an amazing inequality of $L^p$ spaces, how to prove it?

Question

Let $\Omega_1,\Sigma_1,\mu_1$ and $\Omega_2,\Sigma_2,\mu_2$ be $\sigma$-finite complete measure space and $f:\Omega_1\times\Omega_2\longrightarrow [0,+\infty)$ be $\mu_1\otimes\mu_2$ measurable. Then

\begin{eqnarray*} (\int(\int f d\mu_1)^pd\mu_2)^{1/p}\leq \int(\int f^p d\mu_2)^{1/p}d\mu_1 \end{eqnarray*}

How to prove it? For the proof in (http://calclab.math.tamu.edu/~sivan/math663_08a/appendix2.pdf), the author divided an integration on both sides of the inequality. How to deal with the case that the integration he divided is infinite?

Answer 1

Writing $\Omega_i$ as a countable disjoint union of sets of finite measure, we are reduced to the case $\Omega_i$ of finite measure.

Then we can use a truncation argument: take $f_n:=f\chi_{1/n\leqslant f\leqslant n}$, so that the integral we divided by is positive if $n$ is large enough. This proves the identity for $f_n$. Then we use a monotone convergence argument.