I'm currently reading through Folland's Real Analysis, and I am stuck on the proof of one of theorems which I will now include (with proof) for the sake of completeness.
(Theorem 6.18.) Let $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$ be $\sigma$-finite measure spaces, and let $K$ be an $(\mathcal{M}\otimes\mathcal{N})$-measurable function on $X\times Y$. Suppose that there exists $C>0$ such that $\int |K(x,y)|\,d\mu(x)\leq C$ for a.e. $y\in Y$ and $\int |K(x,y)|\,d\nu(y)\leq C$ for a.e. $x\in X$, and that $1\leq p\leq\infty$. If $f\in L^p(\nu)$, then the integral $$ Tf(x)=\int K(x,y)f(y)\,d\nu(y) $$ converges absolutely for a.e. $x\in X$, the function $Tf$ thus defined is in $L^p(\mu)$, and $\|Tf\|_p\leq C\|f\|_p$.
Proof. Suppose that $1<p<\infty$. Let $q$ be the conjugate exponent to $p$. By applying H$\ddot{\text{o}}$lder's inequality to the product $$ |K(x,y)f(y)|=|K(x,y)|^{1/q}\big(|K(x,y)|^{1/p}|f(y)|\big) $$ we have \begin{align*} \int |K(x,y)f(y)|\,d\nu(y) &\leq \left[\int|K(x,y)|\,d\nu(y)\right]^{1/q}\left[\int|K(x,y)||f(y)|^p\,d\nu(y)\right]^{1/p}\\ &\leq C^{1/q}\left[\int |K(x,y)||f(y)|^p\,d\nu(y)\right]^{1/p} \end{align*} for a.e. $x\in X$. Hence, by Tonelli's theorem, \begin{align*} \int\left[\int |K(x,y)f(y)|\,d\nu(y)\right]^{p}\,d\mu(x)&\leq C^{p/q}\iint |K(x,y)||f(y)|^p\,d\nu(y)\,d\mu(x)\\ &\leq C^{(p/q)+1}\int |f(y)|^p\,d\nu(y). \end{align*} Since the last integral is finite, Fubini's theorem implies that $K(x,\cdot)f\in L^1(\nu)$ for a.e. $x$, so that $Tf$ is well defined a.e., and $$ \int |Tf(x)|^p\,d\mu(x)\leq C^{(p/q)+1}\|f\|_p^p. $$ Taking $p$th roots, we are done.
For $p=1$ the proof is similar but easier and requires only the hypothesis $\int |K(x,y)|\,d\mu(x)\leq C;$ for $p=\infty$ the proof is trivial and requires only the hypothesis $\int |K(x,y)|\,d\nu(y)\leq C$. Details are left to the reader (Exercise 26).
My issue with the proof above is presumably a minor detail; I don't understand why the emboldened statement is true. I see how Fubini's theorem would apply if we could show $K(x,y)f(y)\in L^1(\mu\otimes\nu)$, but that's not is shown in the proof. Furthermore it seems that showing $$ \int\left[\int |K(x,y)f(y)|\,d\nu(y)\right]^{p}\,d\mu(x)<\infty\hspace{.4 in}\text{implies}\hspace{.4 in}\iint |K(x,y)f(y)|\,d\nu(y)\,d\mu(x)<\infty $$ would be similar to showing that $L^p\subset L^1$, which definitely does not hold in general (only on finite measures).
What am I missing here? Why does Fubini's theorem apply here? I'm sure it's something trivial I'm overlooking. Any help is greatly appreciated.
Hopefully there are no copyright issues with posting the full proof here.
The author refers to two displayed formulas, not just the one immediately preceding the bolded statement. The logic is: