Limit of Lebesgue integral $\int_F \; \frac{dx}{2 - \sin nx}$ 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.
 A: 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
$${\mu(F) \over 2\pi}\int_0^{2\pi}{1 \over 2 - \sin(t)}\,dt$$
A: First note that
$$
\left|\frac{1}{2-\sin(nx)}\right|\le1\tag{1}
$$
Next note that for $z=\tan{\frac{x}{2}}$
$$
\begin{align}
\int_0^{2\pi}\frac{\mathrm{d}x}{2-\sin(x)}
&=\int_{-\infty}^\infty\frac{\frac{2\;\mathrm{d}z}{1+z^2}}{2-\frac{2z}{1+z^2}}\\
&=\int_{-\infty}^\infty\frac{\mathrm{d}z}{1-z+z^2}\\
&=\int_{-\infty}^\infty\frac{\mathrm{d}z}{\frac{3}{4}+(z-\frac{1}{2})^2}\\
&=\frac{2\pi}{\sqrt{3}}\tag{2}
\end{align}
$$
Thus, the average of $\frac{1}{2-\sin(x)}$ over one of its periods is $\frac{1}{\sqrt{3}}$.
Approximate $F$ by a finite set of intervals $\{I_k\}$ so that $|\cup I_k\;\Delta\;F|<\epsilon$. Suppose $I_k=[a,b]$, then
$$
\begin{align}
\lim_{n\to\infty}\int_{I_k}\frac{\mathrm{d}x}{2-\sin(nx)}
&=\lim_{n\to\infty}\int_a^b\frac{\mathrm{d}x}{2-\sin(nx)}\\
&=\lim_{n\to\infty}\frac{1}{n}\int_{na}^{nb}\frac{\mathrm{d}x}{2-\sin(x)}\\
&=\lim_{n\to\infty}\frac{1}{n}\left(\int_{na}^{na+k2\pi}\frac{\mathrm{d}x}{2-\sin(x)}+\int_{na+k2\pi}^{nb}\frac{\mathrm{d}x}{2-\sin(x)}\right)\\
&=\lim_{n\to\infty}\frac{1}{n}\left(k\frac{2\pi}{\sqrt{3}}+J_n\right)\\
&=\frac{b-a}{\sqrt{3}}\\
&=\frac{|I_k|}{\sqrt{3}}\tag{3}
\end{align}
$$
In $(3)$, $k=\lfloor\frac{n(b-a)}{2\pi}\rfloor$ so that $|nb-(na+k2\pi)|<2\pi$ and therefore by $(1)$, we get $|J_n|<2\pi$. Thus, $\lim\limits_{n\to\infty}\frac{k}{n}=\frac{b-a}{2\pi}$ and $\lim\limits_{n\to\infty}\frac{J_n}{n}=0$.
Thus, again using $(1)$,
$$
\begin{align}
\lim_{n\to\infty}\left|\int_F\frac{\mathrm{d}x}{2-\sin(nx)}-\int_{\cup I_k}\frac{\mathrm{d}x}{2-\sin(nx)}\right|
&\le\lim_{n\to\infty}\int_{\cup I_k\;\Delta\;F}\frac{\mathrm{d}x}{|2-\sin(nx)|}\\
&\le\epsilon\tag{4}
\end{align}
$$
Summarizing,
$$
\begin{align}
\lim_{n\to\infty}\left|\int_F\frac{\mathrm{d}x}{2-\sin(nx)}-\frac{|F|}{\sqrt{3}}\right|
&\le\lim_{n\to\infty}\left|\int_F\frac{\mathrm{d}x}{2-\sin(nx)}-\int_{\cup I_k}\frac{\mathrm{d}x}{2-\sin(nx)}\right|\\
&+\lim_{n\to\infty}\left|\int_{\cup I_k}\frac{\mathrm{d}x}{2-\sin(nx)}-\frac{|\cup I_k|}{\sqrt{3}}\right|\\
&+\lim_{n\to\infty}\left|\frac{|\cup I_k|}{\sqrt{3}}-\frac{|F|}{\sqrt{3}}\right|\\
&\le\epsilon+0+\frac{\epsilon}{\sqrt{3}}\tag{5}
\end{align}
$$
therefore,
$$
\lim_{n\to\infty}\int_F\frac{\mathrm{d}x}{2-\sin(nx)}=\frac{|F|}{\sqrt{3}}\tag{6}
$$
