Can we weaken the hypothesis of theorem 6.18? It states that there exists $C>0$ s.t. $\int\left|K(x,y)\right|d\mu(x)\leq C$ for a.e. $y\in Y$ and $\int\left|K(x,y)\right|d\upsilon(y)\leq C$ for a.e. $x\in X$, and that $1\leq p \leq\infty$. What if we drop the requirement that there exists $C>0$ s.t. $\int\left|K(x,y)\right|d\mu(x)\leq C$ for a.e. $y\in Y$ and $\int\left|K(x,y)\right|d\upsilon(y)\leq C$ for a.e. $x\in X$. And we only discuss $1<p<\infty$?
For example, we consider the case when $K(x,y)=\frac{1}{x+y}$, $1<p<\infty$, and probably $f\in L^p(0,\infty)$, can we still get the result? To prove that I think $||Tf||_p = \left\{\int \left|\int \frac{1}{x+y}f(y) d\upsilon(y)\right|^p d\mu(x)\right\}^{1/p} \leq \left\{\int \left(\int \left|\frac{1}{x+y}f(y)\right| d\upsilon(y)\right)^p d\mu(x)\right\}^{1/p}$.
Then we apply Holder's inequality?
I don't think this is correct for your $K=\frac{1}{x+y}$. Just assume $K\cdot f$ would be integrable for every $f\in L^p$. Let's put $\nu(A)=\mu(A)=\mathcal{L}^1(A\cap[-1,1])$ and $f=1$. Then $\|f\|_p<\infty$. But by the integrability we would have $$\int_\mathbb{R}\frac{1}{x+y}f(y)\, dy=\int_{-1}^{-x}\frac{1}{x+y}dy+\int_{-x}^1\frac{1}{x+y}dy = [\ln(|y+x|)]_{-1}^{-x}+[\ln(|y+x|)]_{-x}^{1}=-\infty$$ for every $x\in[-1,1]$. This seems to be a contradiction.