Central limit theorem as pointwise convergence.

Question

I stumbled accross this in my lecture notes and I am not sure why this is true? Let $S_n$ be the sum of $n$ i.i.d random variables with mean $\mu$ and standard deviation $\sigma$. Then $$ \mathbb{P} \left( \frac{S_n - n\mu}{\sigma\sqrt{n}} \leq \sqrt{15}x + \frac{3\sqrt{5}}{2\sqrt{n}}x^2 \right) \to \mathbb{P}(Z\leq \sqrt{15}x) = \Phi(\sqrt{15}x) $$ I know that $\frac{3\sqrt{5}}{2\sqrt{n}}x^2 \to 0 $. However, central limit theorem states that $\mathbb{P}(\frac{S_n - n\mu}{\sigma\sqrt{n}} \leq x) \to \mathbb{P}(Z\leq x)$ converges pointwise. I am not sure then why the above expression is true? In general, if $x_n\to x$ is it true that $$ \mathbb{P}(\frac{S_n - n\mu}{\sigma\sqrt{n}} \leq x_n) \to \mathbb{P}(Z \leq x) = \Phi(x)? $$

Answer 1

As I understand this, this uses two results.

Define $F(x) = P(Z \leq x)$ and $F_n(x) = P(Z_n \leq x)$.

If $F$ is continuous (which is the case when $Z \sim N(0,1))$ and $Z_n \stackrel{d}\to Z$, then we have $F_n \to F$ uniformly, i.e. $\sup_{x \in \mathbb{R}} |F_n(x)- F(x) | \to 0$.

Now, use the following lemma:

If $f_n \to f$ uniformly and $x_n \to x$, then $f_n(x_n) \to f(x)$.