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.