Difficulty in proving that the sum of two measurable function is a measurable function

Question

Let $(\Omega, \Sigma)$, $(\mathbb{R},\mathcal{B})$ be the two measurable space. Let $f:\Omega\rightarrow \mathbb{R}$, and $g:\Omega\rightarrow \mathbb{R}$ be the two measurable functions. In order to prove that $f+g$ is a measurable map, I need to show that $$\{\omega\in\Omega: f(\omega)+g(\omega)< x\}\in \Sigma, \forall x\in\mathbb{R}$$

I was reading a proof, which says that: $$\{\omega\in\Omega: f(\omega)+g(\omega)< x\}= \bigcup_{r\in\mathbb{Q}}\Big[\{\omega : f(\omega)<r\}\cap\{\omega: g(\omega)< x-r\}\Big].$$

I have no clue about how the above two quantities are equal. Can somebody simplify it?

Answer 1

\begin{align*} f(\omega )+g(\omega )<x&\iff f(\omega )<x-g(\omega )\\ &\iff \exists r\in \mathbb Q:f(\omega )<r<x-g(\omega ), \end{align*} where the last equivalence comes from the density of $\mathbb Q$ in $\mathbb R$.

Answer 2

In the formula $$\{\omega\in\Omega: f(\omega)+g(\omega)< x\}= \bigcup_{r\in\mathbb{Q}}\Big[\{\omega : f(\omega)<r\}\cap\{\omega: g(\omega)< x-r\}\Big] \tag{*}$$ Clearly the RHS is contained in the LHS.

For the converse, suppose $\omega$ is in the LHS. Write $\delta=x-f(\omega)-g(\omega)>0$ and find a rational number $r$ such that $f(\omega)<r<f(\omega)+\delta$. Then $$ g(\omega)=x-f(\omega)-\delta<x-r \,,$$ so $\omega$ is in the RHS of $(*)$, as needed.