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Let $(\Omega,\mathcal{A},\mu)$ be a non-atomic, $\sigma$-finite measure space. If $f \in L^{\infty}(\Omega)$ satisfies that "for every sequence $(\Gamma_n)_{n=1}^{\infty} $ of measurable subsets of $\Omega$ for which $\chi_{\Gamma_n} \rightarrow 0 \ \mu$-a.e. then $\|f \cdot \chi_{\Gamma_n}\|_{\infty} \rightarrow 0$" then $f=0 \ \mu$-a.e.

I read about this result in the book "Interpolation of operators" by Bennett and Sharpley (p.30) but I wasn't able to find or produce a proof. Some help or reference would be appreciated.

The measure $\mu$ is called non-atomic if for every $A \in \mathcal{A}$ with $\mu(A)>0$ there exists a measurable subset $B$ of $A$ such that $0<\mu(B)<\mu(A)$.

[Edit] An attempt has been made to produce a proof in the comments but it is still unclear if that proof really works and if it does (it seems that it doesn't), how non-atomicity is used in order to get that $\|f \cdot \chi_{\Gamma_n}\|_{\infty}>\varepsilon$ .

Walter White
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  • If $|f| >\epsilon$ on $E$ with $\mu (E)>0$ consider $\Gamma_n={x \in E: \epsilon <|f|<\epsilon+\frac 1n)$. – Kavi Rama Murthy Oct 16 '21 at 06:25
  • @KaviRamaMurthy So if I calculated this right, $\limsup\Gamma_n=\bigcap \Gamma_n= \emptyset$ so that $\mu(\limsup(\Gamma_n))=0 \Rightarrow \chi_{\Gamma_n} \rightarrow 0 \ \mu$-a.e. but $|f \cdot \chi_{\Gamma_n}|{\infty}>\varepsilon$ for all $n$ which leads to a contradiction because of our assumption $|f \cdot \chi{\Gamma_n}|_{\infty} \rightarrow 0$. Where is the assumption "non-atomic" and "$\sigma$-finite" used though? – Walter White Oct 16 '21 at 07:16
  • @KaviRamaMurthy But the result is not true for atomic measure spaces. If $\Omega$ consists of a finite number of atoms, then every sequence $(\Gamma_n)$ such that $\chi_{\Gamma_n}\to 0$ is eventually empty. But then of course $|f\chi_{\Gamma_n}|_\infty\to 0$ for every $f\in L^\infty(\Omega)$. – MaoWao Oct 16 '21 at 08:20
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    @WalterWhite $|f\chi_{\Gamma_n}|\infty\geq\epsilon$ only holds if $f\chi{\Gamma_n}\geq \epsilon$ on a set of positive measure, which is not true without further assumptions. – MaoWao Oct 16 '21 at 08:25
  • @MaoWao So do you suggest that the proof in the comment above doesn't work (because the sets $\Gamma_n$ can have zero measure)? Or that it works because of the assumption of non-atomic measure? – Walter White Oct 16 '21 at 09:59
  • @WalterWhite In the current form it is also not true for non-atomic measure spaces. As it is written now, $f$ could be constantly $\epsilon+2$ on $E$, which would imply $\Gamma_n=\emptyset$ for all $n\in\mathbb N$. But I think one can make something similar work (for non-atomic measure spaces), although I haven't really tried to figure out the details. – MaoWao Oct 16 '21 at 11:10

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If $f\neq 0$ on a set of positive measure, then there exists $\epsilon>0$ and a measurable set $\Gamma$ with positive measure such that $|f|\geq \epsilon$ on $\Gamma$. Since $\mu$ is $\sigma$-finite, we can assume that $\Gamma$ has finite measure. By Simpler proof - Non atomic measures, there exists a descending sequence $(\Gamma_n)$ of measurable subsets of $\Gamma$ such that $\mu(\Gamma_n)=2^{-n}$. Then $\chi_{\Gamma_n}\to 0$ a.e. and $\|f\chi_{\Gamma_n}\|_\infty\geq \epsilon$.

MaoWao
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