Does $z$ depend on $x$ in the proof of Theorem 6.20 in Folland? The theorem states:

Let $K$ be a Lebesgue measurable function on $(0, \infty) \times (0, \infty)$ such that $K(\lambda x, \lambda y) = \lambda^{-1}K(x,y)$ for all $\lambda > 0$ and $\int_0^\infty |K(x,1)|x^{-1/p}\,dx = C < \infty$ for some $p \in [1, \infty]$, and let $q$ be the conjugate exponent to $p$. For $f \in L^p$ and $g \in L^q$, let
$$Tf(y) = \int_0^\infty K(x,y)f(x)\,dx, \quad Sg(x) = \int_0^\infty K(x,y)g(y)\,dy.$$
Then $Tf$ and $Sg$ are defind a.e., and $\|Tf\|_p \leq C\|f\|_p$ and $\|Sg\|_q \leq C\|g\|_q$.

The proof proceeds as follows:

Setting $z = x/y$, we have
$$\int_0^\infty |K(x,y) f(x)|\,dx = \int_0^\infty |K(yz,y) f(yz)|y\,dz = \int_0^\infty |K(z,1) f_z(y)|\,dz$$
where $f_z(u) = f(yz)$; moreover
$$\|f_z\|_p = \Bigg[\int_0^\infty |f(yz)|^p\,dy\Bigg]^{1/p} = \Bigg[\int_0^\infty |f(x)|^pz^{-1}\,dx\Bigg]^{1/p} = z^{-1/p}\|f\|_p.$$

The doubt I have is rather elementary. In the very last equality, how can we pull out $z$ if $z$ itself depends on $x$?
The proof then proceeds:

Therefore, by the Minkowski inequality for integrals $Tf$ exists a.e. and
$$\|Tf\|_p \leq \int_0^\infty |K(z,1)\|f_z\|_p\,dz = \|f\|_p \int_0^\infty |K(z,1)|z^{-1/p}\,dz = C\|f\|_p.$$

How does the Minkowski inequality for integrals imply the existence of $Tf$?
I understand the rest of the proof so I have omitted it.
 A: In
$$\|f_z\|_p = \Bigg[\int_0^\infty |f(yz)|^p\,dy\Bigg]^{1/p} = \Bigg[\int_0^\infty |f(x)|^pz^{-1}\,dx\Bigg]^{1/p} = z^{-1/p}\|f\|_p.$$
the change of variable $x=yz$ is performed. After that, $x$ is the variable of integration, and $z$ is a constant. Therefore, it call be pulled out of the integral.
The existence of $Tf$ is just a consequence of the bounding of the map $x \mapsto \lvert K(x,1) f(x) \rvert$ by a map which is integrable.
A: When estimating $\|f_z\|_p$, the $z$ is fixed, and $x$ is there used as a variable of integration unrelated to $z$.
As for the use of Minkowski's inequality, it's in the inequality below:
\begin{align}
\|Tf\|_p
&=\bigg[\int_0^\infty\bigg|\int_0^\infty |K(z,1) f_z(y)|\,dz\bigg|^p\,dy\bigg]^{1/p}\\[0.3cm]
&\leq
\int_0^\infty\bigg(\int_0^\infty|K (z,1)|^p\,|f_z(y)|^p\,dy \bigg)^{1/p}\,dz \\[0.3cm]
&=
\int_0^\infty |K (z,1)|\,\bigg(\int_0^\infty\,|f_z(y)|^p\,dy \bigg)^{1/p}\,dz \\[0.3cm]
&=
\int_0^\infty |K (z,1)|\,\|f_z\|_p\,dz\\[0.3cm]
&=
\int_0^\infty |K (z,1)|\,\|f\|_p\,z^{-1/p}\,dz. \\[0.3cm]
&=C\,\|f\|_p.
\end{align}
