 Solution

The Martingale convergence theorem gives an almost-sure limit $M_{\infty }$, because $|M_n|\le 1$. The dominated convergence theorem also gives $M_n\to M_{\infty }$ in $L^1$.

To identify the limit, take $A\in {\bigcup }_n{\mathcal{F}}_n$. For some $N$, $A\in {\mathcal{F}}_N$, and for every $n\ge N$,Passing to the $L^1$ limit preserves this equality. The sets for which $\mathbb{E}[M_{\infty }{1}_A]=\mathbb{E}[Z{1}_A]$ form a monotone class containing the algebra ${\bigcup }_n{\mathcal{F}}_n$, so the equality holds throughout ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. Since $M_{\infty }$ is ${\mathcal{F}}_{\infty }$-measurable, it is $\mathbb{E}[Z\mid {\mathcal{F}}_{\infty }]$. This proves the conditional-expectation convergence along a filtration both almost surely and in $L^1$.