For the sake of brevity, let S be the set of such x. If S has positive measure, then it contains a subset of positive measure on which liminf[sin(nkx)] is bounded below by some positive constant; ie, the integral of liminf[sin(nkx)] over S will be positive. If S has zero measure, then the same integral will be zero. Thus, we must only compute an appropriate integral to show that m(S)=0.
Since we do not know that S has finite measure, take a sequence of functions fk(x)=2-|x|*sin(nkx), each of which is clearly integrable. By Fatou's lemma, the integral of liminf[fk(x)] over S <= the liminf of the integrals of fk(x) over S. However, each fk is the L1 function 2-|x|*sin(nkx). By the Riemann-Lebesgue Lemma, this goes to 0 as nk goes to infinity.
Hence our original integral of a strictly positive function over S is bounded above by 0, so m(S)=0.
We can justify bringing the limit inside the integral. This is because for every , . Hence, our integrand is bounded by and hence is for all . Then by Lebesgue Dominated Convergence, we can take the pointwise limit of the integrand. to get
It is easy to show that is bounded (specifically by ) which implies that is Lipschitz continuous which implies that it is absolutely continuous.