probability - $X$ and $Y$ are independent and follow $U(0,1)$. Show $P(f(X) > Y) = int_0^1 f(x) dx$

Let $X$ and $Y$ be two independent uniformly distributed r.v. on $[0,1]$, and $f$ is a continuous function from $[0,1]$ to $[0,1]$. Show that $P(f(X) > Y) = \int_0^1 f(x) dx$.

I tried to prove it by change of variable but failed. I can only reach the step

$$P(f(X) > Y) = \int_{\{(x,y): f(x) > y\} }I_{[0,1]}(x) I_{[0,1]}(y) dx dy$$
How can I proceed with the proof? Thanks for any help in advance.

Answer

$$\int_{(x,y):f(x)>y} \mathbf{1}_{[0,1]}(x)\mathbf{1}_{[0,1]}(y)\,\mathrm dx\,\mathrm dy=\int_\mathbb{R}\int_\mathbb{R} \mathbf{1}_{[0,1]}(x)\mathbf{1}_{[0,f(x))}(y)\,\mathrm dy\,\mathrm dx$$