The Probabilistic Method, Section 2.5. Unbalancing Lights I am working through Alon's and Spencer's Probabilistic Method book. In Section 2.5, Unbalancing light, in the proof of Theorem 2.5.1, it is mentioned that $R_i$ has distribution $S_n$, the distribution of the sum of $n$ independent uniform $\{-1,1\}$ random variables, and so
$$
E[|S_n|]=\left(\sqrt{\frac{2}{\pi}}+o(1)\right)\sqrt{n},
$$
with asymptotics found by estimating $S_n$ by $\sqrt{n}N$, where $N$ is standard normal.
It is easy to see from the CLT that if $S_n=\sum_{i=1}^n X_i$ with $X_i$ all i.i.d., $\mathcal{U}(\{-1,+1\})$, then $S_n/\sqrt{n}\to G\sim \mathcal{N}(0,1)$ in distribution. By the continuous mapping theorem, $|S_n|/\sqrt{n}\to |G|$, which is where the $\sqrt{2/\pi}$ comes from. However, how is
$$
|S_n|/\sqrt{n}\to |G|~\text{in distribution}
\qquad\Longrightarrow\qquad
E|S_n|/\sqrt{n}\to E|G|
$$
justified? It seems to me that the proof is incomplete, as integration to the limit would require for instance uniform integrability, but I have not been able to show it. Is it even the case? Is there any other way to prove this rigorously?
Any hint or help much appreciated!

Use the definition of weak convergence that shows the cdfs converge pointwise  and then use the formula for expectation that is in terms of the cdf that works for nonnegative random variables:
$\int_0^\infty P(X\geq t) dt = E[X]$
if $X\geq 0$ (justified by Fubini Tonelli)
Then use dominated convergence to finish.
