To show that a function defined by integral is absolutely continuous

Question

Let $$ F(x)=\int_{[0,x]\times[0,x]}f,\quad x\in[0,1] $$ Here f is a Lesbegue-integrable on the unit square $[0,1]\times[0,1]$. I need to show that $F$ is absolutely continuous and express the derivative of $F$ in terms of $f$. I guess that the derivative of $F$ is $f(x,x)$. But, I can't get an idea to rigorously show that $F$ is absolutely continuous and calculate its derivative. Could anyone help me?

Answer 1

Nice problem. Sketch: Define $$(1)\,\,\,\,g(x) = \int_0^x f(s,x)ds + \int_0^x f(x,t)dt.$$ The definition makes sense by Fubini: for a.e. $x$ we have both integrands in $L^1([0,1]).$ Check that both functions of $x$ on the right of (1) are in $L^1([0,1]).$ Hence $g\in L^1([0,1]).$ Now you want to verify that $\int_0^y g(x)\,dx = F(y)$ for all $y\in [0,1].$