## Set of real numbers with positive measure containing no midpoints

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Does there exists a subset E of R with positive measure and without containing any midpoints (i.e. x,y distinct in E, (x+y)/2 not in E)?

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## 3 Answers

According to James Foran, Non-averaging sets, dimension, and porosity, Canad Math Bull 29 (1986) 60-63, "It follows from the Lebesgue Density Theorem that a measurable, non-averaging subset (of $(0,1]$) cannot have positive measure."

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Link to the article (open access): cms.math.ca/cmb/v29/p60 – Yemon Choi Jul 16 2010 at 6:56

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As already said by Gerry, the answer to your question is negative. However, it becomes positive if you only ask your set to have Hausdorff dimension 1 instead of positive Lebesgue measure, see

Salem, R.; Spencer, D. C. On sets which do not contain a given number of terms in arithmetical progression. Nieuw Arch. Wiskunde (2) 23, (1950). 133--143.

For a more general recent result see

Tamás Keleti Construction of one-dimensional subsets of the reals not containing similar copies of given patterns, Analysis and PDE Vol. 1 (2008), No. 1, 29-33

(if you do not know this journal, you should have a look at it and more generally to the web site of the Mathematical Science publishers, by the way.)

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No, such a set cannot exist and one can prove this using Lebesgue Density Theorem and a simple pegionhole argument. Infact all points $x$ which are density points of $E$ will be a midpoint for some $y,z \in E$ i.e., $x=\frac{y+z}{2}$.

Let $F \subseteq E$ be the set of density points of E, and $x \in F$.

Then there exists a $\epsilon > 0$ such that $m( B_{\epsilon}(x)\cap F) > \epsilon$. Now if $x$ is not a midpoint of $E$ then $\forall d \in (0,\epsilon)$, atleast one of $x-d$ or $x+d$ does not belong to $F$.

But then $m( B_{\epsilon}(x)\cap F)= \int_0^{\epsilon} |F\cap \lbrace x-t,x+t\rbrace| dt < \epsilon$, a contradiction !!

A set $A$ of real number is called Universal if every measurable set of positive measure necessarily contains an affine image of $A$. A simple variation of the above argument will give that all finite set $A$ are infact Universal. However, no example of an infinite Universal set is knwon and its a conjecture of Erdos that no infinite universal sets exists.

This paper has a nice discussion and references to this problem

M. Kolountzakis: Infinite Patterns That Can Be Avoided by Measure, Bull. London Math. Soc. 29 (1997), 4, 415-424. http://fourier.math.uoc.gr/~mk/ps/universal.pdf

As Gerry and Benoît Kloeckner has mentioned the problem becomes interesting when one considers Hausdroff measure instead of Lebesgue measure.

Recently I. Laba and M. Pramanik proved existence of 3 term arithmetic progression even in closed sets which has Hausdroff dimension close to 1, `under the condition that E supports a probability measure obeying appropriate dimensionality and Fourier decay conditions'

I. Laba and M. Pramanik: "Arithmetic progressions in sets of fractional dimension",, Geom. Funct. Anal. 19 (2009), 429-456. http://www.math.ubc.ca/~ilaba/preprints/progressions-may15.pdf

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