$L^1$ convergence of a martingale, conditional expectation

Question

I am trying to prove the following:

Let our space be $(\Omega, \mathcal{F}, P, \{ \mathcal{F}_n \}_{n \in \mathbb{N}})$.

Let $\{X_n \}_{n \in \mathbb{N}}$ be a martingale (adapted to the filtration ) such that $X_n \to X$ in $L^1$. Then $X_n = \mathbb{E}(X| \mathcal{F}_n)$

So our assumptions are that $\mathbb{E} |X_n| < \infty, \ \ \mathbb{E}(X_{n+1} | \mathcal{F}_n) = X_n$ and $\mathbb{E}|X_n - X| \to 0$.

And we need to show that $X_n = \mathbb{E}(X| \mathcal{F}_n)$.

We could take the conditional expectation with respect to $\mathcal{F}_{n-1}$ on both sides:

$$\mathbb{E}(X_n | \mathcal{F}_{n-1}) = \mathbb{E}(\mathbb{E}(X| \mathcal{F}_n)| \mathcal{F}_{n-1})$$

$$X_{n-1} = \mathbb{E}(X| \mathcal{F}_{n-1})$$ due to the tower property, because $ \mathcal{F}_{n-1} \subset \mathcal{F}_n$

Is this ok? Didn't I cheat somewhere here?

Answer 1

Fix a positive integer $n$. By the martingale property and $L^1$ convergence of $\{X_n\}$ we have for $m\geqslant n$, $$\mathbb E[|X_n - \mathbb E\left[X\mid \mathcal F_n]|\right] = \mathbb E\left[| \mathbb E[X_m-X\mid \mathcal F_n]|\right]\leqslant \mathbb E\left[|X_m-X|\right]\stackrel{m\to\infty}\longrightarrow 0, $$ so that $$X_n=\mathbb E[X\mid \mathcal F_n]. $$