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