Folland theorem 6.19: $p=\infty$

Question

Consider the following theorem from Folland's book "Real analysis: Modern techniques and their applications":

In the book, it can be found on p194 (second edition). I understand the proof of the theorem for part (a) completely. However if $p = \infty$, I don't understand why (b) is true.

In Folland's book, the following is written in the proof:

This is a simple application of the monoticity of the integral.

Can someone explain me why this is true? We have to prove that $$\left\|\int_Y f(-,y) d\nu(y)\right\|_\infty\le \int_Y \|f(-,y)\|_\infty d\nu(y)$$ so for example it suffices to show that $$\left|\int_Y f(x,y)d\nu(y)\right|\le \int_Y \|f(-,y)\|_\infty d\nu(y)$$ for $\mu$-almost every $x \in X$.

Intuitively, we would want to do the following: If $x$ is in some set $A$ with complement contained in a set of measure $0$, then $$\left|\int_Y f(x,y)d\nu(y)\right|\le \int_Y |f(x,y)|d\nu(y) \le \int_Y \|f(-,y)\|_\infty d\nu(y)$$ but to justify the last inequality we need to show that $$\int_Y |f(x,y)|d\nu(y) \le \int_Y \|f(-,y)\|_\infty d\nu(y)$$ for all $x \in A$. However, why is this the case? Am I missing something obvious here?

Answer 1

The case $p=\infty$ follows by a carefully application of Fubini-Tonelli's theorem.

• Without loss of generality assume $y\mapsto \|f(\cdot,y)\|_\infty$ is finite. Define the set $$B=\{(x,y)\in X\times Y: |f(x,y)|>\|f(\cdot,y)\|_\infty\}$$ This is a measurable set and, by Fubini's theorem, $$\mu\otimes\nu(B)=\int_Y\mu(\{x:|f(x,y)|>\|f(\cdot,y)\|_\infty\})\,\nu(dy)$$ Since $\mu(\{x:|f(x,y)|>\|f(\cdot,y)\|_\infty\})=0$ for all $y$, we conclude that $$|f(x,y)|\leq \|f(\cdot,y)\|_{L_\infty(X,\mu)}\qquad\mu\otimes\nu-\text{a.s.}$$ Hence, for any $A\in\mathcal{M}$, $\mathbb{1}_A(x)|f(x,y)|\leq\mathbb{1}_A(x)\|f(\cdot,y)\|_\infty$ $\,\,\mu\otimes\nu$-a.s.
• Another application of Fubini's theorem yields $$\begin{align}\int_A\int_Y|f(x,y)|\,\nu(dy)\,\mu(dx)\leq\mu(A)\int_Y\|f(\cdot,y)\|_{L_\infty(X,\mu)}\,\nu(dy)\tag{1}\quad\forall A\in\mathcal{M},\, \mu(A)<\infty\label{one}\end{align}$$
• If $\mu(A)<\infty$ define $A^b=\{x\in A: \int_Y|f(x,y)|\,\nu(dy)>\int_Y\|f(\cdot,y)\|_{L_\infty(X,\mu)}\,\nu(dy)\}$ ( $b$ stands for "bad"). If $\mu(A_b)>0$ then, by the positivity property of the Lebesgue integral ($\int_X|\phi|\,d\mu=0$ iff $|\phi|=0$ $\mu$-a.s.) we have that $$\int_{A^b}\int_Y|f(x,y)|\,\nu(dy)\,\nu(dx)>\mu(A_b)\int_Y\|f(\cdot,y)\|\,\nu(dy)$$ in contradiction to \eqref{one}; hence $\mu(A^b)=0$.
• To conclude, let $\{A_n:n\in\mathbb{N}\}$ be an $\mathcal{M}$-partition of $X$ with $0<\mu(A_n)<\infty$. Each $A^b_n$ has $\mu$-measure $0$; hence, for $x\in X\setminus\bigcup_nA^b_n$ one has $$\int_Y|f(x,y)|\,\nu(dy)\leq \int_Y\|f(\cdot,y)\|_{L_\infty(X,\mu)}\,\nu(dy)$$