Convergence in a Measurable set

Question

Let $E$ the set of all $x\in[0,2\pi]$ at which $\{\sin (nx)\}$ converges.

This implies that $E$ is measurable?

Thanks you all.

Answer 1

It can be shown, that $\sin{nx}$ diverges for all irrational $x/\pi$ (We can easy deduce it from this question Prove the divergence of the sequence $\left\{ \sin(n) \right\}_{n=1}^{\infty}$., Bruno Joyal's answer [with the link to equidistribution]). Hence, even without checking for which $x$ it converges, we can say that $E$ is measurable, because it is countable.

Answer 2

Since the $\limsup$ and $\liminf$ of a sequence of measurable functions is measurable, and $E=f^{-1}(\{0\})$ where $f$ is given by $f(t)=\limsup_n \sin(nt)-\liminf_n \sin(nt)$, it follows that $E$ is measurable.