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.


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Commented Aug 19, 2023 at 15:32
  • $\begingroup$ @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 $n_0\geq m$". $\endgroup$
    – M W
    Commented Aug 19, 2023 at 15:55
  • $\begingroup$ Oh! Sorry. I missed the point for all $n \geq m$ in the definition of $A_{m, k}.$ $\endgroup$ Commented Aug 19, 2023 at 16:17
  • $\begingroup$ Very elegant answer. Thanks for investing time on my question. $\endgroup$ Commented Aug 19, 2023 at 16:26

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