Proof that convergence in probability implies convergence in distribution

Proof

We sandwich $F_n(x)$ between values of $F$ near $x$.

1. Let $F, F_1, F_2, \dots$ be the cumulative distribution functions of $X, X_1, X_2, \dots$. Fix a point $x \in \R$ at which $F$ is continuous, and fix $\varepsilon > 0$.
2. Claim: $\set{X_n \le x} \subseteq \set{X \le x + \varepsilon} \cup \set{\abs{X_n - X} > \varepsilon}$. If $X_n \le x$ and $X > x + \varepsilon$, then $X - X_n > \varepsilon$.
3. By step 2 and Boole's inequality, $F_n(x) \le F(x + \varepsilon) + P(\abs{X_n - X} > \varepsilon)$.
4. Claim: $\set{X \le x - \varepsilon} \subseteq \set{X_n \le x} \cup \set{\abs{X_n - X} > \varepsilon}$. If $X \le x - \varepsilon$ and $X_n > x$, then $X_n - X > \varepsilon$.
5. By step 4 and Boole's inequality, $F(x - \varepsilon) \le F_n(x) + P(\abs{X_n - X} > \varepsilon)$.
6. By convergence in probability, $P(\abs{X_n - X} > \varepsilon) \to 0$ as $n \to \infty$.
7. Steps 3, 5 and 6 give $F(x - \varepsilon) \le \liminf_{n} F_n(x) \le \limsup_{n} F_n(x) \le F(x + \varepsilon)$.
8. Step 7 holds for every $\varepsilon > 0$. Since $F$ is continuous at $x$, $F(x - \varepsilon) \to F(x)$ and $F(x + \varepsilon) \to F(x)$ as $\varepsilon \to 0^+$.
9. So $\liminf_n F_n(x) = \limsup_n F_n(x) = F(x)$, that is, $F_n(x) \to F(x)$.
10. This holds at every continuity point $x$ of $F$. So $X_n \to X$ in distribution.