$X_n-Y_n\rightarrow_{a.s.} 0$ and $Y_n\rightarrow_{a.s.} Z$ imply $X_n\rightarrow_{a.s.} Z$?

Consider two sequences of random variables $\{X_n\}_n, \{Y_n\}_n$ and a random variable $Z$, all defined on the same probability space. Let $\rightarrow_{a.s.}$ denote almost sure convergence. Suppose

(1) $X_n-Y_n\rightarrow_{a.s.} 0$

(2) $Y_n\rightarrow_{a.s.} Z$

Do (1) and (2) imply $X_n\rightarrow_{a.s.} Z$? If yes, which result I'm using?

• Use the face that $X_n=X_n-Y_n+Y_n$ and the definition of almost sure convergence. – Augustin Mar 2 '16 at 13:12
• Is this like a Slutsky's Lemma for almost sure convergence? – TEX Mar 2 '16 at 13:15
• It's even simpler than that. Almost sure convergence is just pointwise convergence on a set of probability $1$. And the intersection of two such sets has probability $1$. – Augustin Mar 2 '16 at 13:17

Yes it´s true. Use the fact that if $A_n\rightarrow_{a.s.}A$ and $B_n\rightarrow_{a.s.}B$, then $A_n+B_n\rightarrow_{a.s.}A+B$.