An identity for double integrals Let $f$ be a real valued continuous function on $\mathbb{R}_+^2$ and $F(x,y)=\int_0^x\int_0^y f(s,t)\,ds\,dt$. 
How to show that
$$\frac{1}{uv}\int_0^u\int_0^vF(x,y)\,dx\,dy=\int_0^u\int_0^v\left(1-\frac{x}{u}\right)\left(1-\frac{y}{v}\right)f(x,y)\,dx\,dy$$
is satisfied.

Let's look at the one variable situation first: If
$$G(x):=\int_0^x g(s)\>ds$$
then
$$\int_0^u G(x)\>dx=\int_0^u\int_0^x g(s)\>ds\>dx=\int_0^u\int_s^u g(s)\>dx\>ds=\int_0^u(u-s)g(s)\>ds$$
(draw a sketch of the $(x,s)$-plane!), and therefore
$${1\over u}\int_0^uG(x)\>dx=\int_0^u\left(1-{s\over u}\right)g(s)\>ds\ .$$
Now do this with your $f(x,y)$  "first with respect to $x$, then with respect to $y$", or use the Stone-Weierstrass approximation theorem: Your $(x,y)\mapsto f(x,y)$ can be approximated by polynomials, hence by linear combinations of functions $\psi(x,y):=g(x)h(y)$. For such functions $\psi$  we of course have
$$\Psi(x,y)=\int_0^x g(x)\>dx \ \int_0^y h(y)\>dy\ ,$$
etcetera.
