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Exercise:

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 that there is a measurable set $F$ contained in $E$ such that $f$ is bounded on F and $m(E \sim F)<\epsilon$.

Idea:

I can extract a set $E_0$={set of points such that f is note finite}. Thus $m(E_0)=0$. I can redefine $E$ as $E \sim E_0$ and that the measure of $E$ is equal to the measure of $E \sim E_0$. Then since $E \sim E_0$ is measurable, there exists an $F_\sigma$ type $F$ contained in $E$ such that $m(E \sim F)< \epsilon$ and $f$ is bounded on $F$.

Is this the right idea?

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What is $\sim$? –  user10444 May 11 at 0:19
    
'~' stands for 'without' –  user0430 May 11 at 0:20
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1 Answer 1

Note that your idea doesn't holds always, since it could be that the domain doesn't have a topology defined. However, you can define the sets $E_i=\{x\in X: |f(x)|\geq i\}$, $i\geq 0$. (Where $X$ is the domain of $f$).

Note that $\mu(E_0)=\mu(E)<\infty$, $E_{i+1}\subset E_i$ and that $\mu(\bigcap_{n=0} ^\infty E_i)=\mu(\{x\in X:|f(x)|=\infty\})=0$ (by hypothesis), so $\lim\limits_{n\to\infty}\mu(E_i)=0$. It then follows that given $\varepsilon>0$ there exists a positive integer $n_0$ such that $\mu(E_{n_0})<\varepsilon$. If $F=X\setminus E_{n_0}$, then $\mu(X\setminus F)=\mu(E_{n_0})<\varepsilon$ and $F=\{x\in X:|f(x)|<n_0\}$ so $f$ is bounded in $F$.

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