1
$\begingroup$

Let $(X,\mathfrak{M},\mu)$ be a measure space where $\mu$ is complete and countably additive, and let $f_n:X\rightarrow \mathbb{R}$ be measurable and $f_n\rightarrow0$ a.e. Prove that functions $\sin{f_n(x)}$ are Lebesgue integrable and $\lim\limits_{n\to\infty}\int\limits_{X}{\sin{f_n(x)}d\mu} = 0$

My reasoning is as follows: We note that a composition of a continuous function preserves a.e. convergence, hence $\sin{f_n(x)}\rightarrow\sin{0}$

Since $|\sin{f_n(x)}|\leq1$, we can apply the Lebesgue dominated convergence theorem and conclude that the limit equals $0$. Is that correct?

The second part of the problem has $\cos{f_n(x)}$ instead of $\sin$. It follows from the problem that the value of the limit should also equal $0$. Why is that?

Isn't it true in the case of $\cos$ that the sequence of functions $\cos{f_n(x)}$ converges a.e. to $\cos{0}$?

$\endgroup$
  • $\begingroup$ First part: the result and your proof are false unless $\mu $ is a finite measure. How do you apply DCT? $\endgroup$ – Kavi Rama Murthy Jan 28 at 12:13
  • $\begingroup$ I see. My initial idea was wrong. What is the right approach then? Do we need to explicitly show the equality by expressing the integral as a supremum? $\endgroup$ – Don Draper Jan 28 at 12:29
  • $\begingroup$ @DavidC.Ullrich, my bad. Thank you for correcting. $\endgroup$ – Don Draper Jan 28 at 12:43
  • $\begingroup$ The first statement, about $\sin(f_n)$, is very easily seen to be false. The second part, about $\cos(f_n)$, is so absurd that it seems likely you have nnot stated the problem correctly. $\endgroup$ – David C. Ullrich Jan 28 at 13:11
  • $\begingroup$ Let $(X,\mathfrak{X},\mu)$ be a measure space with a complete countably additive measure and let $f_n:X\rightarrow\mathbb{R}, n \in \mathbb{N}$ be measurable and $f_n \rightarrow 0$ a.e. Prove that functions $ \sin{f_n(x)} $ are Lebesgue integrable and the following statement is true: $\lim\limits_{n\to\infty}\int\limits_{X}{\sin{f_n(x)}d\mu} = 0$. $\endgroup$ – Don Draper Jan 28 at 13:15
2
$\begingroup$

What you're trying to prove is false. Consider the line with Lebesgue measure. Let $f_n=1/n$. Then $f_n\to0$ ae but $\sin(f_n)$ is not integrable.

Or let $f_n=\frac1n\chi_{[0,n^2]}$: then $\sin(f_n)$ is integrable but $\lim\int\sin(f_n)=\infty$.

$\endgroup$
  • $\begingroup$ Do you mean that my idea is wrong or the statement itself? $\endgroup$ – Don Draper Jan 28 at 12:46
  • $\begingroup$ @DonDraper Read what I wrote! (i) The first sentence is pretty clear. (ii) Regardless, look at the two examples: Do they just show that the idea is wrong or do they show that the statement itself is wrong? $\endgroup$ – David C. Ullrich Jan 28 at 12:59
  • $\begingroup$ Another one in $\mathbb{R}$ with finite limit: $f_n = \chi_{[2n\pi, (2n+1)\pi]}$. Then $f_n \to 0$, and $\int \sin{f_n} = 2$, which is finite and constant in $n$. $\endgroup$ – dafinguzman Jan 30 at 18:05
0
$\begingroup$

Be careful: nothing guarantees that your dominating function, i.e. $g(x) \equiv 1$, is actually integrable on X. This is only true if X has finite measure, so you must find another function in order to apply the DCT.

For the cosine exercise: be careful that your a.e. limit is the constant function $h(x) \equiv 1$, which is not - in general - integrable. This should be a hint that the DCT can't be applied here, since as a consequence we would have integrability of the limit function.

$\endgroup$
  • $\begingroup$ Thank you. This exercise reveals my misunderstanding of DCT. In any case, is DCT applicable here or should another approach be taken? $\endgroup$ – Don Draper Jan 28 at 12:30
  • $\begingroup$ You could use that $\vert sin(f(x)) \vert \leq f(x)$, but you would need more hypothesis on the $f_n(x)$, e.g. their integrability. $\endgroup$ – Simone Ramello Jan 28 at 12:46
  • $\begingroup$ I don't quite understand how to continue (or rather start) the proof. Above, David claims that the statement is wrong at all. $\endgroup$ – Don Draper Jan 28 at 13:08
  • $\begingroup$ With this general hypotheses, it is false. You must add something if you wish to get a true result. $\endgroup$ – Simone Ramello Jan 28 at 13:10
  • $\begingroup$ The only thing I should have specified in my initial post is that $\mu$ is complete and countably additive. But I took that for granted. Would integrability of $f_n(x)$ make the problem plausible? $\endgroup$ – Don Draper Jan 28 at 13:25

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.