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Suppose $f \in L_1(\mathcal{R})$ satisfies for every measurable $A \subset \mathcal{R}$

$$ |\int_A f| \leq [m(A)]^{(1+\epsilon)} $$

for some $\epsilon >0$. Prove $f=0$ a.e.

This is a problem on my analysis qual review sheet. I've been trying a proof by contradiction with no avail... Help?

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    $\begingroup$ Amusing little exercise... Assuming $m$ denotes the Lebesgue measure, decompose each measurable set $A$ of finite measure into $n$ subsets of measure $m(A)/n$ and apply the inequality, then, by the triangular inequality, the absolute value of the integral of $f$ on $A$ is at most $n\cdot(m(A)/n)^{1+\epsilon}=m(A)^{1+\epsilon}/n^\epsilon$. Now, consider the limit when $n\to\infty$ to deduce that the integral of $f$ on $A$ is zero, for every $A$ with finite measure. Can you end this? $\endgroup$ – Did Apr 10 '16 at 21:59
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Here is a hint: by the Lebesgue differentiation theorem, the limit $$\lim_{r\to 0} \frac1{m(B_r(x))} \int_{B_r(x)} f(y) dy$$ where $B_r(x)=\{y\,:\,|x-y|<r\}$ exists and equals $f(x)$ for almost every $x$. Can you take it from there?

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