Almost everywhere pointwise convergence I am trying to solve this problem
Let $a_n$ be a sequence of numbers so that $\lim_{n \to \infty} \sin(a_nx)$ exists pointwise almost everywhere on  $\mathbb{R}$. Show that $\lim_{n \to \infty}a_n$ exists.
I tried to use Egoroff's theorem and other things, but I could not solve it.
 A: There is a stronger result: Suppose $\sin(a_nx)$ converges pointwise on a set of positive measure. Then $a_n$ converges to a finite limit.
Proof: Let $E$ be a set of positive and finite measure where $\sin(a_{n})$ converges pointwise.
We first prove $(a_n)$ must be bounded. If not, then WLOG there exist $0<a_{n_1} < a_{n_2} < \cdots \to \infty.$ By DCT we have
$$\tag 1 \int_E \sin^2(a_{n_k}x)\,dx \to \int_E f(x)^2\,dx.$$
The left side of $(1)$ equals
$$\int_E \frac{1-\cos(2a_{n_k}x) }{2}\, dx = m(E)/2-\frac{1}{2}\int_E \cos(2a_{n_k}x)\, dx.$$
Because $a_{n_k}\to \infty,$ the Riemann-Lebesgue lemma shows the last integral $\to 0.$ Hence $m(E)/2=\int_E f(x)^2\,dx.$
On the other hand,
$$\int_0^1f(x)^2\,dx = \lim \int_0^1f(x)\sin(a_{n_k}x)\,dx.$$
The limit on the right is $0,$ again using RL.
We therefore have $m(E)/2=0,$ contradiction. This proves $(a_n)$ must be bounded.
So now assume the bounded sequence $(a_n)$ has the given limit property but $\lim a_n$ fails to exist. Then
$$L=\liminf a_n< \limsup a_n = M.$$
There exist subsequences $a_{l_k}\to L$ and $a_{m_k}\to M.$ It follows that
$$\sin (a_{l_k}x) \to \sin (Lx),\,\, \sin (a_{m_k}x) \to \sin (Mx)$$
for $x\in E.$ We conclude $\sin (Lx)=\sin(Mx)$ on $E.$ Since these are analytic functions, we have $\sin (Lx)=\sin(Mx)$ everywhere. Differentiating then gives$$L\cos (Lx) = M\cos (Mx)$$everywhere. Now plug in $x=0$ to get $L=M,$ contradiction.
Therefore $\lim a_n$ exists.
