I am having trouble following a proof where Young's Inequality is being used to derive Hölder's Inequality. More precisely, there is a particular and final step that utilizes integration in order to establish the desired inequality.
I am using R. Schillers book: Measures, Integrals and Martingales. Firstly, it is defined that:
$$\lVert u\rVert_p:=\left(\int\lvert u(x)\rvert^p\ \mu(dx)\right)^{1/p}$$
and then Young's Inequality is proved for the use as a Lemma:
Lemma 13.1 (Young's inequality)
Let $p,q\in(1,\infty)$ be conjugate numbers i.e. $\frac{1}{p}+\frac{1}{q}=1$ hence $q=\frac{p}{p-1}$. Then: $$AB\leq\frac{A^p}{p}+\frac{B^q}{q}$$ holds for all $A,B\geq0$; equality occurs if, and only if, $B=A^{p-1}$.
I can follow the proof, and I understand this inequality (I hope). In the next paragraph Hölders inequality is stated:
Theorem 13.2 (Hölders inequality)
Assume that $f\in\mathcal{L}^p(\mu)$ and $g\in\mathcal{L}^q(\mu)$, where $p,q\in[1,\infty]$ are conjugate numbers. Then $fg\in\mathcal{L}^1(\mu)$, and the following inequality holds: $$\left\vert\int fg\ d\mu\right\vert\leq\int \vert fg\vert\ d\mu\leq\lVert f\rVert_p\cdot\lVert g\rVert_q$$
In the proof, we see that the first inequality holds per the definitions on the integral. Then for the second inequality we let: $$A:=\frac{\lvert f(x)\rvert}{\lVert f\rVert_p}\quad \textrm{and} \quad B:=\frac{\lvert g(x)\rvert}{\lVert g\rVert_q}$$
so that we get: $$\frac{\lvert f(x)g(x)\rvert}{\lVert f\rVert_p\lVert g\rVert_q}\leq \frac{\lvert f(x)\rvert^p}{p\lVert f\rVert_p^p}+ \frac{\lvert g(x)\rvert^q}{q\lVert g\rVert_q^q} $$
So far so good. I am on board with this until the next step, where we integrate both sides of the inequality over $x$ and obtain: $$\frac{\int\lvert f(x)g(x)\rvert\ \mu(dx)}{\lVert f\rVert_p\lVert g\rVert_q}\leq \frac{\lVert f\rVert_p^p}{p\lVert f\rVert_p^p}+ \frac{\lVert g\rVert_q^q}{q\lVert g\rVert_q^q}= \frac{1}{p}+\frac{1}{q}=1 $$
Multiplying by $\lVert f\rVert_p\lVert g\rVert_q$ yields Hölders inequality.
My question is then, why are only the numerators integrated? As $\lVert u\rVert_p^p={\left(\left(\int\lvert u(x)\rvert^p\right)^{1/p}\right)}^p=\int\lvert u(x)\rvert^p$, it is the only way I can explain to myself, what happened during the integration. But as we have the norm in the denominators, should they not also be included in the integration when they too depend on $x$?
Or is it something else that happens?