , \infty \right\}$ is a closable martingale.
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