If $X_n$ converges to $X$ in $L_p$ and $Y_n$ converges to $Y$ in $L_p$ then $X_n + Y_n $ converges to $X + Y$ in $L_p$

Question

I want to show that if $X_n \xrightarrow{L^p} X$ and $Y_n \xrightarrow{L^p} Y$ then $X_n + Y_n \xrightarrow{L^p} X + Y$ ($p \geq 1)$.

My idea is to use the following facts (whose proofs I won't give here):

1. $L_p$ convergence implies convergence in probability
2. If $X_n$ converges to $X$ in probability and $Y_n$ converges to $Y$ in probability then $X_n + Y_n $ converges to $X + Y$ in probability

I also want to use my hunch that $L_p$ convergence implies domination, i.e. $X_n \xrightarrow{L^p} X$ $\implies$ there exists some random variable $Z_x$ s.t. $|X_n|$< $Z_x$.

Finally I want to use

Show that convergence in probabiltiy plus domination implies $L_p$ convergence

on the sum $X_n+Y_n$, which, if my facts and my hunch are correct, converges in probability and is dominated by $Z_x+Z_y$ and thus converges to $X+Y$ in $L_p$.

Problem is: is my hunch even correct? And if yes, how to prove it rigorously?

Answer 1

Hint.

$\|\cdot \|_{L^p}$ is a norm. So $$\|X_n+Y_n-X-Y\|_{L^p}\leq \|X_n-X\|_{L^p}+\|Y_n-Y\|_{L^p}.$$