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If a real measurable function $f : E \rightarrow \mathbb{R}\ \cup \{\infty\}$ is finite almost everywhere (i.e. $f^{-1}(\{\infty\})$ has measure zero) and $E \subset \mathbb{R}$ is of finite Lebesgue measure, then $\forall \varepsilon > 0\ \exists\ F \subset E$ with $|E \setminus F| \leq \varepsilon$ and $f$ bounded in $F$.

I supposed that no such $F$ exists, and I tried to prove that $f = \infty$ in a set with positive measure, but without success. Any help? :)

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    $\begingroup$ What does "finite almost everywhere" mean here? Are you assuming that $f:\mathbb R\to\mathbb R\cup\{\infty\}$ and simply saying that $f^{-1}(\{\infty\})$ is null? $\endgroup$
    – hmakholm left over Monica
    May 14, 2016 at 18:44
  • $\begingroup$ not null, but of measure zero. I'll edit the question though, I noticed it's not very clear. $\endgroup$
    – Guillermo Mosse
    May 14, 2016 at 20:26
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    $\begingroup$ Using the result of this question which I asked for the purpose, we can see that the assumption that $f$ is measurable is crucial. Its answer shows that, assuming the Axiom of Choice, $\mathbb R$ can be partitioned into continuum many sets $(X_t)_{t\in\mathbb R}$ such that every $X_t$ intersects every set of positive measure. Then let $f(x)=$the $t$ such that $x\in X_t$, and then $f$ is not bounded in any $F$. $\endgroup$
    – hmakholm left over Monica
    May 15, 2016 at 14:56

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For each $n\in\mathbb{N}$, define $F_n=\{x\in E:|f(x)|> n\}$. Then $F_1\supset F_2\supset\cdots$ and $$\bigcap_{n=1}^{\infty}F_n=\{x\in E:|f(x)|=\infty\}$$

Since $E$ has finite measure, we can apply continuity from above to conclude that $$ 0=|\{x\in E:|f(x)|=\infty\}|=\lim_{n\to\infty}|F_n|$$ Therefore given $\varepsilon>0$ there exists $n$ such that $|F_n|<\varepsilon$, so the set $F=E\setminus F_n$ has the desired properties.

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  • $\begingroup$ How do you know $|F_n|$ exists? It doesn't seem to be assumed that $f$ is measurable. $\endgroup$
    – hmakholm left over Monica
    May 14, 2016 at 19:29
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    $\begingroup$ I suppose I assumed $f$ is measurable, because it's not clear how to interpret the statement that $f$ is finite a.e. otherwise. $\endgroup$
    – carmichael561
    May 14, 2016 at 19:35
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    $\begingroup$ Even if $f$ is not measurable, it still seems to make sense to ask whether $f^{-1}(\{\infty\})$ happens to be a measurable set with measure $0$. $\endgroup$
    – hmakholm left over Monica
    May 14, 2016 at 19:37
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    $\begingroup$ I guess. While OP should clarify, I'm reasonably confident that $f$ is measurable. $\endgroup$
    – carmichael561
    May 14, 2016 at 19:38
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    $\begingroup$ @SeñorBilly: Non-measurable functions are elusive things, difficult to construct. One needs to use the Axiom of Choice to show that they even exist. $\endgroup$
    – hmakholm left over Monica
    May 14, 2016 at 20:53

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