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 $\cos{0}$?

  • $\begingroup$ 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

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 $\lim\int\sin(f_n)=\infty$.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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