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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?

  • Why do we need to use Fubini's theorem? I could not see the need of swapping differential operators here.

  • How is the monotonicity of the integral used to obtain the result for $p = \infty$ here?

Akira
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    How do you show that $f(\cdot,y)\in L^p$ for almost every $y$ and $f(x,\cdot)\in L^1$ for almost every $x$??? – David C. Ullrich Oct 31 '21 at 15:22
  • Maybe you can start with Hölder's inequality and then you use the theory of conjugate exponent ! I mean in $(a)$ for $p=1$ it is clearly the Fubini-Tonelli theorem. Then if you have $\infty > p>1$ you can think about Hölder's inequality with conjugate exponent ! $(b)$ is the same but with Fubini-Lebesgue as $f$ has not constant sign ! – Maman Oct 31 '21 at 15:37

1 Answers1

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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 $|f(x,y)|\le \|f(\cdot,y)\|_{L^\infty(d\mu)}$.

Alex Ortiz
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  • Hi Alex, I have moved my answer to here. I hope you help me clarify my mistake there. – Akira Oct 31 '21 at 17:14
  • @Alex Ortiz: Thanks for your answer. Could you elaborate on the $p = \infty$ case? And seeing as $|f(\cdot,y) |{L^{\infty}(\mu)}$ is a function on $Y$ (right?!), shouldn't we be integrating $|f(x,y)|$ and $|f(\cdot,y) |{L^{\infty}(\mu)}$ with respect to $\nu$? I'm confused though because I don't see how this would lead us to the desired result... – Leonidas Mar 03 '22 at 23:29
  • @Leonidas: Here is some more detail for the $p = \infty$ case. For every $x$, $|\int f(x,y),d\nu(y)|\le \int|f(x,y)|,d\nu(y)\le \int|f(\cdot,y)|{L^\infty(\mu)},d\nu(y)$. Therefore, $|\int f(\cdot,y),d\nu(y)|{L^\infty(\mu)} \le \int|f(\cdot,y)|_{L^\infty(\mu)},d\nu(y)$, which is the desired result. – Alex Ortiz Mar 04 '22 at 02:34
  • @Alex Ortiz: Thank you! I had forgotten that the $L^{\infty}$ norm is an infimum. Now it makes sense! – Leonidas Mar 06 '22 at 16:43