Consider $Y_1, \ldots, Y_n$ random variables such that $n^{1/2}(\bar{Y}_{n} - \mu ) \xrightarrow{D} \text{iid}~ \mathcal{N}(0,\,\sigma^{2}).$
It holds that $$g_{n}(x_n(c))=1_{[n^{1/2}\frac{(\bar{Y}_{n} - \mu )}{\sigma} \leq c]} \xrightarrow{D} 1_{[Z \leq c]}, \quad Z \sim \mathcal{N}(0,1)$$ where $c$ is some deterministic value.
Show: \begin{align} E[g_{n}(x)] &\to P[Z \leq c] = \Phi(c)\quad (n \rightarrow \infty) \\ V[g_{n}(x)] &\to \Phi(c)(1-\Phi(c)) \quad(n \rightarrow \infty) \end{align}
Problem: How do I arrive at those expressions? Heuristically it is somehow clear due to the indicator function (Bernoulli variable). I am interested in the exact formal derivation.
My Attempt: \begin{align} \lim_n E[g_{n}(x)] &= \lim_n E\left[ 1_{[n^{1/2}\frac{(\bar{Y}_{n} - \mu )}{\sigma} \leq c]} \right] = \lim_n P\left[ n^{1/2}\frac{(\bar{Y}_{n} - \mu )}{\sigma} \leq c \right] = P[Z \leq c] = \Phi(c) \\ \lim_n V[g_{n}(x)] &= \lim_n E \left[ 1^2_{[n^{1/2}\frac{(\bar{Y}_{n} - \mu )}{\sigma} \leq c} \right] - \lim_n E\left[ 1_{[n^{1/2}\frac{(\bar{Y}_{n} - \mu )}{\sigma} \leq c} \right]^2 \\ &= \lim_n P\left[ n^{1/2}\frac{(\bar{Y}_{n} - \mu )}{\sigma} \leq c \right] - \lim_n P\left[ n^{1/2}\frac{(\bar{Y}_{n} - \mu )}{\sigma} \leq c \right]^2 \\ &= P(Z \leq c)-P(Z \leq c)^2= \Phi(c)(1- \Phi(c)) \end{align}
(I think I cannot decompose the variance and look at each term separately, can I?)
For clarification, I am interested if I have used the convergence in distribution property right. In particular, looking at the variance, I decompose it and apply to each term separately (3rd equality) that the distribution functions converge (due to convergence in distribution of the predictor). Can I do this (if yes why): $\lim_n P[n^{1/2}\frac{(\bar{Y}_{n} - \mu )}{\sigma} \leq c]^2 = P(Z \leq c)^2$?
Random variable $1_{[Z\leq z]}$ only takes values in $\{0,1\}$ so that $$\mathbb E1_{[Z\leq z]}=0\cdot P(1_{[Z\leq z]}=0)+1\cdot P(1_{[Z\leq z]}=1)=0\cdot P(Z>z)+1\cdot P(Z\leq z)=P(Z\leq z)$$
Further - because it only takes values on $\{0,1\}$ - we have $1_{[Z\leq z]}=(1_{[Z\leq z]})^2$ so that:$$\mathbb E1_{[Z\leq z]}^2=\mathbb E1_{[Z\leq z]}=P(Z\leq z)$$
Then:$$\mathsf{Var}1_{[Z\leq z]}=\mathbb E1_{[Z\leq z]}^2-(\mathbb E1_{[Z\leq z]})^2=P(Z\leq z)-P(Z\leq z)^2=P(Z\leq z)[1-P(Z\leq z)]$$