Show $X_n {\buildrel p \over \rightarrow} X$ and $X_n \le Z$ a.s., implies $X \le Z$ a.s.  Suppose $X_n  {\buildrel p \over \rightarrow} X$ and $X_n \le Z,\forall n \in \mathbb{N}$. Show $X \le Z$ almost surely.
I've try the following, but I didn't succeed.  
By the triangle inequality, $X=X-X_n+X_n \le |X_n-X|+|X_n|$. Hence, $P(X \le Z) \le P(|X_n-X| \le Z) + P(|X_n| \le Z)$. I know that, since $X_n  {\buildrel p \over \rightarrow} X$ then $P(|X_n-X| \le Z) \to 1$, and we have $P( |X_n| \le Z)=1$.
I can't go further.

$X_n  {\buildrel p \over \rightarrow} X$ implies that there is a subsequence $X_{n(k)}$ with $X_{n(k)}\to X$ almost surely. 
