Define $Z_n=\sup_{m\geq n}|X_m-X|$. Then $0\leq Z_n\leq2$ and $Z_n\downarrow0$ almost surely. The <dominated convergence theorem> gives $\mathbb E Z_n\to0$.

Set $W_n=\mathbb E[Z_n\mid\mathcal F_n]$. Since $Z_{n+1}\leq Z_n$, the <tower property of conditional expectation> gives
$$
\mathbb E[W_{n+1}\mid\mathcal F_n]
=\mathbb E[Z_{n+1}\mid\mathcal F_n]
\leq W_n,
$$
so $(W_n)$ is a nonnegative <supermartingale>. The <almost sure supermartingale convergence theorem> gives $W_n\to W_\infty$ almost surely, and <Fatou lemma> yields $\mathbb EW_\infty\leq\liminf_n\mathbb EW_n=0$. Hence $W_n\to0$ almost surely; because $\mathbb EW_n=\mathbb EZ_n\to0$, convergence also holds in $L^1$.

Finally,
$$
\left|\mathbb E[X_n\mid\mathcal F_n]-\mathbb E[X\mid\mathcal F_\infty]\right|
\leq W_n
+\left|\mathbb E[X\mid\mathcal F_n]-\mathbb E[X\mid\mathcal F_\infty]\right|.
$$
The second term tends to zero almost surely and in $L^1$ by part b. The first does so by the preceding argument, proving the <moving-variable conditional-expectation convergence>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 201 / 1 / c / Solution

If bounded random variables $X_n$ converge almost surely to $X$, then
$$
\mathbb E[X_n\mid\mathcal F_n]
\longrightarrow
\mathbb E[X\mid\mathcal F_\infty]
$$
almost surely and in $L^1$. For $Z_n=\sup_{m\geq n}|X_m-X|$, the variables $\mathbb E[Z_n\mid\mathcal F_n]$ form a nonnegative supermartingale with expectations tending to zero.


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 Incoming links: Moving-variable conditional-expectation convergence