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Fix a measurable $g$ and let $f$ be any simple measurable function that vanishes outside a set of finite measure and satisfies $||f||_p=1$ where $p>1$. If $|\int fg|<M$ for all such $f$, why must $$ \{x:|g(x)|>\epsilon\} $$ have finite measure for any $\epsilon>0$? (This should be very simple but I can't see it right now.)

EDIT: As Fan pointed out below, it is necessary to assume that $\mu$ is semifinite!

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  • $\begingroup$ What sort of measure are you using? $\endgroup$ – copper.hat Mar 8 '15 at 3:11
  • $\begingroup$ @copper.hat It turns out that it has to be semifinite for the conclusion to hold, thanks. $\endgroup$ – Aubrey Mar 8 '15 at 3:22
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Suppose for some $\epsilon > 0$, $\{x: |g(x)|>\epsilon\}$ has infinite measure. Then for any $A>0$, there is $E\subset\{x: |g(x)|>\epsilon\}$ such that $A<\mu(E)<\infty$. Let $f$ be $(\mu(E))^{-1/p}\ \overline{\text{sgn}\ g}$ on $E$, and $0$ elsewhere. Then $\|f\|_p=1$, but

$$\int fg>\int_E \epsilon(\mu(E))^{-1/p}=\epsilon\mu(E)^{1-1/p}>\epsilon A^{1-1/p}.$$

Since $A$ can be arbitrarily large, no bound of the form $|\int fg|<M$ is possible, which is a contradiction.

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  • $\begingroup$ Could you please explain the existence of $E$ such that $A<\mu(E)<\infty$? (You're not assuming that $\mu$ is semifinite.) $\endgroup$ – Aubrey Mar 8 '15 at 3:01
  • $\begingroup$ If we don't assume $\mu$ is semifinite, then the result is false. Consider an infinite delta mass at $0$ imposed on the Lebesgue measure on $\mathbb R$. Then any $f$ with finite $L^p$ norm must vanish at $0$. Now let $g(0)=1$ and $g$ vanishes elsewhere. Then $\int fg=0$ for any $f$ with finite $L^p$ norm, but $\{x:g(x)>\epsilon\}$ has infinite measure. $\endgroup$ – Fan Zheng Mar 8 '15 at 3:11
  • $\begingroup$ Great, thank you! I basically couldn't solve it because I was not assuming that it was semifinite! I should add that in the assumptions above. $\endgroup$ – Aubrey Mar 8 '15 at 3:18

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