Proving $P[\{X_n ≤ x\} ∩ \{|X_n − X| < \epsilon\}] \leq P[X ≤ x + \epsilon]$ I'm studying Theorem 5.2.1 from Hogg, Craig Introduction to Mathematical Statistics 8th Edition. I'm adding a few intermediate steps to the proof for clarity and want to know if I'm doing it correctly. Thanks in advance!
My goal is to prove:
$P[\{X_n ≤ x\} ∩ \{|X_n − X| < \epsilon\}] \leq P[X ≤ x + \epsilon]$
where $X_n$ is a sequence of random variables.
My attempt:
$P[\{X_n ≤ x\} ∩ \{|X_n − X| < \epsilon\}]$
 $= P[\{X_n ≤ x\} \cap (\{X_n − X < \epsilon\} \cup \{X \lt X_n + \epsilon\} )]$
 $= P[(\{X_n ≤ x\} \cap (\{X_n − X < \epsilon\}) \cup (\{X_n ≤ x\} \cap \{X \lt X_n + \epsilon\})]$
 $\leq P[\{X_n ≤ x\} \cap \{X \lt X_n + \epsilon\}]$
 $=P[X \lt x + \epsilon]$
 $\leq P[X \leq x + \epsilon]$

Note that
$$
X_n\leqslant x\implies X_n+\epsilon \leqslant x+\epsilon \implies X_n+|X-X_n|\leqslant x+\epsilon \implies X\leqslant x+\epsilon 
$$
Now, as a probability measure is an increasing function, we have that
$$
\{X_n\leqslant x\}\subset \{X\leqslant x+\epsilon \}\implies P(X_n\leqslant x)\leqslant P(X\leqslant x+\epsilon )
$$
∎
