Problem: Let $\mu\in\mathbb R,\sigma_n$ a sequence in $(0,\infty)$ such that $\sigma_n\to0$ and $X_n\thicksim N(\mu,\sigma_n^2)$. Let $h:\mathbb R\to\mathbb R$ be a bounded continuous function. Show that $E[h(X_n)]\rightarrow h(\mu)$ as $n\to\infty.$
Attempted Proof: Let $\varepsilon>0$ be given. Note that $h$ is uniformly continuous in any compact interval. So choose $\delta>0$ such that for all $x\in[\mu-\delta,\mu+\delta]$ we have that $\vert h(x)-h(\mu)\vert<\varepsilon/3.$ Using this we have that $$\mathcal I_1(n)=\frac{1}{\sqrt{2\pi\sigma_n^2}}\int_{\mu-\delta}^{\mu+\delta}\vert h(x)-h(\mu)\vert\exp\left[-\frac{(x-\mu)^2}{2\sigma_n^2}\right]\,dx<\frac{\varepsilon}{3\sqrt{2\pi\sigma_n^2}}\int_{-\infty}^\infty\exp\left[-\frac{(x-\mu)^2}{2\sigma_n^2}\right]\,dx=\frac{\varepsilon}{3}.$$ Since $h$ is bounded, we see that $$\mathcal I_2(n)=\frac{1}{\sqrt{2\pi\sigma_n^2}}\int_{-\infty}^{\mu-\delta}\vert h(x)-h(\mu)\vert\exp\left[-\frac{(x-\mu)^2}{2\sigma_n^2}\right]\,dx\leq\frac{2\|h\|_\infty}{\sqrt{2\pi\sigma_n^2}}\int_{-\infty}^{\mu-\delta}\exp\left[-\frac{(x-\mu)^2}{2\sigma_n^2}\right]\,dx.$$ Now observe that $$\frac{2\|h\|_\infty}{\sqrt{2\pi\sigma_n^2}}\exp\left[-\frac{(x-\mu)^2}{2\sigma_n^2}\right]\to0\quad\text{as }n\to\infty.$$ Next, the property of the exponential that $\exp[-x^k]\leq x^{-k}$ for all $x\in\mathbb R$ and $k\in\mathbb N$ yields that $$\frac{2\|h\|_\infty}{\sqrt{2\pi\sigma_n^2}}\exp\left[-\frac{(x-\mu)^2}{2\sigma_n^2}\right]\leq\frac{2\|h\|_\infty}{(x-\mu)^2}=g(x)\quad\text{for all }x\in\mathbb R.$$ Since the function $g$ is integrable, it follows from Lebesgue's Dominated Convergence Theorem that $\mathcal I_2(n)\to0$ as $n\to\infty.$ The same exact analysis, with the same dominating function $g$, yields that $$\mathcal I_3(n)=\frac{1}{\sqrt{2\pi\sigma_n^2}}\int_{\mu+\delta}^{\infty}\vert h(x)-h(\mu)\vert\exp\left[-\frac{(x-\mu)^2}{2\sigma_n^2}\right]\,dx\to0\quad\text{as }n\to\infty.$$ To finish, choose $N\in\mathbb N$ such that for all $n>N$ both $\mathcal I_2(n)<\varepsilon/3$ and $\mathcal I_3(n)<\varepsilon/3$ hold. It follows that for all $n\in\mathbb N$ we have $$\left\vert E[h(X_n)]-h(\mu)\right\vert\leq\mathcal I_1+\mathcal I_2+\mathcal I_3<\varepsilon.$$ Therefore, we conclude that $E[h(X_n)]\rightarrow h(\mu)$ as $n\to\infty.\quad\blacksquare$
Could anyone please have a look at my attempt at a proof above?
Thank you for your time and appreciate any feedback and help.
For an alternative and much faster solution:
Note that the value of $E[h(X_n)]$ only depends on the distribution of $X_n$, so if we replace $X_n$ by any random variable $Y_n$ that also has an $N(\mu, \sigma_n^2)$ distribution, we have $E[h(X_n)] = E[h(Y_n)]$. A very convenient choice is to fix some $Z \sim N(0,1)$ and let $Y_n = \mu + \sigma_n Z_n$. This way we clearly have $Y_n \to \mu$ almost surely. Since $h$ is continuous, this implies $h(Y_n) \to h(\mu)$ almost surely, and since $h$ is bounded, dominated convergence yields $E[h(Y_n)] \to h(\mu)$.