Absolute continuity of $F=\int_{[0,x]\times [0,x]} f$. For $f\in L^1([0,1]\times [0,1])$ with the Lebesgue measure, define $F(x)=\int_{[0,x]\times[0,x]}f$. How can I prove $F$ is absolutely continuous, and express $F'$ with respect to $f$? I can't figure out.

Put
$$\Phi(x,y) = \int_0^x \int_0^y f(s,t)\,dt\,ds.$$
Then
$$\partial_x\Phi(x,y) = \int_0^y f(x,t)\,dt$$
and
$$\partial_y\Phi(x,y) = \int_0^x f(t,x)\,dt$$
Invoke the chain rule to get
$$F'(x) = \partial_x\Phi(x,x) + \partial_y\Phi(x,x) = \int_0^x f(x,t) + f(t,x)\,dt$$
