representation of integral of the indicator function

Question

I'm trying to see whether $$ f(y) =\int_0^1 I\{ f(u)\leq y \}\,du $$ where $f$ is measurable. and $I$ is the indicator function. It should be true but I can't see why.

Any idea? Thank you!

Answer 1

This is not true. Take for $f$ the constant function equal to $1$.

Then $\{ u \in [0,1] ; f(u)\leq 1/2 \} = \emptyset$ and $I\{ f(u)\leq 1/2 \} = 0$ for $u \in [0,1]$. However $$1 = f(1/2) \neq \int_0^1 I\{ f(u)\leq y \}\,du = 0$$