Lebesgue Measurable function with domain that is finite a.e. and has finite measure, then $\exists$measurable and bounded subset.

Question

Problem

Let $f$ be a (Lebesgue) measurable function on $E$ that is finite a.e. on $E$ and $m(E)<\infty .$ For each $\varepsilon>0$, show that there is a (Lebesgue) measurable set $F$ contained in $E$ such that $f$ is bounded on $F$ and $m(E \sim F)<\varepsilon$.

Question

How can I prove it? I can't get a clue. Thanks!

Answer 1

Let $\epsilon >0$. Consider the sets $F_n = \{x \in E : |f(x)| > n\}$. Notice that $F_{n+1} \subset F_n$ for all $n\in \mathbb{N}$ and $$ \{x \in E: |f(x)| = \infty \} = \bigcap_{n \geq 1} F_n. $$ Since $m(E) < \infty$ and $f$ is finite a.e. on $E$, using continuity from above we obtain \begin{align*} \lim_{n\to\infty} F_n & = m \left( \bigcap_{n \geq 1} F_n\right)\\ & = m(\{x \in E: |f(x)| = \infty \})\\ & = 0. \end{align*} Choose $N \in \mathbb{N}$ such that $n \geq N$ implies $m(F_n) < \epsilon$. Now take $F = E \setminus F_N$. Notice that $F = E \setminus F_N = \{x \in E: |f(x)| \leq N\}$, so $f$ is bounded on $F$ and $m(E\setminus F) = m(F_N) < \epsilon$.

The trick of writing the set as the intersection of a decreasing sequence of sets and then applying continuity of the measure is an extremely common tool in measure theoretic proofs.