Let $f, g : \mathbb{R^n} \to [0, \infty]$ are measurable. Then, prove that $f+g$ is measurable.
My proof is here, but I think it has the problem.
My proof
For all $c\in \mathbb{R},$ $\{ f+g>c \}=\bigcup_{a+b\in \mathbb{Q}, a+b>c} \bigg(\{f>a\}\cap \{g>b \} \bigg)$. The reason is as follows.
If $x\in \bigcup_{a+b\in \mathbb{Q}, a+b>c} \bigg(\{f>a\}\cap \{g>b \} \bigg)$, there exist $a,b \in \mathbb{Q}$ s.t. $f(x)>a$, $g(x)>b$ and $a+b>c$ hold so $f(x)+g(x)>a+b>c$.
If $x\in \{ f+g>c \}$, $f(x)+g(x)>c,$ i.e. $f(x)>c-g(x)$ and there exists $a\in \mathbb{Q}$ s.t. $f(x)> a>c-g(x)$ from the denseness of $\mathbb{Q}$. Then, $f(x)> a$ and $g(x)>c-a$. There exists $b\in \mathbb{Q}$ s.t. $g(x)>b>c-a$.
Then, $f(x)>a, g(x)>b, a+b>c$, so $x\in \bigcup_{a+b\in \mathbb{Q}, a+b>c} \bigg(\{f>a\}\cap \{g>b \} \bigg)$.
I think this proof has a problem.
The value of $f,g$ is in $[0,\infty]$ so $f$ and $g$ can be $\infty.$
I said that "$f(x)>c-g(x)$ and there exists $a\in \mathbb{Q}$ s.t. $f(x)> a>c-g(x)$ from the denseness of $\mathbb{Q}$", but if $g(x)=\infty,$ $c-g(x)=-\infty$ and I can't say there exists $a\in \mathbb{Q}$ s.t. $f(x)> a>-\infty$ since $-\infty$ is not a real number.
So, doen't this proof work? And if this proof isn't correct, how can I prove the statement?
HINT: Rather try to show that $$\{f + g > c\} = \bigcup_{r \in \mathbb Q} (\{f > r\} \cup \{g > c - r\}).$$