If $f : [0,1] \to [0,\infty)$ is a Lebesgue measurable function such that $\int_0^1 f(x) \, dx < \infty$ and $$\int_0^1 f(x)^n \, dx =\int_0^1 f(x) \, dx$$ for all positive integers $n \ge 1$, how is $f$ equal to the characteristic function on almost all of some subset of $[0,1]$?
For me, this problem looks a lot like this question here. Can I adapt the proof of that question to this problem?
For all $x \in f^{-1}((1,\infty))$, you have $f(x)^n \nearrow +\infty$. Thus if the integral is to remain finite, you must have $\lvert f^{-1}((1,\infty)) \rvert = 0$. Next, for all $x \in f^{-1}((0,1))$, you have $f(x)\searrow 0$ and so by the dominated convergence theorem $$\int_{ f^{-1}((0,1))} f(x)^n dx \to 0.$$ Hence if $$\int_{ f^{-1}((0,1))} f(x)dx > 0$$ you will eventually have $$\int^1_0 f(x) dx > \int^1_0 f(x)^ndx$$ which means $$\int_{ f^{-1}((0,1))} f(x)dx = 0$$ and so $\lvert f^{-1}((0,1))\rvert = 0$ and we can conclude that $f = 1_{f^{-1}(\{1\})}$ a.e.