# If $\int f^n=\int f$then $f=\chi_E$

This is a problem in Real Analysis by Bruckner and Thomson:

Let $f$ be a nonnegative lebesque integrable in the interval $[0,1]$, and suppose that for every integer $n=1,2,3,...$ $$\int_0^1[f(x)]^ndx=\int_0^1f(x)dx$$ show that $f$ must be $a.e.$ equal to the characteristic function $\chi_E$ of some measurable set $E\subset [0,1]$.

And there is a Hint: apply Fatou's lemma.

My try: We can pick a sequence of simple functions$\phi_k$ such that $\phi_k\to f$ uniformly on $[0,1]$. since we can assume that $f$ is bounded(since it is integrable) so $\phi_k^n\to f^n$ uniformly on $[0,1]$ Then by Fatou's lemma(or uniform convergence theorem) $\lim_{k\to\infty}\int_{[0,1]}\phi_k^n=\int_{[0,1]}f^n$. we can show that for all $n\in\mathbb N$ we have $\lim_{k\to\infty}\int_{[0,1]}|\phi_k-f|^n=0$.

However I am not sure we can conclude from last equation the result.

• Apply Fatou to the pointwise limit of the sequence $(f^n)$. From this, you can conclude that the measure of the set of points $x$ with $f(x)>1$ is zero. A different argument will then show the set of points $x$ with $f(x)<1$ is zero as well. – David Mitra Nov 28 '14 at 17:39
• @DavidMitra: could you please elaborate? – mac Nov 28 '14 at 18:11
• The pointwise limit $g$ takes on only the values $0$, $1$, and $\infty$. Fatou gives $\int g\le \int f$. As $f$ is integrable and $g(x)=\infty$ if $f(x)>1$, it follows that $|\{x\mid f(x)>1\}|=0$. So, $f\le1$ a.e.. Now show that $|\{ x \mid f(x)<1\}|=0$. (You can argue by contradiction. Note we know $f^n\le f$, now. If $|\{ x \mid f(x)<1\}|>0$, then we'd have $\int f^n<\int f$.) – David Mitra Nov 28 '14 at 18:19
• Yes. Thank you. :) – mac Nov 28 '14 at 18:31

For all $x\in X$, $f(x)^n$ converges to $0$ $1$ or $\infty$ when $f(x)<1$, $f(x)=1$ or $f(x)>1$, respectively. By Fatou's Lemma, $$\int \lim f^n\leq\liminf\int f^n=\int f<\infty$$ So $\int \lim f^n<\infty$, hence $\lim f^n<\infty$ a.e., that is $f\leq 1$ a.e. But then $f^2\leq f$ a.e. and $\int f^2=\int f$, which means that $f^2=f$ a.e., so $f=0$ or $1$ a.e., and this means precisely that $f$ is a.e. equal to a characteristic function.