# Why does this set have finite measure?

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!

• What sort of measure are you using? – copper.hat Mar 8 '15 at 3:11
• @copper.hat It turns out that it has to be semifinite for the conclusion to hold, thanks. – Aubrey Mar 8 '15 at 3:22

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.
• Could you please explain the existence of $E$ such that $A<\mu(E)<\infty$? (You're not assuming that $\mu$ is semifinite.) – Aubrey Mar 8 '15 at 3:01
• 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. – Fan Zheng Mar 8 '15 at 3:11