Does $\lim_{n \to \infty} P(X_n \le z) = \lim_{n \to \infty} P(Y_n \le z)$ imply $\lim_n \sup_z |P(X_n \le z) - P(Y_n \le z)| = 0 $?

Question

Let $X_n$, $Y_n$ : sequence of real-valued continuous random variables. Say I have the following condition.

$\lim_{n \to \infty} P(X_n \le z) = \lim_{n \to \infty} P(Y_n \le z)$, for any $z \in \mathbb{R}$.

Then, $\lim_n \sup_z |P(X_n \le z) - P(Y_n \le z)| = 0 $?

It is well-known that if the 'target' of convergence in distribution does not dependent on $n$, uniform convergence is satisfied. i.e.

$$\lim_{n \to \infty} P(X_n \le z) = P(Y \le z) $$ then

$$\lim_n \sup_z |P(X_n \le z) - P(Y \le z)| = 0 $$

But I'm not sure this can be applied to my case.

Answer 1

Let $X_n=n$ and $Y_n=3n$ for every $n$

Then: $$\lim_{n\to\infty}P(X_n\leq z)=0=\lim_{n\to\infty}P(Y_n\leq z)\text{ for every }z\in\mathbb R$$

However: $$\sup_z|P(X_n\leq z)-P(Y_n\leq z)|\geq|P(X_n\leq 2n)-P(Y_n\leq 2n)|=1\text{ for every }n$$