What integral theorem can justify this limit of probabilities, where we have a limit in the integrand and the limits of integration?

Question

Let $(Y_n)$ be a sequence of random variables with finite mean and variance such that $Y_n$ converges in distribution to a random variable $Z$ with finite mean and variance (so that in particular, $P(|Z|<\infty)=1$). I want to justify the following line:

$$\lim_{n\to\infty}P(|Y_n|<\sqrt{n})=P(|Z|<\infty)=1.$$

If $f_n$ is the PDF of $|Y_n|$ and $f_{|Z|}$ is that of $|Z|$, then this statement can be rewritten as $$\lim_{n\to\infty} \int_{-\infty}^{\sqrt{n}} |f_n|\mathrm{d}x=\int_{-\infty}^{\infty} f_{|Z|}\mathrm{d}x=1.$$

If this is not true with the information I'm providing, then relax it as follows: Let $Y_n$ be a sequence of random variables such that $\sqrt{n}(Y_n-\theta)\to n(0,\sigma^2)$ in distribution, i.e., the normal distribution with mean 0 and variance $\sigma^2.$ Take $Z$ to be such a random normal variable. I want to show that

$$\lim_{n\to\infty} P(|Y_n-\theta|<\epsilon)=P(|Z|<\infty).$$

What integral limit theorems can be used here?

Answer 1

It is true, but it is an unusual line because finite mean and variance have nothing to do with it.

1. Since $Y_n\rightarrow Z$ in distribution then $|Y_n|\rightarrow |Z|$ in distribution.

2. The CDF of $|Z|$ is a non-decreasing real-valued function $F_{|Z|}:\mathbb{R}\rightarrow\mathbb{R}$ and hence has at most countably many points of discontinuity.

3. For any $M\geq 0$ at which the CDF of $|Z|$ is continuous we have:

$$ P[|Y_n|\leq \sqrt{n}] \geq P[|Y_n|\leq M] \quad \forall n \geq M^2$$ and so \begin{align} \liminf_{n\rightarrow\infty} P[|Y_n|\leq \sqrt{n}] &\geq \liminf_{n\rightarrow\infty} P[|Y_n|\leq M] \\ &= P[|Z|\leq M] \end{align} where the final equality is where we use the assumption that $F_{|Z|}(z)$ is continuous at $z=M$. Now take $M\rightarrow \infty$, which is possible because we only need $M$ to avoid an at-most-countable number of points of discontinuity of the CDF of $|Z|$.