Let $F\subset\mathbb{R}$ be a measurable set of finite Lebesgue measure.
Find the limit
$$ \lim_{n \to \infty} \int_{F} \frac{dx}{2-\sin nx}.$$
Let $f_{n}(x)=\frac{1}{2-\sin nx}$. This function is bounded and
continuous. Thus, it is integrable. I even use Mathematica to compute the Riemann integral.
However, since we don't really know what is $F$. So we don't know $\int_{F}\frac{dx}{2-\sin nx}$. Or this could be seen as a hint, suggesting that the limit does not exist.
I want to use the dominated convergence theorem to show that since $\lim \limits_{n \to \infty}f_{n}(x)$ does not exist, $\lim \limits_{n \to \infty}\int_{F}\frac{dx}{2-\sin nx}$
does not exist.
However, the condition of that theorem is that $\lim \limits_{n \to \infty}f_{n}(x)$ exists. So I guess I can not use this theorem.
Best Answer
Srivatsan's guess is correct. One way to see this is to write $${1 \over 2 - \sin(nt)} = {1 \over 2} \bigg({1 \over 1 - {\sin(nt) \over 2}}\bigg)$$ $$= \sum_{k=0}^{\infty} {\sin^k(nt) \over 2^{k+1}}$$ If this sum is truncated at some $k = k_0$, the remainder is bounded by ${1 \over 2^{k_0}}$ in absolute value, and thus the integral of the remainder over $F$ is bounded by ${1 \over 2^{k_0}}\mu(F)$ in absolute value. As $k_0$ goes to infinity this goes to zero. Thus it suffices to show the following limit exists and determine its value. $$\lim_{k_0 \rightarrow \infty} \lim_{n \rightarrow \infty}\int_F \sum_{k = 0}^{k_0} {\sin^k(nt) \over 2^{k+1}}\,dt$$ This in turn will equal the following, if the limit in each term exists and the sum of the limits converges. $$= \sum_{k=0}^{\infty} \lim_{n \rightarrow \infty}\int_F {\sin^k(nt) \over 2^{k+1}}\,dt$$ Now we examine a given term of the above. The function $\sin^k(nt)$ is the sum of $2^k$ exponentials when you write $\sin(nt) = {e^{iknt} - e^{-iknt} \over 2i}$. As $n$ goes to infinity, the integral over $F$ of any such term other than a constant term (a "middle term") will go to zero by the Riemann-Lebesgue lemma, and the middle term, if it exists, integrates to $\mu(F){1 \over 2\pi}\int_0^{2\pi}\sin^k(t)\,dt$ because $\sin^k(t)$ has the same middle term. If the middle term doesn't exist, corresponding to $k$ odd, then the limit is still $\mu(F){1 \over 2\pi}\int_0^{2\pi}\sin^k(t)\,dt$, which is now zero. So we have $$\lim_{n\rightarrow \infty} \int_F\sin^k(nt)\,dt = \mu(F){1 \over 2\pi}\int_0^{2\pi}\sin^k(t)\,dt$$ So the overall limit is $${\mu(F) \over 2\pi}\sum_{k=1}^{\infty} \int_0^{2\pi}{\sin^k(t) \over 2^{k+1}} \,dt$$ Adding up this geometric series (i.e. reversing what we did before gives us the desired sum $${\mu(F) \over 2\pi}\int_0^{2\pi}{1 \over 2 - \sin(t)}\,dt$$