I am trying to prove: if $X_n \to X$ a.s. and $Y_n \to Y$ a.s. then $X_n + Y_n \to X + Y$ a.s. I saw the proofs here in stackexchange and I understand them so far but I am curious about my "proof". It feels like I missed something important or did a stupid mistake but I can't figure out where (or maybe its right? :D). Thank you so far guys! Here the proof:
Let there be an $\varepsilon$ such that $\frac{\varepsilon}{2} > \vert X_n -X \vert$ and $\frac{\varepsilon}{2} > \vert Y_n -Y \vert$ for some sufficient big $n = n(\varepsilon) \in\mathbb{N}$. This $\varepsilon$ exist due to the a.s. convergence of our assumptions further it holds:
$\varepsilon > \vert X_n -X \vert + \vert Y_n -Y \vert \geq \vert (X_n +Y_n) - (X +Y) \vert$
Further:
$0 \leq P (\{ \vert (X_n +Y_n) - (X +Y) \vert \geq \varepsilon\}) \leq P (\{ \vert X_n -X \vert + \vert Y_n -Y \vert \geq \varepsilon \}) \leq P (\{ \vert X_n -X \vert \geq \frac{\varepsilon}{2} \} \cup \{\vert Y_n -Y \vert \geq \frac{\varepsilon}{2} \}) \leq P (\{ \vert X_n -X \vert \geq \frac{\varepsilon}{2} \}) +P( \{\vert Y_n -Y \vert \geq \frac{\varepsilon}{2} \})$
And this equals $0$ due to the assumptions and it follows $X_n + Y_n \to X + Y$ a.s. (where the last step comes from $P(\limsup \limits_{n\rightarrow\infty} \{ \vert X_n -X \vert > \varepsilon) = 0 $ which is equivalent to a.s.).
I don't think the proof is totally right and I miss something. Especially on the last step. I am not sure that I can say that this also holds for the $\limsup \limits_{n\rightarrow\infty}$ case but I also don't find an argument why not. Maybe you guys can help me out.