$\{x:f(x)\le g(x)\}$ measurable?

Question

Let $(X,\mathcal{F},\mu)$ and $(\mathbb{R},\mathcal{B},\lambda)$ be two measurable spaces (the second one being the Borel space). Let $f,g : X\to \mathbb{R}$ be measurable maps, meaning that $f^{-1}(B)\in \mathcal{F}$ and $g^{-1}(B)\in\mathcal{F}$ for every $B\in\mathcal{B}$. I was wondering if there are examples when the set $\{x:f(x)\le g(x)\}$ is not $\mathcal{F}$-measurable.

Answer 1

Since $f>g$ iff $\exists q \in \mathbb{Q}:f>q>g$ we have $$\begin{aligned}\{x:f(x) > g(x)\}&=\bigcup_{q\in \mathbb{Q}}\{x:f(x)>q>g(x)\}=\\ &=\bigcup_{q\in \mathbb{Q}}(\{x:f(x) >q\}\cap\{x:g(x)<q\}) \in \mathcal{F} \end{aligned}$$ because both $f$ and $g$ are measurable therefore $\{x:f(x)>g(x)\}^c=\{x:f(x)\leq g(x)\}\in \mathcal{F}$.

Answer 2

you can see your set as

\begin{eqnarray} \{x: (f-g)(x)\le 0\}&=&\cup_{r\in \mathbb{Q}_-}\{x : f(x)+g(x) <r\}\\&=&\cup_{r\in \mathbb{Q}_-}\{x : f(x) <r-g(x)\}\\&=&\cup_{r\in \mathbb{Q}_-}\cup_{s\in \mathbb{Q}}(\{x : f(x) \le s\}\cap \{x : s \le r-g(x)\})\\&=&\cup_{r\in \mathbb{Q}_-}\cup_{s\in \mathbb{Q}}(f^{(-1)}(]-\infty,s])\cap (g^{(-1)}(]-\infty,s+r]). \end{eqnarray}