The set $E=\{t\in [-\pi, \pi] : \exists \lim_{n\rightarrow+\infty}\sin(n_kt)\}$ is Lebesgue-measurable and $|E|=0$ Let $\{n_k\}_{k \geq 1}$ be a strictly increasing sequence of integers. Prove that this set:
$$E=\{t\in [-\pi, \pi] : \exists \lim_{n\rightarrow+\infty}\sin(n_kt)\}$$
is Lebesgue-measurable and $|E|=0$.
So, the exercise suggests to consider the function $(t)= \lim_{n\rightarrow+\infty}\sin(n_kt) ,\  t\in E$ and $u(t)=0, \ t \notin E$ and to prove that this function is $0$ almost-everywhere using the Rienman-Lebesgue theorem. I've seen from other solutions that this leads to prove that $|E|=0$  (if these solutions are correct, which I doubt), but I struggle to see why $E$ is measurable. Any help?
 A: As per comments let me put an answer to clear the confusion.
First, we note that if $f_n$ is a sequence of (bounded) measurable functions on $[-\pi, \pi]$ (any topological space with the Borel sigma-algebra works) then the functions $g=\limsup f_n, h=\liminf f_n$ are measurable and the set where $\lim f_n$ exists is precisely the zero set of $g-h$ so is measurable (usual manipulations show that the boundness restriction is unneccesary but here we do not need that); also a direct proof of the measurability of the limit set of $f_n$ can be obtained by noting that $x \in E$ iff $f_n(x)-f_m(x)$ is Cauchy which means that for any $k>0$ there is $k(x)$ st $f_n(x)-f_m(x) \in (-1/k,1/k), n,m \ge k(x)$ which leads to the usual description of $E$ obtained through countable unions and intersections from the measurable sets $E_{n,m,k}=(f_n-f_m)^{-1}(-1/k,1/k)$
Now let $u(t)=\lim \sin n_kt, t \in E, u(t)=0$ otherwise and note that $u(t)\sin n_kt \to u(t)^2, t \in E, 0$ otherwise, so applying Riemann-Lebesgue $\int_{-\pi}^{\pi}u(t)\sin n_kt dt\to 0$ and by boundness etc we can switch limit and integral so $\int_E u(t)^2dt=0$ which means that if $E=E_0 \cup E_1$ where $E_0$ is the subset of $E$ where $u(t)=0$ and $E_1$ the set where $u(t) \ne 0$ we have that the measure of $E_1$ is zero (as $E_1=\cup E_{1,n}$ where $u^2(t) \ge 1/n$ on $E_{1,n}$ so $E_{1,n}$ are null sets, hence $E_1$ is)
But now we notice that if $\sin n_kt \to 0$, then $\cos 2n_kt \to 1$ so we apply the same reasoning to $v(t)=\lim \cos 2n_kt$ on the set $F$ where the limit exists and $0$ elsewhere (so using now $\int_{-\pi}^{\pi}v(t)\cos 2n_kt dt\to 0$), to get as before that if $F=F_0 \cup F_1$, the measure of the set $F_1$ where $v(t) \ne 0$ is zero; since $E_0 \subset F_1$ we are done!
