# Show that $S$ has Lebesgue measure zero.

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.
• Do you mean that $A_{m, k} \cap S$ is a nowhere dense set for every $m$ and $k\$? For otherwise if you take $x_0 \in [a,b] \subseteq A_{m,k}$ then it does not necessarily imply that $x_0 \in S$ and hence we cannot always conclude that $x_0 \in A_{n_0, k}$ i.e. $\left |\sin (n_0! \pi x_0) \right | \leq \frac {1} {k + 1}$ for all $n_0 \geq m.$ Feel free to correct me if I am wrong. Commented Aug 19, 2023 at 15:32
• @AkiroKurosawa I think you may be missing the fact that $A_{n_0,k}\supseteq A_{m,k}$ whenever $n_0\geq m$, and so we can indeed directly conclude $x_0\in A_{n_0,k}$, with no need to worry about $S$. Your "i.e." clause should read "$x_0\in A_{n_0,k}$, i.e. $\left |\sin (n! \pi x_0) \right | \leq \frac {1} {k + 1}$ for all $n\geq n_0$, which follows immediately from $n_0\geq m$".
• Oh! Sorry. I missed the point for all $n \geq m$ in the definition of $A_{m, k}.$ Commented Aug 19, 2023 at 16:17