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!
 A: Uniform integrability does the job. We have 
$$\mathbb E(S_n^2)=n\mathbb E[X_1^2],$$
hence defining $Y_n:=n^{-1/2}|S_n|$, we have 
$$\mathbb E[Y_n; Y_n\geqslant R]\leqslant \sqrt{\mathbb E[Y_n^2]}\cdot \sqrt{\mathbb P\{Y_n\geqslant R\}}\leqslant \sqrt{R^{-2}\mathbb E[Y_n^2]}\leqslant \frac 1R.$$
For non-negative random variables $Z_n,Z$ such that $Z_n\to Z$ in distribution we have the equivalence 
$$\mathbb E[Z_n]\to \mathbb E[Z]\Leftrightarrow \{Z_n,n\geqslant 1\}\mbox{ is uniformly integrable.}$$
Hence proving uniform integrability is a natural way to deal with such a problem. 
A: 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.
A: Alternatively, you could use the exact formula for $\mathbb{E}(|S_{2n}|)$ followed by the asymptotics as in Mike Spivey's answer here:
$$\mathbb{E}(|S_{2n}|)=(2n) \binom{2n}{n}{1\over 4^n} = 2{\sqrt{n\over \pi}} \left(1 - \frac{1}{8n} + \frac{1}{128n^2} + \frac{5}{1024n^3} - \frac{21}{32768 n^4} + O(n^{-5})\right).$$
