Conditional-expectation convergence along a filtration

If ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$ and $Z$ is bounded, thenalmost surely and in $L^1$. Testing the almost-sure martingale limit on the algebra ${\bigcup }_n{\mathcal{F}}_n$ identifies it with the terminal conditional expectation.

Let $({\mathcal{F}}_n)$ be an increasing filtration and ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. Suppose $X_n\to X$ almost surely and $|X_n|\le Y$ for one integrable random variable $Y$. Thenalmost surely and in L1 space. To prove this, put $Z_n={\sup }_{m\ge n}|X_m-X|$. Then $Z_n\downarrow 0$, $Z_n\le 2Y$, and $\mathbb{E}[Z_n\mid {\mathcal{F}}_n]$ is a nonnegative supermartingale whose expectation tends to zero. Its almost-sure limit is therefore zero. This bounds the moving-variable error; the conditional-expectation convergence along a filtration, extended from bounded to integrable variables by truncation, handles $X$ itself. Dominated convergence also makes the error tend to zero in L1 space.

A separate finite bound for each $X_n$ is insufficient. For a uniform random variable $U$ on $(0,1)$, take $X_n=n{1}_{\{U<1/n\}}$ and trivial ${\mathcal{F}}_n$. Then $X_n\to 0$ almost surely, but $\mathbb{E}{X}_n=1$ for every $n$.