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

Let $f : E \to \Bbb{R}$ and $E$ be as stated in the hypothesis. Since $f$ is finite almost everywhere, there exists an $E_0 \subseteq E$ for which $m(E_0) = 0$ and $f$ is finite over $E-E_0$. Let $\epsilon > 0$. Then there exists a closed set $C$ contained in $E$ such that $m(E-C) < \epsilon$. Note that $E-(C-E_0) = E_0 \cup (E-C)$ which implies $m(E-(C-E_0)) \le m(E_0) + m(E-C) < \epsilon$. Let $F = C-E_0$. Then $f$ will not be infinite over $C$, but it still could be unbounded; so it remains to show that it is bounded on $C$. By way of contradiction, suppose that $f$ is unbounded. Then for every $n \in \Bbb{N}$, there exists an $x_n \in C$ such that $|f(x_n)| \ge n$. Since $C$ and $E_0$ are measurable subsets of $E$, over which $f$ is a measurable function, $f$ is also measurable over $C-E_0 = F$. Moreover, since $|~|$ is continuous, the composition $|~| \circ f$ is measurable over $F$. Therefore, for every $n$, the measurable set $\{x \in F \mid |f(x)| \ge n \}$ is nonempty.

This is where I get stuck...I feel like there is a contradiction lurking somewhere but I cannot spot it. I was trying to argue that the intersection $\bigcap_{n \in \Bbb{N}} \{x \in C \mid |f(x)| \ge n \} = \{x \in C \mid |f(x)| = \infty \}$ is not empty, which would give the desired contradiction, but this is very unlikely...I could use a hint.

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Hint: Let $G=\{x \in E: |f(x)| \lt \infty\}$. Let $E_n=\{x\in E: |f(x)| \le n\}$. Then $$E_1 \subset E_2\ldots E_n \subset E_{n+1}\ldots\text{ and } G=\cup_{n \in \mathbb{N}}E_n$$

Then $$\mu(G)=\lim_{n \to \infty}\mu(E_n)$$

Let $\epsilon \gt 0$. Then there exists a $n_0 \in \mathbb{N}$ such that $\mu(G-E_{n_0}) \lt \epsilon$. Let $F=E_{n_0}$.

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