How to prove that $\int_0^1 dx \int_0^x f(x,y) dy = \int_0^1 dy \int_y^1 f(x,y) dx$ for a continuous function $f$ the given interval

Question

In the book of Mathematical Analysis II by Zorich, at page 136 (Question 7), it is asked that

Let $f : E → R$ be a function that is continuous on the set $E = \{ (x,y) ∈ \mathbb{R}^2 | 0 ≤ x ≤ 1 ∧ 0 ≤ y ≤ x \}$. Prove that

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

I'm thinking how can I prove it for 2 days, but I have almost no idea how to even start, so I would appreciate any help or hint.

Answer 1

Hint: You can apply Zorich's theorem without worrying about Lebesgue integration.

Using the indicator function $\mathbf{1}_{\{0 \leqslant y \leqslant x \leqslant1\}}$, we have

$$\int_0^1 \left(\int_0^x f(x,y) \, dy \right)\, dx = \int_0^1 \left(\int_0^1 f(x,y)\mathbf{1}_{\{0 \leqslant y \leqslant x \leqslant1\}} \, dy \right)\, dx = \ldots$$

Show that the integrand on the right-hand side is Riemann integrable over $[0,1]^2$? Also show that for every fixed $x \in [0,1]$ we have existence of the single Riemann integral

$$\int_0^1 f(x,y)\mathbf{1}_{\{0 \leqslant y \leqslant x \leqslant1\}} \, dy $$