If $M_n \to M_{\infty}$ in $\mathscr L^{2}$, then inequality holds with equality

Question

From Williams' Probability with Martingales

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I tried rewriting the RHS to:

$$\sum_{k=n+1}^{\infty} E[(M_k - M_{k-1})^2] = \sum_{k=n+1}^{n+r} E[(M_k - M_{k-1})^2] + \sum_{k=n+r+1}^{\infty} E[(M_k - M_{k-1})^2]$$

$$ = E[(M_{n+r} - M_n)^2] + \sum_{k=n+r+1}^{\infty} E[(M_k - M_{k-1})^2]$$

Now let $r \to \infty$.

How is $f$ used? Is it used in justifying

$$\lim_r E[(M_{n+r} - M_n)^2] = E[\lim_r(M_{n+r} - M_n)^2]$$

$$\big( = E[(M_{\infty} - M_n)^2] \big)$$

Answer 1

For any $r\ge 0$ $$ \mathbb{E}(M_{\infty}-M_n)^2=\mathbb{E}(M_{\infty}-M_{n+r})^2+\mathbb{E}(M_{n+r}-M_n)^2 \\\overset{(d)}=\mathbb{E}(M_{\infty}-M_{n+r})^2+\sum_{k=n+1}^{n+r}\mathbb{E}(M_k-M_{k-1})^2. $$

Taking $r\to\infty$, the first term converges to $0$ by (f).