# Using the Lebesgue dominated convergence theorem

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 $$\cos{0}$$?

• First part: the result and your proof are false unless $\mu$ is a finite measure. How do you apply DCT? – Kavi Rama Murthy Jan 28 at 12:13
• 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? – Don Draper Jan 28 at 12:29
• @DavidC.Ullrich, my bad. Thank you for correcting. – Don Draper Jan 28 at 12:43
• 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. – David C. Ullrich Jan 28 at 13:11
• 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: $\lim\limits_{n\to\infty}\int\limits_{X}{\sin{f_n(x)}d\mu} = 0$. – 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 $$\lim\int\sin(f_n)=\infty$$.

• Do you mean that my idea is wrong or the statement itself? – Don Draper Jan 28 at 12:46
• @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? – David C. Ullrich Jan 28 at 12:59
• 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 $n$. – 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.

• Thank you. This exercise reveals my misunderstanding of DCT. In any case, is DCT applicable here or should another approach be taken? – Don Draper Jan 28 at 12:30
• 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. – Simone Ramello Jan 28 at 12:46
• I don't quite understand how to continue (or rather start) the proof. Above, David claims that the statement is wrong at all. – Don Draper Jan 28 at 13:08
• With this general hypotheses, it is false. You must add something if you wish to get a true result. – Simone Ramello Jan 28 at 13:10
• 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? – Don Draper Jan 28 at 13:25