Convergence in probability implies convergence in distribution

Statement

Let $X, X_1, X_2, \dots$ be real random variables on one probability space. If $X_n \to X$ in probability, then $X_n \to X$ in distribution.

Remarks

Proof: for $\varepsilon > 0$, $F_n(x) \le F(x + \varepsilon) + P(|X_n - X| > \varepsilon)$ and $F(x - \varepsilon) \le F_n(x) + P(|X_n - X| > \varepsilon)$. Let $n \to \infty$ and then $\varepsilon \to 0$. At a continuity point $x$ of $F$, this gives $F_n(x) \to F(x)$. The converse is false. Let $X \sim \mathcal{N}(0, 1)$ and $X_n = -X$ for all $n$. Then $X_n \to X$ in distribution, but $P(|X_n - X| > 1) = P(|X| > 1/2)$ does not tend to $0$. The converse holds when the limit is a constant: if $X_n \to c$ in distribution for a real $c$, then $X_n \to c$ in probability.

Explanations

L1 human

If $X_n$ is very likely to be close to $X$, then the probabilities for $X_n$ are close to those for $X$. But two variables can have the same law and still be far apart, so the converse fails.