$\int_0^1[ \int_0^x$ $ f(x,y)\,dy]dx$ = $\int_0^1[ \int_y^1 f(x,y)\,dx]dy$?

Question

Let $f$ be integrable on $[0, 1] × [0, 1]$.

Show that

$\int_0^1[ \int_0^x f(x,y)\,dy]dx =\int_0^1[ \int_y^1$ $ f(x,y)\,dx]dy$

This is my first problem on the double Lebesgue integral so if someone could explain a bit about what's going on that would be much appreciated. Thanks

Answer 1

Big Hint:

For $x,y \in [0,1]$ you have $$ 1_{[0,x)}(y) = \begin{cases} 1 & 0 \leq y < x \leq 1 \\ 0 & \text{otherwise} \end{cases} $$ and $$ 1_{(y,1]}(x) = \begin{cases} 1 & 0 \leq y < x \leq 1 \\ 0 & \text{otherwise} \end{cases} $$ Hence, for $x,y \in [0,1]$, you find $1_{[0,x)}(y) = 1_{(y,1]}(x)$. Now, notice that $$ \int_0^1 \int_0^x f(x,y) \,dy\,dx = \int_0^1 \int_0^1 1_{[0,x)}(y)\,f(x,y)\,dy\,dx = \int_0^1 \int_0^1 1_{(y,1]}(x)\,f(x,y)\,dy\,dx $$

Use Fubini-Tonelli.