3
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – David Mitra Nov 28 '14 at 17:39
  • 1
    $\begingroup$ @DavidMitra: could you please elaborate? $\endgroup$ – mac Nov 28 '14 at 18:11
  • 3
    $\begingroup$ 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 $\int f^n<\int f$.) $\endgroup$ – David Mitra Nov 28 '14 at 18:19
  • $\begingroup$ Yes. Thank you. :) $\endgroup$ – mac Nov 28 '14 at 18:31
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.