Why is this set measurable?

Question

I'm trying to understand the proof of Theorem 16.10, Probability and Measure, Patrick Billingsley, I put part of it here exactly as presented in the book

Theorem: If $f,g$ are nonnegative and $\int_Afd\mu=\int_Agd\mu$ for all $A$ in $\mathscr{F}$, and $\mu$ is $\sigma$-finite, then $f=g$ almost everywhere

Proof: Suppose that $f$ and $g$ are nonnegative and that $\int_Afd\mu\leq\int_Agd\mu$ for all $A$ in $\mathscr{F}$. If $\mu$ is $\sigma$-finite, there are $\mathscr{F}$-sets $A_n$ such that $A_n\uparrow\Omega$ and $\mu(A_n)<\infty$. If $B_n = [0\leq g<f, g\leq n]$, then the hypothesized inequality applied to $A_n\cap B_n$ implies $\int_{A_n\cap B_n}fd\mu\leq\int_{A_n\cap B_n}gd\mu< \infty$ (finite beacuse $A_n\cap B_n$ has finite measure and $g$ is bounded there) and hence $\int I_{A_n\cap B_n}(f-g)d\mu=0 \ldots$ (the proof continues)

I understand everything that follows except from one part when the author uses the fact that $B_n = [0\leq g<f, g\leq n]$ is a measurable set. Why is this a measurable set? Thanks in advance

Answer 1

I'm not quite familiar with your notation, but I think this is another way writing $B_n$: $$B_n:= \{ x\in X \mid 0 \leq g(x) < f(x) \text{ and } g \leq n\}$$ If that's correct, then note that: $$B_n = \{x \in X \mid g(x) \geq 0\} \cap \{x \in X \mid (f-g)(x) > 0\} \cap \{x \in X \mid g(x) \leq n\} \\ = g^{-1}([0,\infty)) \cap (f-g)^{-1}((0,\infty)) \cap g^{-1}((-\infty,n])$$ But all of the sets in the last expression are measurable as $g, f-g$ are both measurable (difference of measurable functions is measurable) and hence $B_n$ is measurable.

Answer 2

The set $B_n$ can be written as $B_n = \{g \gt 0\}\cap\{g \lt f\} \cap\{g \le n\}$. The first and third sets are measurable by definition. For the second one, notice that $g < f \iff \exists r \in \mathbb Q$ such that $g < r < f$, so $$\{g \lt f\} = \bigcup_{r\in\mathbb Q} \{g < r\}\cap\{f > r\}$$ Which is a countable union of measurable sets.