Convergence in distribution and convergence in Kolmogorov distance Let $X, Y$ be two random variables with laws $F$ and $G$ respectively. The Kolmogorov distance between these two laws is defined as:
$$
d_{Kol}(F, G) = \sup_{x \in \mathbb R} |\mathbb P(X \leq x) - \mathbb P(Y \leq x)|
$$
Now consider a sequence of random variables $X_n \sim F_n$ and a random variable $X \sim F$. It is easy to see that $d_{Kol}(F_n, F) \rightarrow 0$ implies $X_n \rightarrow_d X$ ($\rightarrow_d$ is for convergence in distribution). My questions are:

*

*Under what assumptions, the other direction is also true? (that is, convergence in distribution implies convergence in Kolmogorov distance). This paper (the first couple of paragraphs) claims this to be true when $F$ is continuous, but I got lost when trying to prove it when using the second Dini's theorem.

*In general, can convergence in probability imply convergence in Kolmogorov distance (without assuming continuous target law)?

Any ideas and insights would be really appreciated.

For 1 with $F_{\infty}$ absolute continuous to the Lebesgue measure, you could check out exercise 3.2.9 in Cyrus Maz's solution to Durrett's Probability. The basic idea is that since $F_{n}$ and $F_{\infty}$ are monotone and their range $[0,1]$ is compact, you can cover $[0,1]$ by finitely many $\epsilon$-balls centered at $F(x_i)$'s for some carefully chosen $x_i$'s, and control the convergence at these points to get a small Komogorov distance. In the end, pointwise convergence + monotonicity of functions + compact ranges implies uniform convergence, I think this is why the paper mentions Dini's Theorem.
2 won't hold without assuming continuous target law. For a counterexample, consider $X_n$ with point mass at $\frac{1}{n}$ and $X_{\infty}$ with point mass at $0$. Then $X_n$ converge to $X_{\infty}$ in probability. But for each $n$, $|\mathbb{P}(X_n \leq 0) - \mathbb{P}(X_{\infty}\leq 0)| = |0 - 1| = 1$ for all $n$, so $d_{Kol}(F_n, F) \geq 1$ for all $n$.
