Chi function with a set of finite measure

Question

Let $E$ be a set of finite measure and $f:E \rightarrow \mathbb{R}$ be integrable. Suppose further that there exists $c$ real number such that $$\int_Ef^n=c\ \ \ \forall\ \ n=2,3,4,...$$ The question is show that there exists a measurable set $A \subseteq E$ such that $f=\chi_{_{A}}.$

$\underline{My\ \ attempt}:$

I honestly have no idea to start here, the only thing I see might help is $$\int_Ef^n=c\ \ \ \forall\ \ n=2,3,4,...$$ but even though I do not know what is $f$ exactly. So I will appreciate any hint or help for that.

Thank you.

Answer 1

This is simpler than you expect. Note that $$0 \le \int (f^2 - f)^2 = \int f^4 - \int 2f^3 + \int f^2 = c -2c + c = 0$$ so that $f^2 - f = f(f-1) = 0$ almost everywhere. Thus $f = 0$ or $f=1$ almost everywhere.

Answer 2

Let $g =f^{2}$. Then $\int g^{n} =c$ for all $n$ and so $c \geq \int_C g^{n}$ where $C=\{|f| >1\}$. By Monotone Convergence Theorem we get $m(\{|f| >1\})=0$. Now $\int g^{n} (1-f) =\int f^{2n}-\int f^{2n+1}=0$ and the integrand is non -negative. Hence $g^{n} ((1-f)=0$ almost everywhere which gives $f=0$ or $1$ almost everywhere. Hence $f=I_A$ where $A=\{x: f(x)=1\}$.