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

Question

Can you prove or disprove that the sequence $\{\sin (nx)\}$ has a pointwise almost everywhere convergent subsequence with respect to the Lebesgue measure on $\mathbb{R}$ ?

Edit:

I am adding my thoughts here as a motivation, because otherwise this question will hit the bottom. So initially I thought whether $\{\sin{nx}\}$ is convergent in $L_1([0,\pi])$ or on other finite interval in $\mathbb R$. Then I computed the integral $\|\sin{nx}\|_1=\int\limits_{0}^{\pi}{|\sin{nx}|dx}=n\int\limits_{0}^{\pi/n}{\sin{nx}dx}=2$. Therefore $\|\sin{nx}\|_1\rightarrow 2$ but this doesn't tell me to which function converges. Then I thought about convergence in $L_2([0,\pi])$, but obviously it is not convergent there. This is because it is part of the orthonormal basis in $L_2$ (up to a constant), so it is not even Cauchy sequence. But still it is weakly convergent in $L_2$, because from Bessel's inequality it follows that $\langle f,\sin{nx}\rangle\rightarrow 0$ for each $f\in L_2([0,\pi])$. Finally, I decided to check whether $\{\sin{nx}\}$ has a convergent a.e subsequence. If there is no such sequence, then it can not be convergent in $L_1$. And if there is such subsequence, then by Lebesgue DCT it will follow that it converges in $L_1$ (and probably it might be the limit of the whole sequence). I just didn't see that LDCT will work again for $L_2$ : if there is a convergent a.e subsequence, then it should converge in $L_2$ also, but this is impossible since $\{\sin{nx}\}$ is orthonormal (up tp a constant) and no subsequence of $\{\sin{nx}\}$ is Cauchy in $L_2$. This is actually the answer of @Julián Aguirre.

Now I have another Question: Is it possible to prove that there is no convergent pointwise a.e subsequence of $\{\sin{nx}\}$ using only first year calculus and knowing of course what is a set with Lebesgue measure $0$ in $\mathbb R$ ?

What is obvious is that for each $x$ there is a convergent subsequence since $\{\sin{nx}\}$ is bounded. But all $x\in [0,\pi]$ are uncountable set, so we can not apply for example Cantor diagonal argument.

Answer 1

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 $$ \int_0^\pi(\sin(n_{k+1}\,x)-\sin(n_k\,x))^2\,dx=\pi. $$

Answer 2

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

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

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