Invert Integral I need to invert I need to reverse the integration order $$\int_{0}^{1}[\int_{0}^{x}f(x,y) dy]dx$$
I know how to reverse the order through the graph, which will look like this:
$$\int_{0}^{1}[\int_{y}^{1}f(x,y) dx]dy$$
but how can I formalize it step by step through writing?
 A: This type of "inversion" can be carried out without resorting to a graph by a direct application of Fubini's theorem after an indicator function is introduced. Let $S = \{(x,y)\subset [0,1]^2:x \geqslant y\}$ and define the indicator function
$$\mathbb{1}_S(x,y) = \begin{cases}1,&x\geqslant y\\ 0, &x < y \end{cases}$$
Note that even if we are working with Riemann integrals here, the product function  $(x,y) \mapsto f(x,y)\mathbb{1}_S(x,y)$ is integrable since a discontinuity is introduced only on a set of measure zero where $x = y$.
The inner integral becomes
$$\int_0^x f(x,y) dy = \int_0^1f(x,y)\mathbb{1}_S(x,y)  \, dy $$
Now we can apply Fubini's theorem to the iterated inetgral over the fixed region $[0,1]\times [0,1]$ to obtain
$$\int_0^1\left(\int_0^x f(x,y)\, dy\right) \, dx \\=  \int_0^1\left(\int_0^1 f(x,y)\mathbb{1}_S(x,y) \,dy\right) \, dx \underbrace{=}_{\text{Fubini}}\int_0^1\left(\int_0^1 f(x,y)\mathbb{1}_S(x,y) \,dx\right) \, dy\\ =  \int_0^1\left(\int_y^1 f(x,y)\,dx\right) \, dy$$
A: Let $T=\{(x,y)\in\Bbb R^2|0\leqslant x\leqslant1\wedge0\leqslant y\leqslant x\}$. Then$$\int_0^1\int_0^xf(x,y)\,\mathrm dy\,\mathrm dx=\iint_Tf(x,y)\,\mathrm dx\,\mathrm dy.$$But you have $T=\{(x,y)\in\Bbb R^2\mid0\leqslant y\leqslant1\wedge y\leqslant x\leqslant1\}$, and therefore$$\iint_Tf(x,y)\,\mathrm dx\,\mathrm dy=\int_0^1\int_y^1f(x,y)\,\,\mathrm dx\,\mathrm dy.$$
