Problem in the measurability of $f+g$.

Question

Let $f$ and $g$ measurable. In my course, to prove the measurability of $f+g$, they do like following. Let $$E_\alpha =\{x\mid f(x)+g(x)>\alpha \}.$$ Then, there is $r_n\in\mathbb Q$ s.t. $f(x)>r_n>\alpha -g(x)$. Then, if $x\in E_\alpha $, then $x\in \{y\mid f(y)>r_n\}\cap \{y\mid g(y)>\alpha -r_n\}$. Therefore, $$E_\alpha \subset \bigcup_{n\in\mathbb N}\{y\mid f(y)>r_n\}\cap \{y\mid g(y)>\alpha -r_n\}.$$

I really don't understand what is this sequence $(r_n)_n$. I agree that there is a $r\in \mathbb Q$ s.t. $f(x)>r_n>\alpha -g(x)$, but what is this sequence $(r_n)_n$ ?

Answer 1

As written your proof looks indeed strange. First to prove that $E_\alpha$ is measurable you have to find an equality s.t. the rhs is measurable (an inclusion is not enough). Furthermore the usual to prove that $f+g$ is measurable is to notice that:

$$x \in E_\alpha \iff \exists r \in \mathbf{Q} \; f(x)>r>\alpha-g(x)$$

Thus:

$$E_\alpha = \bigcup_{r\in \mathbf{Q}}\{y \; \vert \; f(y)>r\} \cap \{y \; \vert \; g(y)>\alpha-r\}$$

Since $\mathbf{Q}$ is countable, the rhs is measurable.

I don't know exactly what the sequence $(r_n)_{n\in \mathbf{N}}$ is, maybe an enumeration of $\mathbf{Q}$?