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Statement

A Cantor set is a compact, totally disconnected, perfect subset of $\mathbb R$ (p. 9). A set $X\subset\mathbb R$ is full measure universal if for every Lebesgue measurable $F$ with $m(\mathbb R\setminus F)=0$ there are $\lambda\in\mathbb R\setminus\{0\}$ and $t\in\mathbb R$ with $\lambda X+t\subset F$; a set that is not full measure universal is not measure universal (pp. 9--10). The Newhouse thickness $\tau_N(K)$ of a Cantor set $K$ is the infimum, over the steps of the construction that removes the largest remaining gap of each interval in turn, of the ratio of the shorter of the two child intervals to the removed gap (p. 11).

Theorem 3.6 (Gallagher--Lai--Weber; p. 12). Cantor sets in $\mathbb R$ with positive Newhouse thickness are not full measure universal, and therefore not measure universal.

Source. Yeonwook Jung, Chun-Kit Lai and Yuveshen Mooroogen, Fifty years of the Erdős similarity conjecture, arXiv:2412.11062v2 (1 January 2025), whose labels and page numbers are cited here; the edition is identified on the source card. The theorem is from John Gallagher, Chun-Kit Lai and Eric Weber, On a topological Erdős similarity problem, Bull. Lond. Math. Soc. 55 (2023), no. 3, 1104--1119.

Read depth. Claims checked: the statement and definitions were read clause by clause on pp. 9--12, and the survey's proof (p. 12) was read; nothing here is independently reviewed, and the paper of Gallagher, Lai and Weber was not read.

Proof pointer

Page 12. By Proposition 3.2 (p. 10) it suffices to find a null set $M$ that meets every $\lambda X+t$ with $\lambda\ne0$. Take $N>1/\tau(X)$ and the symmetric Cantor set $K$ of measure zero obtained by repeatedly removing the middle $1/(2N+1)$ of each interval, whose thickness is $N$, and set $M=\bigcup_{(n,l)\in\mathbb Z^2}2^n(K+l)$. For given $\lambda,t$ choose $(n,l)$ with $|\lambda|\in(2^{n-1},2^n]$ and $t\in(l2^n,(l+1)2^n]$; then $\lambda X+t$ and $2^n(K+l)$ have product of thicknesses above $1$ and neither lies in a gap of the other, so the Newhouse gap lemma (Lemma 3.5, p. 12) makes them intersect.

Dependencies

Proposition 3.2 (p. 10), the equivalence of full measure universality with a sumset condition, and the Newhouse gap lemma (Lemma 3.5, p. 12).

Bears on

• Problem 120: answers the question affirmatively for every set containing a Cantor set of positive Newhouse thickness; these are uncountable sets. It does not reach countable sets such as decreasing sequences, nor Cantor sets of zero thickness, and does not settle the problem.