Problem: Assume that $X_n\overset{P}{\longrightarrow}X,\,Y_n\overset{P}{\longrightarrow}Y$, and all random variables in question are defined on the same probability space. Show that $X_n+Y_n\overset{P}{\longrightarrow}X+Y$.
My Attempt: First we show that if $Z_1,Z_2$ are random variables then $$P(\vert Z_1+Z_2\vert\geq\varepsilon)\leq P(\vert Z_1\vert\geq\varepsilon/2)+P(\vert Z_2\vert\geq\varepsilon/2).$$ Note that the inequality above is a consequence of using the subadditivity of the probability measure and $$\{\vert Z_1+Z_2\vert\geq\varepsilon\}\subseteq\{\vert Z_1\vert\geq\varepsilon/2\}\cup\{\vert Z_2\vert\geq\varepsilon/2\},$$ so we prove the above set containment. Note that $$(\{\vert Z_1\vert\geq\varepsilon/2\}\cup\{\vert Z_2\vert\geq\varepsilon/2\})^\complement=\{\vert Z_1\vert<\varepsilon/2\}\cap\{\vert Z_2\vert<\varepsilon/2\}.$$ If $\omega\in\{\vert Z_1\vert<\varepsilon/2\}\cap\{\vert Z_2\vert<\varepsilon/2\}$, then $\vert Z_1(\omega)\vert<\varepsilon/2$ and $\vert Z_2(\omega)\vert<\varepsilon/2$, so the triangle inequality implies that $$\vert Z_1(\omega)+Z_2(\omega)\vert\leq\vert Z_1(\omega)\vert+\vert Z_2(\omega)\vert<\varepsilon,$$ so that $\omega\notin\{\vert Z_1+Z_2\vert\geq\varepsilon\}.$ Therefore, we have $$\{\vert Z_1+Z_2\vert\geq\varepsilon\}\subseteq\{\vert Z_1\vert\geq\varepsilon/2\}\cup\{\vert Z_2\vert\geq\varepsilon/2\},$$ so $$P(\vert Z_1+Z_2\vert\geq\varepsilon)\leq P(\vert Z_1\vert\geq\varepsilon/2)+P(\vert Z_2\vert\geq\varepsilon/2).$$ With this result, we have for any fixed $\varepsilon>0$ that $$P(\vert X_n+Y_n-X-Y\vert\geq\varepsilon)\leq P(\vert X_n-X\vert\geq\varepsilon/2)+P(\vert Y_n-Y\vert\geq\varepsilon/2)\overset{n\to\infty}{\longrightarrow}0$$ where we used the assumption that $X_n\overset{P}{\longrightarrow}X$ and $Y_n\overset{P}{\longrightarrow}Y$. Since $\varepsilon>0$ was arbitrary, it follows that $X_n+Y_n\overset{P}{\longrightarrow}X+Y.$
Do you agree with my proof above? Any comments are most welcomed and appreciated.
Thank you very much for your time.