I have some question on the proof of Theorem 8.2.2 from durrett's book.
In Theorem 8.2.2, Theorem 8.2.1 which is stated below is used.
If $S_n$ is a sqaure integrable martingale with $S_0=0$,and $B_t$ is a Brownian motion, then there is a sequence of stopping times $0=T_0\leq T_1 \leq T_2\dots$ for the Brownian motion such that $$(S_0,S_1,\dots,S_k) \overset{d}{=} (B(T_0),B(T_1),\dots,B(T_k))$$
Theorem 8.2.2 is as follows,
Let $\mathcal{F}_m = \sigma(S_0,S_1,\dots,S_m)$. $\lim\limits_{n\rightarrow\infty} S_n$ exists and is finite on $\sum\limits_{m=1}^{\infty}E((S_m-S_{m-1})^2|\mathcal{F}_{m-1})<\infty$
The proof is, let $\mathcal{B}_t$ be the filtration generated by Brownian motion, and let $t_m = T_m - T_{m-1}$. Therefore, $E((S_m-S_{m-1})^2|\mathcal{F}_{m-1})=E(t_m|\mathcal{B}(T_{m-1}))$.
Let $M=\inf\{n:\sum\limits^{n+1}_{m=1}E(t_m|\mathcal{B}(T_{m-1}))>A\}$.
After some calculation we can get
$E\sum\limits^M_{t=1}t_m=E\sum\limits^M_{m=1}E(t_m|\mathcal{B}(T_{m-1}))\leq A$
From this it concludes that $\sum\limits^M_{m=1}t_m <\infty$, and as $A\rightarrow \infty$,$P(M<\infty)\downarrow 0$ ,so $T_{\infty}=\sum\limits^{\infty}_{m=1}t_m <\infty$.
The part I can't get is that how is $P(M<\infty)\downarrow$ related to $\sum\limits^{\infty}_{i=0}t_m$ ?Does $P(M<\infty)\downarrow$ means that $M\rightarrow \infty$ almost surely?