proof: $f+g$ is measurable

Question

I am struggling to understand the line "$\{x\in X : (f+g)(x) > a\} = \cup S_r$". How two sets can be equal? Could you elaborate on this?

Thank you in advance.

Answer 1

1. If $x \in \bigcup S_r$, then there is a rational $r$ such that $f(x)>r$ and $g(x)> \alpha -r$. It follows that $(f+g)(x)> \alpha$.

2. Let $(f+g)(x)> \alpha$ and suppose that $x \notin \bigcup S_r$. Now choose a sequence $(r_n)$ of rationals such that

$f(x) > r_n$ for all $n$ and $r_n \to f(x)$. Since $x \notin S_{r_n}$ for all $n$, we have

$g(x) \le \alpha -r_n$ for all $n$. With $n \to \infty$ we get $g(x) \le \alpha-f(x)$, a contradiction !

Answer 2

It is evident that $S_r=\{f>r\}\cap\{g>\alpha-r\}\subseteq\{f+g>\alpha\}$ for every $r\in\mathbb Q$.

If conversely $f(x)+g(x)>\alpha$ or equivalently $f(x)>\alpha-g(x)$ then, because $\mathbb Q$ is dense in $\mathbb R$, we can find some $r\in\mathbb Q$ with $r<f(x)$ (or equivalently $f(x)>r$) but still $r>\alpha-g(x)$ (or equivalently $g(x)>\alpha-r$). So we found an $r\in\mathbb Q$ with $x\in S_r$.