For which $g$ is this integral defined?

Question

Let $f \in L^1(\Omega).$

Then $\int_{\Omega}fg$ is only defined for $g \in L^\infty(\Omega)$, am I right? Holder's inequality tells you the integral is finite. Can I widen the class of $g$ such that the integral is defined? Thanks.

Answer 1

Without more information about $f$, not much can be said; it's certainly not necessary that $g \in L^{\infty}$; as a counterexample, consider Lebesgue measure on $\mathbb{R}$ and

$$f(x) = \chi_{[1, 2]}$$

$$g(x) = \frac{1}{x}$$

Similarly constructed counterexamples show that $g$ need not be in $L^p$ for any given $p$.

For $fg \in L^1$, it is sufficient that $g \in L^\infty$ by Holder, though.