Set of real numbers with positive measure containing no midpoints - MathOverflow

Set of real numbers with positive measure containing no midpoints

2010-07-16T06:13:30Z

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)?

Answer by Gerry Myerson for Set of real numbers with positive measure containing no midpoints

2010-07-16T06:54:23Z

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."

Answer by Benoît Kloeckner for Set of real numbers with positive measure containing no midpoints

2010-07-16T10:42:16Z

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.)

Answer by Vagabond for Set of real numbers with positive measure containing no midpoints

2010-07-16T10:43:30Z

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