Pointwise almost everywhere convergent subsequence of $\{\sin (nx)\}$

A different proof. If $\{\sin(n_k\,x)\}$ converges a.e., then $$ (\sin(n_{k+1}\,x)-\sin(n_k\,x))^2 $$ converges to $0$ a.e. By the dominated convergence theorem $$ \lim_{k\to\infty}\int_0^\pi(\sin(n_{k+1}\,x)-\sin(n_k\,x))^2\,dx=0, $$ but $$ \int_0^\pi(\sin(n_{k+1}\,x)-\sin(n_k\,x))^2\,dx=\pi. $$

In fact $\sin (n_kx)$ diverges a.e. for every subsequence $n_k.$ Proof: Fix a subsequence $n_k$ and let $E= \{x\in \mathbb {R}: \lim_{k\to\infty}\sin (n_kx)\,\,\text {exists}\}.$ Let $f$ be the pointwise limit function on $E.$ Let $a> 0$ and put $E_a=E\cap [-a,a].$ On $E_a$ we have $f\in L^1\cap L^2.$ By the DCT,

$$\int_{E_a}f^2 = \lim_{k\to\infty} \int_{E_a}f\sin(n_kx)\,dx.$$

By the Riemann Lebesgue lemma, this limit is $0.$ But then we get

$$0= \int_{E_a}f^2 = \lim_{k\to\infty} \int_{E_a}\sin^2(n_kx) \,dx = \lim_{k\to\infty} \int_{E_a}\frac{1-\cos (2n_kx)}{2}\, dx = m(E_a)/2.$$

again using DCT and R-L. Therefore $m(E_a) = 0,$ and since $a$ was arbitrary, $m(E)=0.$