 Solution

The martingale property gives $\mathbb{E}[X_n\mid {\mathcal{F}}_m]=X_m$. Both $X_m$ and $X_n$ are square-integrable, so their product is integrable by the Cauchy-Schwarz inequality. Using the ${\mathcal{F}}_m$-measurability of $X_m$ in conditional expectation givesThe pull-out step for the possibly unbounded $X_m$ can be justified by first truncating it, then passing to the limit in the L2 norm and using the Cauchy-Schwarz inequality. Expanding the square now yieldsThe left side is nonnegative. Thus $a_n=\mathbb{E}[X_n^2]$ is nondecreasing and bounded above by $M$, and so $a_n\to a\le M$. For $m\le n$,By symmetry of the norm difference this proves that $(X_n)$ is a Cauchy sequence in the L2 space. The completeness established in part (a) gives an $X_{\infty }\in L^2$ withThis is the L2-bounded martingale convergence theorem, derived from martingale-difference orthogonality and completeness. The same sequence also converges almost surely by part (b), since $\mathbb{E}|X_n|\le \sqrt{M}$; uniqueness of a limit in probability identifies that limit with this $X_{\infty }$.