In Section 1.3 of the book "Martingale Limit Theory and Its Applications" by P. Hall and C. C. Heyde, the authors give a martingale proof for the following result: If $S_n = \sum_{i=1}^n X_i$ is a sum of independent random variables with $S_n$ converging in distribution as $n\rightarrow \infty$, then $S_n$ converges almost surely. See here for an alternative proof.
In the above book, the authors argue that since $Z_n = \dfrac{e^{i t S_n}}{ E[e^{i t S_n}]}$ is a martingale, the martingale convergence theorem quickly yields the asserted result. Here the filtration is defined by $\mathcal{F}_n = \sigma\left(X_1, \ldots, X_n\right)$, and this martingale property of $Z_n$ is a consequence of the independence among $(X_n)_{n \geq 1}$. I initially thought this is a very nice proof, but I was not able to complete the details. So I would like to know how to rigorously prove the claim using the martingale $Z_n$. Thanks!
Below is how I continued with this martingale approach and why I got stuck
I will denote $Z_n$ by $Z_n(t)$ to emphasize its dependence on $t$. Convergence in distr. implies the convergence of characteristic functions. And the continuity of characteristic function at $t = 0$ allows us to prove that, for any sufficiently small $t$, $Z_n(t)$ converges a.s. and thus $e^{i t S_n}$ converges a.s. However, $e^{i t S_n}$ is a periodic function, and, apparently, we can only conclude that almost surely, $e^{i t S_n}$ converges for countably many $t$. I don't know how to conclude that $S_n$ converges a.s.
