If $f\in L^{1}([0,1]\times[0,1])$ and $F(x)=\int_{[0,x]\times[0,x]} f$ for$x\in [0,1]$, prove $F$ is absolutely continuous from the definition.

Question

I've seen there was a similar question.

This question was only for $L^{1}[0,1]$. I want to extend it to $L^{1}([0,1]\times[0,1])$.

The question is:

If $f\in L^{1}([0,1]\times[0,1])$ and for $x\in [0,1]$, $$F(x)=\int_{[0,x]\times[0,x]} f.$$ Prove that $F$ is absolutely continuous and express the derivative of $F$ using $f$.

There must be difference of solution. For example, $$\int_{[0,b_k]\times[0,b_k]} f - \int_{[0,a_k]\times[0,a_k]} f \neq \int_{[a_k,b_k]\times[a_k,b_k]} f$$

But $$\int_{[0,b_k]} f - \int_{[0,a_k]} f = \int_{[a_k,b_k]} f$$

Could someone give me a hint to solve this problem?

Answer 1

\begin{align} F(x) &= \int_{\left[0, x\right]\times \left[0, x\right]} f(t, s) \mathrm dt \mathrm ds\\ &= \int_0^x \int_0^t f(t, s) \mathrm ds \mathrm dt + \int_0^x \int_t^x f(t, s) \mathrm ds \mathrm dt\\ &= \int_0^x \int_0^t f(t, s) \mathrm ds \mathrm dt + \int_0^x \int_0^s f(t, s) \mathrm dt \mathrm ds\\ &= \int_0^x \left(\int_0^t \left(f(t, s) + f(s, t)\right)\mathrm ds\right)\mathrm dt \end{align}

So $F$ is absolute continuous and $$F'(x) = \int_{0}^x \left(f(x, t) + f(t, x)\right)\mathrm d t$$