I am trying to solve this problem:
Let $g:[0,1] \to \mathbb R$ be a non negative integrable function over $[0,1]$. Prove that if there is $\alpha \in \mathbb R$ such that for all $n \in \mathbb N$, $$\int_0^1 g(x)^ndx=\alpha,$$ then $g=\mathcal X_E$ a.e. for some measurable subset $E \subset [0,1]$.
I am pretty lost with this problem. First of all note that since $g$ is non negative, then $\alpha \in \mathbb R_{\geq 0}$. I would like to show that $g(x)=1$ for all $x$ in some measurable set $E$ (or at least almost for all $x$ in $E$ and that $g(x)=0$ almost everywhere else.
We have the equality $$\lim_{n \to \infty} \int_0^1 g(x)^ndx=\alpha$$
I thought of writing $[0,1]=\{x:g(x) \in [0,1)\} \cup \{g(x) \geq 1\}$. If I call $S=\{x:g(x) \in [0,1)\}$, then $$\int_0^1 g(x)^ndx=\int_S g(x)^ndx+\int_{S^c} g(x)^ndx$$
Since $g(x)^n \leq g(x)$ for all $x \in S$, then here I can apply the dominated convergence theorem, so $\lim_n \int_S g(x)^ndx=0$. I got stuck here, I don't know what else to do, I would appreciate some help. Thanks in advance.
Using Markov's inequality you get that for any $\eta>1$,
$$\mu(g\geqslant \eta)\leqslant \frac{1}{\eta^p}\int g^p\stackrel{p\to\infty}\longrightarrow 0$$
So that $\mu(g>1)=0$. This means that $0\leqslant g\leqslant 1$ a.e. -- show this implies that $g=\chi_{\{g=1\}}$ almost everywhere, given the condition.
Markov's inequality is obtained as follows. Given $g$ integrable over a space $X$, for any $\alpha>0,p\geqslant 1$,
$$\mu(|g|\geqslant \eta)=\int_{|g|\geqslant \eta}1d\mu\\=\int_{|g|\geqslant \eta}\frac{\eta^p}{\eta^p}d\mu\\ \leqslant \frac{1}{\eta^p}\int_X |g|^p$$