Why does this hold for all continuous functions?

Question

For all $f$ continuous on $([0,1]^2,\mathcal{B}[0,1]^2,\lambda)$ then $f$ is Lebesgue integrable and $$\int_0^1\int_0^{x^2}f ~dy~dx=\int_0^1\int_{\sqrt{y}}^1f~dx~dy$$

Answer 1

Hint:

Fubini's theorem is applicable if the integrand is absolutely integrable on $[0,1]^2$. It then follows that

$$\int_0^1\int_0^{x^2} f(x,y) \, dy \, dx= \int_0^1\int_0^1 f(x,y) \mathbf{1}_{\{y \leqslant x^2\}} \, dy \, dx \\= \int_0^1\int_0^1 f(x,y) \mathbf{1}_{\{x \geqslant \sqrt{y}\}} \, dx \, dy \\ = \int_0^1\int_{\sqrt{y}}^{1} f(x,y) \, dx \, dy$$

For integrability, note that $f$ is continuous and $|f(x,y) \mathbf{1}_{\{y \leqslant x^2\}} | \leqslant |f(x,y)|$.