If $f\ge0$ and $ \int_{0}^{1} f(x)^kdx$ does not depend on $k\geq 1$ then $f=\mathbf 1_A$ almost everywhere, for some measurable subset $A$ of $[0,1]$

Question

If $ \int_{0}^{1} (f(x))^k dx= C$ for all $k\geq 1$, then there exists a measurable subset A of [0,1] where $f(x)=1_{A}(x)$ for almost every x

Note that $ f:[0,1] -> R^{+}$

I am stuck on this problem. Here is what I have done so far.. We know $\int_{0}^{1} f(x) dx = \int_{0}^{1} f^2(x) dx $ which implies $\int (f-f^2) dx= 0$ which implies $f(x)-f^2(x)= 0$ almost everywhere on [0,1], which implies $f(x)(1-f(x))=0$ a.e. on [0,1] hence f(x)= 0 or 1 for a.e. x on [0,1].

Where do I go from here? Is this correct so far? How do I know that the set of points where $f(x)=1$,call it A, will be measurable necessarily? Also how do I use the other powers of f? I feel I am missing something.

Answer 1

You first have to show that $f$ a.e. takes values in $[0,1]$ only. To see this define for $n\geq 1$ the set $A_n=\{f\geq 1+\frac{1}{n}\}$. Then for any $k$ we have $C=\int_0^1 f^k dx \geq |A_n| (1+\frac{1}{n})^k$ which implies $|A_n|=0$. So the set $\{f>1\}=\cup_n A_n$ has zero measure. And you may proceed as before.