Theorem 6.19. In short, we want to show
$$\left\|\int |f(\cdot, y)|\,d\nu(y)\right\|_p \leq \int \|f(\cdot, y)\|_p \, d\nu(y)$$
for m'ble $f$.
Could we prove this using just the duality of the norm? That is, there exists some $g \in L^q(\mu)$ such that
$$ \begin{align} \left\|\int |f(\cdot, y)|\,d\nu(y)\right\|_p &= \sup_{\|g\|_q = 1} \iint |f(x,y)||g(x)| \, d\mu(x) \, d\nu(y) \\ &= \sup_{\|g\|_q = 1}\iint |f(x,y)||g(x)| \, d\nu(y) \, d\mu(x) \tag{Fubini} \\ &\leq \int \left(\sup_{\|g\|_q = 1} \int |f(x,y)| |g(x)| \, d\nu(y) \right) \, d\mu(x) \tag{$\dagger$}\\ &= \int \|f(\cdot, y)\|_p \, d\nu(y). \end{align} $$
where for $(\dagger)$, I used the fact that for m'ble $h$ we have, $\sup\int h \leq \int \sup h$.
Am I overlooking anything? It seems this is nicer than the approach used in the book.