Show that $\{m\in X:f(m)\ge g(m) \}$ is measurable Show that if $f$ and $g$ are two (extended) measurable functions over ($X,\sigma(X),\mu$), then $\{m\in X:f(m)\ge g(m) \}$ is measurable.
I know that in order to have $f:X\rightarrow \Bbb R \bigcup \{\infty,-\infty \}$ as a measurable function, {$x: f(x)\gt \alpha$} $\in \sigma(x)$
 for all $\alpha \in \Bbb R$. I also know how to prove $af+bg,fg$ ($a,b \in \Bbb R$) and so on are measurable, but not sure how to prove the above condition. Could someone provide a proof please? Thanks.

$$\{m \in X : f(m) \geq g(m)\} = \bigcap_{q\in \mathbb{Q}}(\{m \in X: f(m) \geq q\} \cup \{m \in X: q \geq g(m)\}).$$
As a countable intersection of finite unions of measurable sets, this set is also measurable.
