I've got the following problem... We are given a sequence of integers $n_k\to+\infty$ as $k\to\infty$. Then we define $A:=\{x\in[0,2\pi]\,\colon\,\lim_{k\to\infty}\sin(n_kx) \text{ exists}\}.$
I have to prove now that $\operatorname{meas}(A)=0$, where the measure is the usual Lebesgue measure on the real line.
I've proved the following relation which I assume can be useful for the sequel, namely for any measurable $B\subseteq A$, setting $f(x):=\lim_{k\to\infty}\sin(n_kx),$ it follows $\int_B 2(f(x))^2\mathrm dx=\lim_{k\to\infty}\int_B1-\cos(n_kx)\mathrm dx=\operatorname{meas}(B).$ Then $A$ is measurable (I have proved it), so I want to apply the previous result: my plan is to show that $f(x)$ must be almost everywhere $0$ but I cannot see a neat argument.
Another strategy I've thought about would be to reason by contradiction and see what happen if $A$ has strictly positive measure. At any rate this way seems nice to me because the result in this case would imply that $f$ must be equal to $0$ almost everywhere on $A$ in contrast with the previous way which uses this conclusion as the starting assumption.
In any case, I've spent quite a lot of time on this without success, so I'm asking you to help me...
thanks for your patience
-Guido-