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?

Solution 1:

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

$$\int_0^1\left(\int_0^x f(x,y)\, dy\right) \, dx \\= \int_0^1\left(\int_0^1 f(x,y)\mathbb{1}_S(x,y) \,dy\right) \, dx \underbrace{=}_{\text{Fubini}}\int_0^1\left(\int_0^1 f(x,y)\mathbb{1}_S(x,y) \,dx\right) \, dy\\ = \int_0^1\left(\int_y^1 f(x,y)\,dx\right) \, dy$$