Here if $f$ is a measurable function on [0,1], define $G(x,y)$ $$G(x,y)=f(x)-f(y),$$ If we know $ G(x,y)\in L^1([0,1]\times[0,1])$, then $f$ is integrable on $[0,1]$ or not? If not, can we give a counterexample?
First we know $ G(x,y)\in L^1([0,1]\times[0,1])$, so by Fubini Theorem,$$\int_0^1\int_0^1G(x,y)dxdy=\int_0^1\int_0^1f(x)-f(y)dxdy=\int_0^1f(x)dx-\int_0^1f(y)dy=0.$$ $\int_0^1\int_0^1|G(x,y)|dxdy=2\int_0^1\int_0^y |f(x)-f(y)|dxdy$. But I don't know how to use the above equation to prove $|f|\in L^1([0,1])$. I guess we can find a counterexample?
One part of Fubini's theorem is that if $H \in L^1([0,1]\times[0,1])$, then for almost all $y\in [0,1]$ the function $h_y \colon x\mapsto H(x,y)$ belongs to $L^1([0,1])$.
So we know that $x\mapsto f(x) - f(y)$ is in $L^1([0,1])$ for almost all $y$.
But if $h\in L^1([0,1])$, then $h+c\in L^1([0,1])$ for all $c\in\mathbb{R}$ (or $\mathbb{C}$), and since $f(y)$ is finite when $x\mapsto f(x)-f(y)\in L^1([0,1])$, it follows that indeed $f\in L^1([0,1])$.
So we have the result $G\in L^1([0,1]\times [0,1]) \iff f\in L^1([0,1])$.
We obtain the identity
$$\int_0^1\int_0^1 \lvert G(x,y)\rvert\,dx\,dy = 2\int_0^1\int_0^y \lvert f(x)-f(y)\rvert\,dx\,dy$$
by splitting the integral into two parts, changing the order of integration in one part, and renaming, due to the symmetry of $G$:
$$\begin{align} \int_0^1\int_0^1 \lvert G(x,y)\rvert\,dx\,dy &= \int_0^1 \int_0^y \lvert f(x)-f(y)\rvert\,dx\,dy + \int_0^1\int_y^1 \lvert f(x) - f(y)\rvert\,dx\,dy\\ &= \int_0^1 \int_0^y \lvert f(x)-f(y)\rvert\,dx\,dy + \int_0^1\int_0^x \lvert f(x)-f(y)\rvert\,dy\,dx\\ &= \int_0^1 \int_0^y \lvert f(x)-f(y)\rvert\,dx\,dy + \int_0^1\int_0^x \lvert f(y)-f(x)\rvert\,dy\,dx\\ &= \int_0^1 \int_0^y \lvert f(x)-f(y)\rvert\,dx\,dy + \int_0^1\int_0^y \lvert f(x)-f(y)\rvert\,dx\,dy. \end{align}$$