Sums of independent random variables converging in distribution, and hence in probability?

Question

For fun, I'm working through the exercises in Durrett's "Probability: Theory and Examples", and I'm stuck on the following:

Exercise 3.2.21

Let $X_1,X_2,\ldots$ be independent. If $S_n = X_1 + \ldots + X_n$ converges in distribution then it converges in probability (and hence a.s. by Exercise 2.5.10). Hint: The last exercise implies that if $m, n \to \infty$ then $S_m - S_n \to 0$ in probability. Now use Exercise 2.5.11.

Here's what I have:

There is some random variable $S_\infty$ such that $S_n \Rightarrow S_\infty$. We now want to show $P(|S_n - S_\infty| > \epsilon) \to 0$ for arbitrary $\epsilon > 0$.

Exercise 3.2.20 (the "previous exercise") states that a sequence of r.v.'s $Y_n \Rightarrow 0$ if and only if their characteristic functions $\varphi_n(t) \to 1$ as $n \to \infty$, whenever $|t| \le \delta$ for some $\delta > 0$.

I can see how $S_n - S_m \to 0$ in probability as a result of this. The next part of the hint is completely mystifying. Exercise 2.5.11 states that whenever $S_n / n \to 0$ in probability, we have $(\max_{1 \le m \le n} S_m) / n \to 0$ in probability, where $S_n$ is the sum of an i.i.d. sequence of r.v.'s.

I don't think the $X_i$ being identically distributed is actually needed in the solution of Exercise 2.5.11 (Ottaviani's inequality doesn't require it), but even so, I have no idea what Durrett is getting at.

Answer 1

In the edition I have at my disposal (the second one), he refers (up to a typo) to the following exercise (6.5 page 49):

Show that random variables are a complete space under the metric defined in the previous exercise (that is $d(X,Y)=\mathbb E\left(\left|X-Y\right| /\left(1+\left|X-Y\right|\right)\right)$), i.e. if for any positive $\varepsilon$, there exists $N$ such that for $m,n\geqslant N$, $d\left(X_m,X_n\right)\lt\varepsilon$, then there exists a random variable $X_\infty$ such that $X_n\to X_\infty$ in probability.

The existence of $S_\infty$ is not trivial in general, as discussed here and there.