A convergent sequence of normal random variables Say $\{X_n\}$ is a sequence of normal random variables with means $0$ and variances $\sigma_n^2$. Also suppose that $X_n\to X$ (everywhere) and $\sigma_n^2\to \sigma^2.$ Then, using characteristic functions and the uniqueness theorem, it follows that $X$ is normal with mean $0$ and variance $\sigma^2$.
My question is simply if there is another, more elementary, way of reaching the same conclusion, without using the uniqueness theorem for characteristic functions?
(Of course, I wouldn't consider the highfalutin notions of weak convergence, or convergence in distribution, to be "more elementary.")
 A: An elementary proof need not be easier, it just requires less theorems and techniques. Let's see an elementary proof:
$$X_n \sim N(0, \sigma_n) \Rightarrow P(X_n \in A) = \int_A \frac{1}{(2 \pi \sigma_n^2)^{1/2}} \exp\bigg\{-\frac{x^2}{2 \sigma_n^2}\bigg\}\, dx$$
therefore since $X_n \to X$ everywhere, you have that $P_n \xrightarrow[]{*} \bar{P}$ (where $P_n (A):= P (X_n \in A) $ and $\bar{P}(A) = P(X \in A)$ and convergence means convergence in distribution) now you can use the bounded convergence theorem  for $a>0$ and $A = (a, b)$ and $\bar{A} = (-b, -a)$. You can't use bounded convergence theorem for a set including $0$ since, if $\sigma_n \to 0$ then $\frac{1}{(2 \pi \sigma_n^2)^{1/2}} \exp\bigg\{-\frac{0^2}{2 \sigma_n^2}\bigg\} \to \infty$.
Now you conclude that:
$$\bar{P}(\,(a,b)\,) = \lim_n\int_a^b \exp\bigg\{-\frac{x^2}{2 \sigma_n^2}\bigg\}\, dx = \int_a^b \exp\bigg\{-\frac{x^2}{2 \sigma^2}\bigg\}\, dx$$
since $\bar{P}(\,(-\infty,-\epsilon)\,)=\bar{P}(\,(\epsilon, \infty)\,)\xrightarrow[\epsilon \to 0]{}1/2$
we conclude that $$\bar{P}(\{0\}) = 1 - \lim_{\epsilon \to 0} \big(\bar{P}(\,(-\infty,-\epsilon)\,)+\bar{P}(\,(\epsilon, \infty)\,)\big) = 0$$ 
therefore 
$P(\,X \in (-\infty, x)\,)=\bar{P}(\,(-\infty, x)\,) = \int_{-\infty}^x \exp\bigg\{-\frac{x^2}{2 \sigma^2}\bigg\}\, dx$ allows us to conclude that $X \sim N(0, \sigma)$
