showing that $f=\chi_A$ for some measurable set $A$! hi need some hints with this question:
If $f^n$ is integrable for each $n$ and $\int f^n dµ = c$ for some constant c then
show that $f(x) = \chi_A(x)$ for some measurable set $A ⊂ X$.
I know that we need to start with showing that $f(x)=1$ for $x \in A$ and $f(x)=0$ otherwise. but when I set $x \in A$ do we need to define the set $A$? or how can I attack this problem?
any hints are very much appreciated. thank you.   
 A: Let us assume that $f \geq 0$.
It suffices to show that $N_1 := f^{-1}((1,\infty))$ and $N_2 := f^{-1}((0,1))$ are null sets.
Let us first show that for $N_1$. If this is no null set, then for some $m \in \mathbb{N}$ the set $N := f^{-1}((1 + 1/m, \infty))$ has positive measure. This implies
$$c = \int f^n d\mu \geq \int_N f^n d\mu \geq \int_N (1+1/m)^n d\mu = \mu(N) \cdot (1+1/m)^n \rightarrow \infty  \text{ for } n\rightarrow \infty,$$
a contradiction.
This shows $0 \leq f \leq 1$ almost everywhere and thus $0 \leq f^n \leq f \in L^1$ for all $n \in \mathbb{N}$. Note that because of $0 \leq f \leq 1$ we get $f^n \rightarrow \chi_A$ pointwise, where $A := f^{-1}(\{1\})$. Using the monotone convergence theorem, we conclude
$$c = \int f^n d\mu \rightarrow \int \chi_A d\mu.$$
If $N_2$ had positive measure, we would have $f > \chi_A$ on the set $N_2$ of positive measure and hence $c = \int f d\mu > \int \chi_A d\mu = c$, a contradiction.
This shows that $N_2$ also has measure zero and hence $f = \chi_A$ almost everywhere.
