Given a subset X of unit interval, can we find a subset Y of X of same outer measure as X such that Y does not contain three points of the form x, y and (x+y)/2?
I can do this assuming CH but can we do this without this assumption? Thanks!!
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6$\begingroup$ This is a beautiful question. $\endgroup$– Mohammad GhiasiCommented Apr 9, 2015 at 15:32
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4$\begingroup$ Tangential remark: Note that you can always find a non null $Y \subseteq X$ avoiding mid points. This follows from: (R. Rado) One can partition $\mathbb{R}^n$ into countably many sets such that every piece is mid point free. See P. Komjath: Set theoretic constructions in Euclidean spaces, New Trends in Discrete and Computational Geometry (J. Pach, ed.), Springer, 1993, 303-325 $\endgroup$– AshutoshCommented Apr 9, 2015 at 16:00
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$\begingroup$ First thoughts. For $\varepsilon>0$ call $X$ $\varepsilon$-free if there are no $x,y,z\in X$ with $x+\varepsilon=y$ and $y+\varepsilon=z$. Given $\varepsilon$, is there $Z$ of inner measure $0$ such that $X\setminus Z$ is $\varepsilon$-free? (Note that it could conceivably happen that $m^*(X)=m^*(Z)=m^*(X\setminus Z$), see this answer.) Could we also get that $X\setminus Z$ is $\varepsilon'$-free for every $\varepsilon'\ge\varepsilon$ (this would solve the problem)? $\endgroup$– MirkoCommented Apr 10, 2015 at 0:25
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$\begingroup$ Yes, under the additional assumption that $X$ is measurable. I posted an answer that has a gap, and my proof does not seem to work when $X$ is non-measurable with positive outer measure, but it does work when $X$ is measurable with positive measure. (The case when $X$ has measure zero is trivial, since then we could take $Y$ to be the empty set.) $\endgroup$– MirkoCommented Apr 10, 2015 at 3:41
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2$\begingroup$ @Mirko If X is a Borel and has positive measure, then it has a similar copy of every finite set of reals. You can prove this using Lebesgue density theorem. $\endgroup$– AshutoshCommented Apr 23, 2015 at 19:51
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2 Answers
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I think I can do it if every subset of $\mathbb R$ of cardinality $< \mathcal c$ has measure $0$ (which is a consequence of Martin's axiom).
Let $m$ be the outer measure of $X$. WLOG we can assume $m > 0$. Let $G$ be a $G_\delta$ containing $X$ with outer measure $m$. Index the $G_\delta$ subsets of $G$ with measure $< m$ by ordinals: $G_\alpha: \alpha < A$ where $A$ is the least ordinal of cardinality $\mathcal c$. Proceed by transfinite induction to choose, for each $\alpha < A$, $x_\alpha \in X \backslash G_\alpha$ such that $\{x_\beta: \beta \le \alpha\}$ is midpoint free. This is possible because $X \backslash G_\alpha$ has cardinality $c$, while the set of "forbidden" points ($(x_\beta + x_\beta')/2$ and $2 x_\beta - x_\beta'$ for $\beta, \beta' < \alpha$) has cardinality $< \mathcal c$.
answered Apr 9, 2015 at 15:50
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Warning: The following answer has a gap, and I do not know if it could be fixed. I leave it for consideration, now that I already wrote it. (It does work, in ZFC, but under the additional assumption that $X$ is measurable.)
Repeat the usual construction of a Bernstein set: Let $\{F_\alpha: \alpha < \mathfrak c = 2^{\aleph_0}\}$ be a list of all closed uncountable subsets of the real line. Call a set $S$ free if it contains no three different points $x,y,z$ with $y-x=z-y$ (that is, no points with $y=\frac{x+z}2$). Let $E(S)=\{z\in\mathbb R\setminus S: \exists x,y\in S, z=\frac{x+y}2,\mathrm{\ or\ } z=y+(y-x)\}$, that is $E(S)$ are the excluded (or "forbidden") points for $S$. (Note that $z=y+(y-x)$ could be to the left or to the right of $y$, depending on whether $y-x$ is negative or positive.) For each $\alpha$ pick different $x_\alpha,y_\alpha\in F_\alpha$ such that if $P_\alpha=\{x_\beta:\beta<\alpha\}$ and $Q_\alpha=\{y_\beta:\beta<\alpha\}$ then we have :
$x_\alpha\not\in E(P_\alpha)\cup P_\alpha\cup Q_\alpha$, and $y_\alpha\not\in P_\alpha\cup Q_\alpha$.
(We could do this since $|F_\alpha|=\mathfrak c$, but $|E(P_\alpha)|\le\aleph_0\cdot |P_\alpha|\le\aleph_0\cdot|\alpha|<\mathfrak c$.)
Let $P=\{x_\alpha:\alpha<\mathfrak c\}$ and $Q=\mathbb R\setminus P$. Then each of $P$ and $Q$ is a Bernstein set of inner measure zero and full outer measure. Moreover, by construction $P$ is free. We have that $X\cap P=X\setminus Q$ and since $Q$ is of inner measure zero, the outer measure $m^*(X)=m^*(X\setminus Q)$ (this is the gap in my proof, this is not necessarily the case, unless we assume that $X$ is measurable). That is $m^*(X\cap P)=m^*(X)$, and $X\cap P$ is free since $P$ is.
Well, the gap is, I only know that $X$ has positive outer measure, but this does not rule out the possibility that $X$ is of inner measure zero, so it might happen that $X\subset Q$ and in that case $X\cap P$ would be empty.
I got mislead by the following answer (even if the answer is fine, but I didn't read it right, it only applies when we start with a measurable set, and the set $X$ above is not necessarily measurable):
A set $E$ with positive Lebesgue measure can be decomposed as a union $E=A\cup B$ where each of $A$ and $B$ have zero inner measure, and therefore each of $A$ and $B$ are nonmeasurable with $m^*(A)=m^*(B)=m(E)$.
Non Lebesgue measurable subsets with "large" outer measure
