Let $\mathcal{A}$ a $\sigma$-algebra, and let $f,g\colon X\to [-\infty,+\infty]$ be a function. Let $\mathcal{M}(X,\mathcal{A})$ the family of the measurable function. I must prove that
Theorem. Let $f,g\in\mathcal{M}(X,\mathcal{A})$. Then
$(i)\;\{f<g\}\in\mathcal{A}$;
$(ii)\;\{f\le g\}\in\mathcal{A}$;
$(iii)\;\{f=g\}\in\mathcal{A}.$
Proof.$\;(i)$ We observe that $$\{f<g\}=\{f<g\;|\; g<+\infty\}\cup\{g=+\infty\}.$$ We note that $$\{g=+\infty\}=\bigcap_{n=1}^{\infty}\{g\ge n\}\in\mathcal{A}.$$ If $f(x),g(x)\in\mathbb{R}$ and $f(x)<g(x)$ eists $q\in\mathbb{Q}$ such that $f(x)<q<g(x)$, therefore $$\{f<g\;|\; g<+\infty\}=\bigcup_{q\in\mathbb{Q}}\big(\{f<q\}\cap\{q<g\}\big)\in\mathcal{A}.$$ Then $\{f<g\}\in\mathcal{A}.$
Quesion 1. It's correct?
$(ii)\;$ We have to prove that $\complement\{g<f\}=\{f\le g\}.$ \begin{equation} \begin{split} \complement\{g<f\}=&\complement\{g<f\;|\; f<+\infty\}\cap\complement\{f=+\infty\}\\ =&\{f\le g\;|\;f<+\infty\}\cap\{f<+\infty\}\qquad(1) \end{split} \end{equation}
Question 2. How can I proceed from $(1)$ to prove that $\complement\{g<f\}=\{f\le g\}?$
Thanks!