Consider the set $S \subseteq \mathbb R$ defined by $S : = \left \{x \in \mathbb R\ \bigg |\ \lim\limits_{n \to \infty} \sin (n! \pi x) = 0 \right \}.$ Show that Lebesgue measure of $S$ is zero.
I know that $S$ is an uncountable additive subgroup of $\mathbb R$ containing $\mathbb Q.$ I also note that $e \in S$ as the fractional part $\{n! e \}$ tends to zero as $n \to \infty.$ I am trying to show that Lebesgue outer measure of $S$ is zero. This will prove that $S$ is an uncountable proper additive subgroup of $\mathbb R.$
One way to do this is to show that $S$ is measurable and it doesn't contain any interval (equivalently, $S$ doesn't have any interior point) then this forces $S$ to be a proper Lebesgue measurable additive subgroup of $\mathbb R$ and consequently, the Lebesgue measure of $S$ is $0.$
But the above fact (i.e. $S$ is Lebesgue measurable which doesn't contain any interval) requires a proof which I can't quite get hold of. Could anyone please help me in this regard?
Thanks for your kind attention to this question.
For $k,m \in \mathbb{N}$ let $$ A_{m,k}:=\{x \in \mathbb{R} \mid \forall n \ge m: ~ |\sin(n!\pi x)| \le \frac{1}{k+1}\}. $$ Each $A_{m,k}$ is closed, hence measurable and $$ S=\bigcap_{k\in \mathbb{N}} \bigcup_{m\in \mathbb{N}} A_{m,k}. $$ Thus $S$ is measurable. Moreover each $A_{m,k}$ is nowhere dense: Let $[a,b]$ be an interval contained in some $A_{m,k}$. Then for some $n_0 \ge m$ the length of the interval $n_0!\pi[a,b]$ is $\ge 2\pi$. Thus there is some $x_0 \in [a,b]$ such that $1=|\sin(n_0!\pi x_0)| \le \frac{1}{k+1}$, a contradicton. Thus $\bigcup_{m\in \mathbb{N}} A_{m,k}$ is of first category for each $k \in \mathbb{N}$. Now, $S$ is of first category too. By Baire's Theorem $\mathbb{R} \setminus S$ is dense and therefore $S$ does not contain any interval.
Hope everything is correct, please check.