The specific description is following:
Let {${X_n}$} be a martingale satisfying $EX_1=0$ and $EX_n^2<\infty$ for all $n$. Show that $E(X_{n+m}-X_n)^2=\sum_{j=1}^mE(X_{n+j}-X_{n+j-1})^2$ and {$X_n$} converges almost surely.
P.S.: It's almost trivial to show $E(X_{n+m}-X_n)^2=\sum_{j=1}^mE(X_{n+j}-X_{n+j-1})^2$ and one can move on to show {$X_n$} converges a.s. if it's $L^ 2$-bounded. But the conditions include only $EX_n^2<\infty$ for all $n$, not $L^2$-bounded. How can I reach the a.s. convergence?
Thus is false. Let $\{Y_i\}$ be i.i.d. taking values $\pm 1$ with probability $\frac 1 2$ each and $X_n=Y_1+Y_2+\cdots+Y_n$. Then $\{X_n\}$ is a martingale, $EX_n=0$ for all $n$ and $EX_n^{2} <\infty$ for all $n$ but this martingale does not converge almost surely. If it does converge then the series $\sum Y_n$ converges almost surely which implies $Y_n \to 0$ almost surely. But $|Y_n|=1$ for all $n$.