probability with martingales 12.2 sum of zero-mean independent variables in L^2 I am struggling with the following theorem from David Williams, Probability with Martingales:
THEOREM
Suppose that $(X_{k}:k\in\mathbb{N})$ is a sequence of independent random variables such that, for every $k$, $E(X_{k})=0, \sigma_{k}^2:=Var(X_{k})<\infty$. 
(a) Then
$$
    \sum\sigma_{k}^2<\infty\Rightarrow\sum X_{k}\text{   converges a.s.  .}
$$
(b) If the variables $(X_{k})$ satisfies 
$$
\exists K \in [0,\infty),\forall k, \omega,\\ 
|X_{k}(\omega)|\leq K,
$$
then
$$\sum X_{k}\text{   converges a.s.}\Rightarrow\sum\sigma_{k}^2<\infty.
$$
The proof for the statement (a) is easy to understand, but I cannot get the other one. According to the proof, "since $\sum X_{n}$ converges a.s., the partial sums of $\sum X_{k}$ are a.s. bounded, and it must be the case that for some $c$, $P(T=\infty)>0$." Here $T$ is the stopping time
$$T = \inf\{r: |\sum_{k=1}^r X_k| > c\}.$$
The problem is that I cannot find this $c$. I know that it's trivial if its boundedness is uniform in $\Omega$, but it's not the case, is it? Can anyone figure out which $c$ meets this condition? Thanks in advance.

Using the notation from Williams, $M_n = X_1 + \dots + X_n$, we know that $M_n$ is a bounded sequence almost surely since it converges almost surely. 
I will write $T_n = \inf\{r : |M_r| > n\}$. Notice that we have
$$\{\sup_{r \geq 1} |M_r| < \infty\} = \bigcup_{n \geq 1} \{T_n = \infty\}.$$
So if $\mathbb{P}(T_n = \infty) = 0$ for each $n$ then 
$$1 = \mathbb{P}(\sup_{r \geq 1} |M_r| < \infty) = \mathbb{P}\big(\bigcup_{n \geq 1} \{T_n = \infty\} \big) \leq \sum_n \mathbb{P}(T_n = \infty) = 0$$
which is a contradiction and hence there is an $n$ such that $\mathbb{P}(T_n = \infty) > 0$ as desired.
