Application of Central limit theorem and law of large numbers Let $X_i$ be identically distributed independent random variable, with mean 0 and variance 1. Let $S_n = \frac{X_1+X_2+\dots+X_n}{n}$.
Let $f$ be function such that $f(0^-) = 0$ and $f(0+) = 1$. Prove $f(\frac{S_n}{n})$ converges in distribution to Bernoulli random variable (with $p = 0.5$). ($f(0^-)$ and $f(0^+)$ are left and right limits of f at $0$ respectively)
I know we have to use central limit theorem to get this but I can't figure out how. Any hints will be helpful.
 A: Let $g\colon\mathbb R\to\mathbb R$ be uniformly continuous and bounded and let $Y_n:=f(S_n/n)$. We want to compute
$
\lim_{n\to\infty}\mathbb E\left[g(Y_n)\right]
$. As $g$ is bounded and $\{S_n=0\}$ has a measure converging to $0$,
it suffices to compute
$$
\ell_+:=\lim_{n\to\infty}\mathbb E\left[g(Y_n)\mathbf{1}\{S_n/n>0\}\right],\quad \ell_-:=\lim_{n\to\infty}\mathbb E\left[g(Y_n)\mathbf{1}\{S_n/n<0\}\right].
$$
Note that
$$
\lvert\mathbb E\left[\left(g(Y_n)-g(1)\right)\mathbf{1}\{S_n/n>0\}\right]\rvert
\leqslant \lVert g\rVert_\infty\mathbb P(S_n/n>\delta)+
\lvert\mathbb E\left[\left(g(Y_n)-g(1)\right)\mathbf{1}\{\delta>S_n/n>0\}\right]\leqslant \lVert g\rVert_\infty\mathbb P(S_n/n>\delta)+\sup_{0<s<\delta}\lvert g(f(s))-g(1)\rvert
$$
hence by the law of large number, the following estimate holds for each positive $\delta$:
$$
\limsup_{n\to \infty}\lvert\mathbb E\left[\left(g(Y_n)-g(1)\right)\mathbf{1}\{S_n/n>0\}\right]\rvert\leqslant \sup_{0<s<\delta}\lvert g(f(s))-g(1)\rvert
$$
and by assumption on $f$ and $g$, the last last quantity goes to $0$. Therefore,
$$
\ell_+=\lim_{n\to\infty}g(1)\mathbb P(S_n/\sqrt n>0)=g(1)/2, 
$$
by the central limit theorem.
Similarly, $\ell_-=g(0)/2$ hence $$
\lim_{n\to\infty}\mathbb E\left[g(Y_n)\right]=\frac{g(0)+g(1)}2.$$
