Show that the set $\{x \in \mathbb{R}| \lim_{n \to \infty} \sin(a_n x) \mbox{ exists}\}$ has zero measure

If $\lim_{n\to \infty}\sin (a_n x)$ exists on a set of positive measure, then by Egorov's theorem you can find a set $E$ with $0<m(E)<\infty$ such that $f_n(x)=\sin (a_n x)$ converges uniformly on $E$ to a measurable function $f(x)$.

Then by the Riemann-Lebesgue lemma we must have $\int _F \sin (a_n x) dx \to 0 $ for any measurable subset $F$ of $E$. But also, by uniform convergence we have $\int_F \sin (a_n x)dx \to \int_F f(x) dx$ for any measurable set $F\subset E$. Thus, $\int_F f=0$ for all $F\subset E$, which implies that $f\equiv 0$ on $E$.

Now, observe that $\sin^2 (a_nx)$ also converges uniformly to $0$ on $E$, hence $$0 = \lim_{n\to\infty} \int_E \sin^2(a_nx) dx= \lim_{n\to\infty} \int_E (1-\cos(2a_nx))/2 dx = m(E)/2$$ because $\int_E \cos(2a_nx)dx \to 0$ by the Riemann-Lebesgue lemma.

However, this is a contradiction, since $m(E)$ was assumed to be $>0$.

Note: In fact, instead of using Egorov's theorem and uniform convergence, you can just use the dominated convergence theorem, since $\sin(a_n x)$ is bounded.

For any sequence $f_n$ of measurable functions, the set $\{x: \lim_{n \to \infty} f_n(x) \ \text{exists}\}$ is measurable. For example, you can write it as

$$ \bigcap_{m \in \mathbb N} \bigcup_{N \in \mathbb N} \bigcap_{i,j>N} \{x: |f_i(x) - f_j(x)| < 1/m \}$$