Limsup of a probability I've encountered the following questions. Suppose  $X_n \rightarrow X$ in distribution, and $a<b$
Prove that
$$ P( a \le X\le b) \ge \limsup_{n\to \infty}P(a\le X_n\le b)   $$
I know how to find the limsup of a r.v, but i'm confused on the concept of limsup of a probability.
My idea of approaching this problem is that, we can consider the event of $X$ being within $[a,b]$, and if we can prove that the $\lim\sup$ of this event is a subset of the event itself, the inequality would hold, but I'm worried if I'm mixing the idea of $\lim\sup$ of event and $\lim\sup$ of random variable. Would someone be so kind to give me some hint?
 A: Hint:
If $\:\epsilon>0\:$ and $\:F_{X}\:$ is continuous at $\:a-\epsilon\:$ and $\:b+\epsilon\:$
then: $$\limsup_{n\to\infty}P\left(a\leq X_{n}\leq b\right)\leq\limsup_{n\to\infty}P\left(a-\epsilon<X_{n}\leq b+\epsilon\right)=F_{X}\left(b+\epsilon\right)-F_{X}\left(a-\epsilon\right)$$
You can take $\epsilon$ as small as you like, since the set of elements at wich $F_X$ is not continuous is countable.
A: The probabilities are numbers, so they "live" in $\mathbb R$, actually in $[0,1]$, so $(P(a\le X_n\le b))_{n\in\mathbb N}$ is a sequence of real numbers (so everything you know about such sequences applies here). As you correctly say, the $\lim \sup$ of sets is something different. It is again a set, not a number. 
This question is about a direct application (or proof of ) the Portmanteau Lemma which says that convergence in distribution implies (and is implied by) $$\lim_{n\to \infty}\sup{P(X_n \in C)} ≤ P(X \in C) \quad \text{for all closed sets } C$$ Since, $[a,b]$ is a closed set, then the conclusion can be reached by a direct application of the Portmanteau Lemma. (Otherwise, if you must prove this by yourself, then since the statement is given in full generality (no restrictions on $F$) I suggest you refer to a/every textbook. The proof is not difficult but uses some standard tools: dominated convergence etc, as well as some other equivalences before proving this one.)
