# Integral of a nonnegative Lebesgue-measurable function on $[0,1]$.

Let $f$ be a nonnegative Lebesgue-measurable function on $[0,1]$. Suppose that $f$ is bounded above by $1$ and that $\displaystyle \int_{[0,1]} f = 1$.

Problem. Show that $f(x) = 1$ almost everywhere on $[0,1]$.

I don’t know how to start. Would it help to represent $f$ as the pointwise limit of a sequence $(f_{n})_{n \in \Bbb{N}}$ of simple functions on $[0,1]$?

Let $X=[0,1]$. Consider $1-f$. If $f\ne 1$ a.e. then $1-f\ne 0$ a.e. and indeed there is a positive measure set, $E$ so that $1-f>0$ on $E$. But then
$$0=1-1=\int_X 1-\int_X f=\int_X(1-f)=\int_E (1-f)>0.$$