Prove $\int_{[0,1]}\int_{[0,x]}fdydx=\int_{[0,1]}\int_{[y,1]}fdxdy$

Question

$\int_{[0,1]}\int_{[0,x]}fdydx=\int_{[0,1]}\int_{[y,1]}fdxdy$ for $f$ Lebesgue integrierbar on $\mathbb{R^2}$

It's given a Theorem without proof that states:

Let $a,b\in\mathbb{R},a<b, f,g\in L^1([a,b],\mathbb{R}).$ Then there exist $F(x):=\int^x_af, G(x):=\int^x_ag$ for every $x\in[a,b],$ and moreover $\int^b_aF·g=-\int^b_afG+F(b)G(b)-F(a)G(a)$

I tried using this theorem but it got just more confusing, I don't know if this theorem can be useful, but it seems the most similar to this problem. Is there any other way? With Fubini I don't think I can do something.

Answer 1

Hint: Consider $f(x,y)\cdot \chi_{ \{(x,y)\in[0,1]^2: \;x\geq y \}}(x,y)$ .