Sum of two sequences of functions converging in measure still converges in measure

Suppose $f_n\to f$ in measure and $g_n\to g$ in measure. Can I claim that $(f_n+g_n)\to f+g$ in measure?

Attempt at the proof:

Since we know that $f_n$ and $g_n$ converge in measure respectively, we know that for all $\epsilon>0$, we have that $m \left(x\in D: \left|f_n(x)-f(x)\right|\geq \epsilon/2 \right)\to 0$ and similarly for $g_n$.

Then $\{f_n(x)+g_n(x)-f(x)-g(x):x\in D\}\subset \{f_n(x)-f(x):x\in D\}\cup \{g_n(x)-g(x):x\in D\}$.

And so the measures of these sets would be $\epsilon/2+\epsilon/2=\epsilon$? (I don't know)

I assume some form of the triangle inequality is supposed to be used here? How to continue?

Solution 1:

$|f_n + g_n - (f+g)| \le |f_n - f| + |g_n - g|$, so wherever $|f_n - f| < \epsilon/2$ and $|g_n-g| < \epsilon/2$ we'll have $|f_n + g_n - (f+g)| < \epsilon$. Thus

$$ \eqalign{\{x: |f_n(x) + g_n(x) &- (f(x) + g(x))| \ge \epsilon\}\cr& \subseteq \{x: |f_n(x) - f(x)| \ge \epsilon/2\} \cup \{x: |g_n(x) - g(x)| \ge \epsilon/2\}}$$ so if each of the sets on the right has measure $< \ldots$