0

I wish to prove that there exist meager subsets of $\mathbb{R}$ whose complements have Lebesgue measure zero.

Let us first enumerate the rationals by $\{q_k\}_{k \in \mathbb{N}}$. My intuition was to cover $\mathbb{R}$ with intervals centered at rational points each with radius say $1/(n2^{k})$ (so the radius depends on which rational the interval is centered at) and call this cover $B_n$. Then for all $n \in \mathbb{N}$, $B_n$ is a dense subset of $\mathbb{R}$, hence its complement must be nowhere dense and thus meager. Furthermore, the set $B = \bigcap_{n \in \mathbb{N}} B_n$ must also be meager also be dense in $\mathbb{R}$ as it is a countable intersection of dense sets. So it follows that $B^c$ must be meager, and $m(B) \leq m(B_n) = 1/n$.

It appears that by sending $n \rightarrow \infty$ we acheive the desired result, but I was wondering if conceptually this makes sense. As we approach infinity, at some point in the limit would the set $B_n$ no longer be dense?

Also, how do we know $B$ is not empty?

CBBAM
  • 5,883
  • 2
  • 6
  • 18
  • No. It will always contain the rational numbers. – Umberto P. Jun 08 '22 at 18:25
  • https://math.stackexchange.com/questions/1705057/there-exist-meager-subsets-of-mathbbr-whose-complements-have-lebesgue-measu – Onur Oktay Jun 08 '22 at 18:41
  • @UmbertoP. Even in that case, how do we know the complement of each $B_n$ or $B$ is not empty? – CBBAM Jun 08 '22 at 19:09
  • I’m not sure how helpful this is, but an equivalent problem is finding a meagre subset of $[0, 1]$ with measure $1$. If we had a meagre subset $V$ of $[0, 1]$ with measure $1$, we would see that $\bigcup\limits_{n \in \mathbb{Z}} {x + n \mid x \in V}$ is a meagre set whose complement has measure zero. Conversely, given a meagre $V$ whose complement has measure zero, $V \cap [0, 1]$ is meagre with measure 1. – Mark Saving Jun 08 '22 at 19:39
  • See https://math.stackexchange.com/q/2494705/977780 and https://math.stackexchange.com/q/4463360/977780 and https://math.stackexchange.com/q/1705057/977780 – Sourav Ghosh Jun 09 '22 at 09:56

1 Answers1

2

With some modifications, your procedure can be made rigorous.

Fixed an enumeration $\{q_n\}_{n \in \mathbb{N}}$ of a countable dense set $Q$.

Define $U_m = \bigcup\limits_{n \in \mathbb{N}} B_{2^{-(n + m)}}(q_n)$. Each $U_m$ is an open set containing $Q$, hence dense. We see that $\mu(U_m) \leq \sum\limits_{n = 0}^\infty 2^{1 -(n + m)} = 2^{2 - m}$. Let $U = \bigcap\limits_{m \in \mathbb{N}} U_m$; then for all $m$, we have $\mu(U) \leq \mu(U_m) \leq 2^{2 - m}$. Therefore, $\mu(U) = 0$. Since $U$ is the countable union of dense open sets, it is the complement of a meagre set.

Mark Saving
  • 31,855