Using the Lebesgue dominated convergence theorem

Question

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 $\cos{0}$?

Answer 1

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.

Answer 2

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