Problem understanding a proof (Folland's Real analysis, theorem 6.18) Okay, so as stated here: Issue with proof in Folland&#39;s &#39;Real Analysis&#39; (Theorem 6.18), the text says: 


  (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 $C&gt;0$ 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 $1&lt;p&lt;\infty$. 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) &amp;\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}\\
&amp;\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)&amp;\leq C^{p/q}\iint |K(x,y)||f(y)|^p\,d\nu(y)\,d\mu(x)\\
&amp;\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.


I'm fine with the whole proof except one minor part: Why is $Tf$ a $\mathcal{M}$-measurable function? Could we prove that $K(x,y)f(y)\in L^1(\mu\otimes\nu)$, by Fubini's theorem we would have the desired measurability, but I  don't think this is necessarily true.

(I also don't quite understand where Fubini comes in, as stated in the text. While working out the details, I only used Tonelli's theorem.)
Because the integral is just the limit of serious of sums, so that it must be measurable due to that the limit of a serious of measurable function is measurable.

For Fubini's Theorem. it states that integrable on product space imply a.e.  integrable (as well as measurable) on every original space. So in your problem, $K$ is integrable on product space and so is $f$ because $f$ is a measurable function in one variable so that their product $K(x,y)f(y)$ is integrable on product space. So you could compute its integration. And then, you find it's finite, so by Fubini's Theorem, $\int K(x,y)f(y)dy$ is a.e. defined and integrable.

Is this clear enough? The spirit is to deduce integrability on one variable from integrability on product space.
Because the integral is just the limit of serious of sums, so that it must be measurable due to that the limit of a serious of measurable function is measurable.

For Fubini's Theorem. it states that integrable on product space imply a.e.  integrable (as well as measurable) on every original space. So in your problem, $K$ is integrable on product space and so is $f$ because $f$ is a measurable function in one variable so that their product $K(x,y)f(y)$ is integrable on product space. So you could compute its integration. And then, you find it's finite, so by Fubini's Theorem, $\int K(x,y)f(y)dy$ is a.e. defined and integrable.

Is this clear enough? The spirit is to deduce integrability on one variable from integrability on product space.