Understanding a convergence proof

Question

I've got a proof for the following result : if $X_n \to X$ in probability then $X_n \to X$ in distribution. The proof goes like this :

Let $t \in \mathbb{R}$ and $\varepsilon >0$. We have $ 1 : \{ X \leq t- \varepsilon \} \subset \{ X_n \leq t \} \cup \{ |X_n - X| \geq \varepsilon \} $

Then $2 : F(t-\varepsilon) \leq F_n(t) + P(|X_n - X| \geq \varepsilon ) $ (we are talking here about CDFs)

Then $3 : F(t-\varepsilon) \leq \lim \inf F_n (t) $

In a similar way : $ 4 : F(t+\varepsilon) \geq \lim \sup F_n (t) $

$5 $ Taking the limit for $\varepsilon \to 0 $ we have $\lim F_n(t) = F(t)$ for each point where $F$ is continuous.

I have trouble getting the points $1,3,4,5$ if you can explain them to me.

Thank you for your answer,

Answer 1

1. If $X \leq t-\varepsilon$ then $X_n \leq t$ or $|X_n-X| \geq \varepsilon$ (since if $X \leq t-\varepsilon$ and $X_n > t$ then $X_n > X-\varepsilon$).
2. use the union bound
3. $\lim$ vs $\liminf$ and $\limsup$ in the proof convergence in probability implies convergence in distribution
4. redo 1-3 with $\{X_n \leq t\} \subset \{ X \leq t+\varepsilon \} \cup \{ |X_n - X| \geq \varepsilon \}$
5. $F(t-\varepsilon) \leq \lim \inf F_n (t)$ implies $\lim_{\varepsilon \downarrow 0} F(t-\varepsilon) \leq \lim \inf F_n (t)$ assuming the limit exists (which it does if we assume $F$ is continuous at $t$). This gives $F(t) \leq \lim \inf F_n (t)$. Similarly, $\lim \sup F_n (t) \leq F(t)$. Since liminf $\leq$ limsup, you get liminf = limsup, so the regular limit exists and equals $F(t)$.