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 $m(E \sim F)<\epsilon$.


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?

  • 1
    $\begingroup$ What is $\sim$? $\endgroup$ – user10444 May 11 '14 at 0:19
  • $\begingroup$ '~' stands for 'without' $\endgroup$ – user0430 May 11 '14 at 0:20
  • $\begingroup$ Dear @user0430 I see that, although you have already asked 18 question in this site and received answers in most of them, you have not mark a best answer in any of them. You can do it so by clicking on the checkmark next to the answer that you think is the one that helped you the most. Please read here for more detail. $\endgroup$ – Leo Sera Jul 2 '15 at 21:36

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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