I'm reading Theorem 6.19 in textbook Real Analysis: Modern Techniques and Their Applications by Gerald B. Folland.
Suppose that $(X, \mathcal{M}, \mu)$ and $(Y, \mathcal{N}, \nu)$ are $\sigma$-finite measure spaces, and let $f$ be an $(\mathcal{M} \otimes \mathcal{N})$-measurable function on $X \times Y$.
a. If $f \geq 0$ and $1 \leq p<\infty$, then
$$
\left[\int\left(\int f(x, y) d \nu(y)\right)^{p} d \mu(x)\right]^{1 / p} \leq \int\left[\int f(x, y)^{p} d \mu(x)\right]^{1 / p} d \nu(y)
$$
b. If $1 \leq p \leq \infty, f(\cdot, y) \in L^{p}(\mu)$ for a.e. $y$, and the function $y \mapsto\|f(\cdot, y)\|_{p}$ is in $L^{1}(\nu)$, then $f(x, \cdot) \in L^{1}(\nu)$ for a.e. $x$, the function $x \mapsto \int f(x, y) d \nu(y)$ is in $L^{p}(\mu)$, and
$$
\left\|\int f(\cdot, y) d \nu(y)\right\|_{p} \leq \int\|f(\cdot, y)\|_{p} d \nu(y).
$$
In the proof of (b), he said that
Assertion (a) therefore follows from Theorem 6.14. When $p<\infty$, (b) follows from (a) (with $f$ replaced by $|f|$ ) and Fubini's theorem; when $p=\infty$, it is a simple consequence of the monotonicity of the integral.
Could you please explain?
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Why do we need to use Fubini's theorem? I could not see the need of swapping differential operators here.
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How is the monotonicity of the integral used to obtain the result for $p = \infty$ here?
Best Answer
If $p < \infty$, the suggestion for proving (b) is to replace $f$ by $|f|$ and apply (a) together with Fubini's theorem.
We aren't going to use Fubini to switch the order of integration—we want to use the part of Fubini's theorem which says that if you know $\int(\int|g(x,y)|\,d\nu)\,d\mu < \infty$, then the inner integral $\int|g(x,y)|\,d\nu<\infty$ $\mu$-a.e. By (a), $$(\int(\int |f(x,y)|\,d\nu)^p\,d\mu)^{\frac1p}\le \int(\int|f(x,y)|^p\,d\nu)^{\frac1p}\,d\mu.$$ By assumption, this last expression is finite, so use Fubini to conclude $f(x,\cdot)\in L^1(\nu)$ for a.e. $x$.
As for when $p = \infty$, the suggestion is to use monotonicity of the integral, which says that if $0\le F\le G$, then $\int F\,d\mu\le \int G\,d\mu$. Consider the inequality $|f(x,y)|\le \|f(\cdot,y)\|_{L^\infty(d\mu)}$.