Folland theorem 6.19 proof doubt Consider the following fragment from Folland's real analysis book:

They mention theorem 6.14, which is the following theorem:

However, to apply theorem 6.14, shouldn't we check that if $h(x)= \int_Y f(x,y)d\nu(y)$, then $\int_A |h| d\mu<\infty $ for every $A \in \mathcal{M}$ with $\mu(A) < \infty$? My question is, in other words, how is theorem 6.14 applied? Note that we may assume that the right hand side of the inequality we want to prove is finite, and this ensures that the supremum in theorem 6.14 is finite. Thanks in advance for any help!
 A: You don't have to check this condition since you use theorem 6.14 with $f=g$ from 6.19, $g(x)=\int f(x,y) d\nu(y)$ from 6.19 and $q=p$ from 6.19.
A: Define $$h(x) = \int_{Y}f(x,y)d\nu(y).$$ We claim that we can assume $h \in L^p(\mu).$ Indeed, otherwise we can truncate the function $f$ so that $h$ is in $L^p$ and then use a monotone convergence type argument. Then you can use that the map $$ L : L^p(\mu) \to L^q(\mu)^*$$ is an isometry (basically the content of 6.14) to conclude $$ \Vert h \Vert_{L^p(\mu)} = \Vert L(h) \Vert_{Op} \leq \int_{Y}\bigg(\int_{X}f(x,y)^p d\mu(x)\bigg)^\frac{1}{p} d\nu(y).$$ Some more details on the trucation: define $f_{k} = \min(f,k)$ and pick using sigma-finiteness sets of finite measure $X_{n}$ increasing to $X$, $Y_{n}$ increasing to $Y$, and then we get by the first part: $$\bigg(\int_{X_{n}}\bigg(\int_{ Y_{m} } f_{k}(x,y)d\nu\bigg)^p d\mu \bigg)^\frac{1}{p} \leq \int_{Y_{m}}\bigg(\int_{X_{n}}f_{k}(x,y)^p d\mu(x)\bigg)^\frac{1}{p} d\nu(y).$$ Use monotone convergence to take $k\to \infty$, then $n \to \infty$, then $ m \to \infty.$
