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!
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.