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I am having some troubles with the proof of this: let be $f\in\mathfrak{L}_1(X,S,\mu)$ such that exists $k\geq0$ that satisfies; for all $E\in S$ with $0<\mu(E)<\infty,$ $$\frac{|\int_{E}f \, d\mu|}{\mu(E)}\leq k$$ then $|f|\leq k$ almost everywhere. We can suppose that exists $E\in S$ with the condition.

I have proved the trivial case $k=0,$ but when $k>0$ I tried (with any conlcution) to define $E:=\{x\in X: |f(x)|>k\}$ and prove $\mu(E)=0$ by contradiction, we can prove $\mu(E)<\infty$ using $\,\,f$ is integrable and then we can use the hypothesis.

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Hint: Let $n>0$.

You may proceed by showing that if $\{ |f(x)|>k+\frac{1}{n}\}$ has positive measure then there is $\theta$ real so that $E=\{ {\rm Re\; } e^{-i\theta} f(x) >k \}$ has positive measure. Then use this set $E$ (and note that $n$ was arbitrary).

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  • $\begingroup$ so, the proof is by contradiction? $\endgroup$
    – Twnk
    Commented Apr 3, 2020 at 20:43
  • $\begingroup$ @Twink yes, as well as $\sigma$-additivity of the measure $\endgroup$
    – H. H. Rugh
    Commented Apr 4, 2020 at 21:17

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