Consider a lebesgue measurable function $f:S\rightarrow \mathbb{R}$, where $S$ is a Lebesgue measurable subset of $\mathbb{R}$. I would like to show that there must exist a subset $S_{0}$ of $S$ with $|S_{0}|>0$, $|.|$ denotes Lebesgue measure, such that $f\in L^{\infty}(S_{0})$.

I suggest arguing by contradiction: Suppose the contrary. Then $f$ is not essentially bounded on any subset $U$ of $S$ with $|U|>0$. Then $|f|=\infty$ almost everywhere.

Does this make sense ?

  • $\begingroup$ You consider a function $f: S \to \Bbb R$ hence $f=\infty$ is not possible. If you actually want to consider a function $f: S \to \Bbb R \cup \{-\infty,+\infty\}$ then your proof is ok wit $|f| = \infty$ on the last line. $\endgroup$ – Gono Nov 24 '17 at 12:13
  • $\begingroup$ Hi Gono. That is the contradiction ! I started with assuming $f$ is Lebesgue measurable, real-valued, and there is no subset with positive measure on which $f$ is essentially bounded and ended up with $f$ is infinite almost everywhere (the contradiction). Please comment. Thanks $\endgroup$ – Medo Nov 24 '17 at 12:20
  • $\begingroup$ If you want to prove this statement directly, Lusin's Theorem might come in handy: en.wikipedia.org/wiki/Lusin%27s_theorem $\endgroup$ – humanStampedist Nov 24 '17 at 12:43
  • $\begingroup$ I'm not sure your proof is sufficient, since what you obtain is a bit different from what you infer: If $f$ is not essentially bounded on any subset $U$ of $S$, then for every subset $U$ you obtain a sequence $x_k\in U$ such that $f(x_k)\rightarrow\infty$. Since $f$ is not continuous you are not done yet. $\endgroup$ – humanStampedist Nov 24 '17 at 12:46
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    $\begingroup$ Perhaps you can put $S_n=\{x\in S; |f(x)|\leq n\}$, then $S=\cup S_n$, and $S_n\subset S_{n+1}$, hence $\lambda(S_n)\to \lambda(S)$, and it is easy to finish. $\endgroup$ – Kelenner Nov 24 '17 at 12:52

Here is the answer from (quoted) the link https://en.wikipedia.org/wiki/Lusin%27s_theorem

Let $ (X,\Sigma ,\mu )$ be a Radon measure space and Y be a second-countable topological space, let $f:X\rightarrow Y$ be a measurable function. Given $\epsilon > 0$, for every $ A\in \Sigma$ of finite measure there is a closed set $E$ with $\mu(A \setminus E) < \epsilon$ such that $f$ restricted to $E$ is continuous. If A is locally compact, we can choose E to be compact and even find a continuous function$f_{\varepsilon }:X\rightarrow Y$ with compact support that coincides with $f$ on $E$ and such that $ \sup _{x\in X}|f_{\varepsilon }(x)|\leq \sup _{x\in X}|f(x)|$.


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