Central limit theorem and convergence of means If $Z_1,Z_2,Z_3,\ldots$ are i.i.d. with $P(Z_i=-1) = P(Z_i=+1) = \frac 12,$ then we have by the Central Limit Theorem that $\frac{\sum_{i=1}^n Z_i}{\sqrt{n}}\stackrel{d}{\to} \mathcal{N}(0,1),$ so that for any continuous bounded function $f,$ we have $\mathbb{E}f\left(\frac{\sum_{i=1}^n Z_i}{\sqrt{n}}\right)\to\mathbb{E}f(W)$ where $W\sim\mathcal{N}(0,1).$ Now, $|\cdot|$ is not a bounded function, so it is not necessarily true that
$$\mathbb{E}\left|\frac{\sum_{i=1}^n Z_i}{\sqrt{n}}\right|\to\mathbb{E}|W|.$$
My question is whether the above is true for this specific distribution of $Z_i.$ If not, what does $\mathbb{E}\left|\frac{\sum_{i=1}^n Z_i}{\sqrt{n}}\right|$ converge to?
 A: The following result appears in Billingsley's book Convergence of probability measures (1968, and I guess the 1995 edition).

Theorem. Let $(Y_n)_{n\geqslant 1}$ be a sequence of non-negative random variables defined on a probability space $(\Omega,\mathcal F,\mu)$ such that $Y_n\to Y$ in distribution and $Y_n,Y$ have a finite expectation. Then the following are equivalent:
i) $\mathbb E[Y_n]\to\mathbb E[Y]$;
ii) the sequence $(Y_n)_{n\geqslant 1}$ is uniformly integrable.

Assume that i) holds and fix a positive $\varepsilon$. Fix $R$ such that $\mathbb E[Y\chi_{\{Y>R\}}]\lt\varepsilon$ and $\mu\{|Y|=R\}=0$. The set of discontinuity points of map $t\mapsto \chi_{-(R,R)}(t)$ has a null $\mu_Y$ measure, hence
$$\lim_{n\to\infty}\mathbb E[Y_n\chi_{\{Y_n\lt R\}}]=\mathbb E[Y\chi_{\{Y\lt R\}}].$$
Conversely, we use the map $t\mapsto \chi_{-(R,R)}(t)$ for a well-chosen $R$.
In the context of the question, with $Y_n:=n^{-1/2}\left|\sum_{j=1}^nX_j\right|$, we observe that $\mathbb E[Y_n^2]=\mathbb E[X_0^2]$ (it holds more generally in the case of an iid sequence of zero-mean square integrable functions).
