I'm reading about this inequality from Wikipedia.
Suppose that $\left(S_{1}, \mu_{1}\right)$ and $\left(S_{2}, \mu_{2}\right)$ are two $\sigma$-finite measure spaces and $F: S_{1} \times S_{2} \rightarrow \mathbb{R}$ is measurable. Then $$\left[\int_{S_{2}}\left|\int_{S_{1}} F(x, y) \mu_{1}(\mathrm{d} x)\right|^{p} \mu_{2}(\mathrm{d} y)\right]^{\frac{1}{p}} \leq \int_{S_{1}}\left(\int_{S_{2}}|F(x, y)|^{p} \mu_{2}(\mathrm{d} y)\right)^{\frac{1}{p}} \mu_{1}(\mathrm{d} x)$$ with obvious modifications in the case $p=\infty$. If $p>1$, and both sides are finite, then equality holds only if $|F(x, y)|=\varphi(x) \psi(y)$ a.e. for some non-negative measurable functions $\varphi$ and $\psi$.
If $F$ is non-negative, then the function $$f:y \mapsto \int_{S_{1}} F(x, y) \mu_{1}(\mathrm{d} x)$$ and thus $|f|^p$ are measurable. For the LHS to be well-defined, the function $|f|^p$ should be measurable. In our case, $F$ is not assumed to be non-negative. Could you elaborate on how to prove it?
Best Answer
You are correct that they left out some hypotheses. One hypothesis that works is $F \geq 0$. Another hypothesis that works is that the right hand side is finite. You can find a statement and proof of this result in "Real Analysis" by Folland.