Nonlinear Fubini-Tonelli?

Let me state Minkowski's integral inequality:

Let $1\leq p <\infty$. Let $F$ be a measurable function on the product space $(X,\mu)\times (T,\nu)$, where $\mu,\nu$ are $\sigma$-finite. We have

$\left[\int_{T}\left(\int_{X}\left|F(x,t)\right|d\mu(x)\right)^pd\nu(t)\right]^{\frac{1}{p}}\leq\int_{X}\left[\int_{T}\left|F(x,t)\right|^pd\nu(t)\right]^{\frac{1}{p}}d\mu(x)$.

If I may say so, the following proof is quite elegant (in fact, the above result is Exercise 1.1.6., page 12 of Classical Fourier Analysis by Loukas Grafakos, and the proof below is my solution to this exercise):

Proof. Let $G:T\to [0,\infty]$ be defined by $G\left(t\right)=\int_{X} \left|F\left(x,t\right)\right|d\mu\left(x\right)$. Then Minkowski's integral inequality is nothing but a reformulation of the assertion that $\left\|G\right\|_{L^p\left(T\right)}\leq \int_{X} [\int_{T} \left|F\left(x,t\right)\right|^p d\nu\left(t\right)]^{\frac{1}{p}} d\mu\left(x\right)$. To prove this assertion, we first note the well-known fact that $\left\|G\right\|_{L^p\left(T\right)}=\sup_{g\in L^{q}\left(T\right)} \left\|Gg\right\|_{L^1\left(T\right)}$; i.e., the norm of $G$ as an operator on $L^q\left(T\right)$ equals the $L^p$-norm of $G$. Now compute that for $g\in L^q\left(T\right)$, H\"older's inequality gives

$\left\|Gg\right\|_{L^1(T)} = \int_T \left|\int_X \left|F\left(x,t\right)\right|g\left(t\right)d\mu\left(x\right)\right|d\nu\left(t\right)$ $= \int_X \int_T \left|F\left(x,t\right)\right|\left|g\left(t\right)\right| d\nu\left(t\right)d\mu\left(x\right)$ $\leq \int_X \left\|F\right\|_{L^p\left(T\right)} \left\|g\right\|_{L^q\left(T\right)} d\mu\left(x\right)$ $= \int_X \left(\int_T \left|F\left(x,t\right)\right|^p d\nu\left(t\right)\right)^{\frac{1}{p}} d\mu\left(x\right)$,

where the second equality follows from Fubini's theorem and where we have used the notation $\left\|F\right\|_{L^p\left(T\right)}=\left(\int_T \left|F\left(x,t\right)\right|^p d\nu\left(t\right)\right)^{\frac{1}{p}}$. This proves Minkowski's integral inequality. Q.E.D.

The idea of the proof is to remove the exponents (of course, in the case $p=1$, the result is Tonelli's theorem). This can be done using duality. I hope this helps!

Following up on Willie Wong's comment: $$ \begin{align} & \phantom{= {}} \int_0^1 \left( \int_0^x f(y)\,dy \right)^2\,dx = \int_0^1 \int_0^x \int_0^x f(y)f(z)\,dy\,dz \, dx = \int_0^1\int_0^1 \int_{\max\{y,z\}}^1 f(y)f(z)\, dx \, dy\, dz \\ & = \int_0^1\int_0^1 \left( f(y)f(z) \int_{\max\{y,z\}}^1 1\, dx \right) dy \, dz = \int_0^1\int_0^1 f(y)f(z) \left(1-{\max\{\,y,z\,\}}\right) \, dy \, dz. \end{align} $$

The integration with respect to $x$ now appears as $1-\max\{\,y,z\,\}$ on the inside; the square of the integral with respect to $y$ now appears as $\int_0^1\int_0^1 \cdots\,dy\,dz$ on the outside.

As for "dimensional" correctness, $dy$ and $dz$ are both in the same units as $dy$ in the original integral; $f(y)$ and $f(z)$ are both in the same units as $f(y)$ in the original integral; and $x$ has to be in the same units as $y$ in the original integral; so $x$ is in the same units as $1-\max\{\,y,z\,\}$. So the units are of the form $a^2 b^3$ on both sides of the "$=$".