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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.

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    $\begingroup$ Hint: Call $S'=\{ x: g(x)>1\}$, then $g^n(x)\to \infty$ for every $x\in S'$. Apply Fatou's lemma to conclude that $S'$ must have measure $0$. $\endgroup$
    – Jose27
    May 13 '15 at 3:59
  • $\begingroup$ If $m(S')>0$, then $\infty=\int_S' \lim g(x)^ndx \leq \lim \inf \int_S' g(x)^ndx \leq \lim \int_{(0,1)} g(x)^ndx=\alpha$. It follows $m(S')=0$. So $g(x) \in [0,1]$ a.e.. I don't know how to prove from there the statement of the problem, could you help me with that? $\endgroup$
    – user16924
    May 13 '15 at 13:38
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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 $\alpha>0,p\geqslant 1$,

$$\mu(|g|\geqslant \eta)=\int_{|g|\geqslant \eta}1d\mu\\=\int_{|g|\geqslant \eta}\frac{\eta^p}{\eta^p}d\mu\\ \leqslant \frac{1}{\eta^p}\int_X |g|^p$$

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  • $\begingroup$ Thanks for the answer but I have not read Markov's inequality from the book where I am studying, I'll reread your answer when I get to that point. $\endgroup$
    – user16924
    May 13 '15 at 4:36
  • $\begingroup$ Thanks for the explanation of the inequality. I see why $0 \leq g \leq 1$ a.e., but why the set $S=\{x : 0\leq g(x)<1\}$ has measure zero? I can't see that. $\endgroup$
    – user16924
    May 13 '15 at 5:22
  • $\begingroup$ I mean, from what you've done it is clear that $\lim_n g(x)^n=\mathcal X_E$ a.e. with $E=\{x: g(x)=1\}$, and by the given condition, we have $\alpha=\lim_n \int_{[0,1]}g(x)^ndx=\int_{[0,1]} \lim_n g(x)^ndx=\int_{[0,1]} \mathcal X_E(x)dx=m(E)$. I don't see how to conclude from here that $g(x)=\mathcal X_E$ a.e.. Could you help me with that? $\endgroup$
    – user16924
    May 13 '15 at 15:22
  • $\begingroup$ Does the equality $m(E)=\int_{[0,1]}g(x)dx$ imply $g(x)=\mathcal X_E$ a.e.? If this is the case, I don't see why this is true although I do see why the reverse implication holds. $\endgroup$
    – user16924
    May 13 '15 at 15:27

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