This question is related to the one in https://mathoverflow.net/questions/65322/the-structure-of-certain-maximal-sets-of-means-into-amenable-semigroups. I open a different topic because they can be treated separately.

Let $S$ be a countable amenable semigroup and let $IM(S)$ be the set of the invariant means on $S$.

Definition. For any subset $W\subseteq S$, we define $$ W^-=\inf_{m\in IM(S)}m(\chi_W)\;\;\;\;\;\;\;\;\;\;\;and\;\;\;\;\;\;\;\;\;\;\;\;W^+=\sup_{m\in IM(S)}m(\chi_W) $$

There are many and well-known examples that show that these numbers could be different. I am interested in the case when they are equal. As far as I know it seems to me that this situation has no name in literature, so let me fix the following terminology

Definition. $W$ is said to have the property IM (Intrinsically Measurable) if $W^-=W^+$. In this case $\mu(W)$ denotes the intrinsic measure of $W$.

Basic examples of sets with IM are easy. I need to give such examples in order to formulate my first question. First of all, recall the following (almost) classical

Definition Let $k$ be a positive integer possibly infinite. A subset $W\subseteq S$ is called $k$-tile if there are $s_1, s_2,... s_k$ elements in $S$ such that

  1. $s_iW\cap s_jW=\emptyset$, for all $i\neq j$
  2. $S\setminus\bigcup s_iW$ has the property IM and $\mu(S\setminus\bigcup s_iW)=0$

Tiles obviously have the property IM. Moreover the following operations preserve the property IM: finite union of disjoint sets with IM; if $V\subseteq T$ have the property IM, then also $T\setminus V$ have IM. Let me call elementary those sets with the IM that can be obtained by tiles using the previous operations.

Now I list some basic questions about this property IM. The second question is not just a specification of the first one, but it is interesting for the application that is the motivation for studying this property.

Question 1. Is there an (explicit) example of a non-elementary IM set $W$ with $\mu(W)>0$?

Question 2. Let $S$ be the multiplicative semigroup of positive integers and let $W$ be the subset of those positive integers with first digit $1$, $2$ or $3$. Does $W$ have the property IM?

Note that the property IM is not closed under countable union, but even the answer to the following question is not completely evident to me

Question 3. Is the class of IM sets closed under finite intersection? [This question has been answered in the negative by Ben Willson (see below)]

Thanks in advance for any comment,


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    $\begingroup$ I guess you are interested in subsets $W$ for which $W^-=W^+$ is strictly positive. Otherwise any finite set in an infinite amenable semigroup, or more generally any subset of a coset of an infinite index subgroup in an amenable group, do the job. $\endgroup$ – Alain Valette May 18 '11 at 17:27
  • $\begingroup$ PS: That was for Question 1 of course. $\endgroup$ – Alain Valette May 18 '11 at 17:27
  • $\begingroup$ Yes of course. I've edited. Thanks. $\endgroup$ – Valerio Capraro May 18 '11 at 18:40
  • $\begingroup$ There is also another construction that trivially preserves IM, namely take $V\subseteq T$ tiles and consider $T\setminus V$. This still has IM, but there is no reason why this should be finite union of disjoint tiles.. $\endgroup$ – Valerio Capraro May 18 '11 at 19:12

You should look at

Joseph Rosenblatt and Zhuchang Yang, Functions with a unique mean value, Illinois Journal of Mathematics, Volume 34, Number 4, Winter 1990

where various properties of sets and functions with unique invariant mean a studied. It is a standard fact that Bohr sets in $\mathbb Z$ (or any amenable group) have a unique mean. For example:

$$\mu\left(\left\lbrace n \in \mathbb Z \ \left| \ \min_{m \in \mathbb Z}|\sqrt{2} \cdot n - m| \leq \varepsilon \right.\right\rbrace\right) = 2\varepsilon.$$

This set is non-elementary in your sense.

EDIT: The rough idea is that if there exist $q$ translates of a set such that the union covers each group element at most $p$ times, then the measure is less that $p/q$. Similarily, if the union of the translates covers each group element at least $p'$ times, then the measure is bigger than $p'/q$. Now, translating the set corresponds (up to some $\delta)$ to translating the interval of length $2 \varepsilon$. It is now easy to see that $p/q$ and $p'/q$ can be made to converge to $2 \varepsilon$.

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  • $\begingroup$ Thank you very much, Adreas! This is exaclty what I need. $\endgroup$ – Valerio Capraro May 21 '11 at 16:00
  • $\begingroup$ I have actually some trouble to prove that 1) any Bohr set has a unique invariant mean; 2) your set is not elementary (I guess this would follow from the fact that elementary sets must have measure $\in\mathbb Q$ - am I wrong?) Maybe I will think about more, but if you want to include some detail, it would be great! $\endgroup$ – Valerio Capraro May 23 '11 at 21:54
  • $\begingroup$ Thank you, ANdreas! But I'm not going to accept the answer.. I want Question 2 :P $\endgroup$ – Valerio Capraro May 24 '11 at 18:46

In answer to question 3, consider the integers with addition.

Let $W_1$ be the set of odd integers. Then it is a $2$-tile, hence IM.

Let $W_2$ be the set of even negative integers and positive odd integers. $$W_2=\{ \ldots, -6,-4,-2,1,3,5,\ldots\}$$

$W_2$ is not quite a $2$-tile, but using the same justification, it is IM.

$W_1\cap W_2$ is just the positive odd integers which I'm fairly certain is not IM.

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  • $\begingroup$ $W_2$ has intuitively the property IM, but I can't see a proof at the moment. I'm probably missing something very stupid. About $W=W_1\cap W_2$, it seems to me that the following argument shows that $W^-=0$ and $W^+=1$, isn't true? Take $k>0$ and define the $k$-density of $A\subseteq\mathbb Z$ to be $$ d_k=\lim_{n\rightarrow\infty}\frac{|A\cap[-kn,n]|}{(k+1)n+1} $$ This density extends to an invariant mean in a classical way and one has $d_k(W)=\frac{1}{k+1}$. To get measure close to 1 one can use Folner sequence barycentered on the positive numbers.. Does it work? $\endgroup$ – Valerio Capraro May 19 '11 at 12:30
  • $\begingroup$ I think that works to show that $W$ is not IM. $\endgroup$ – Ben Willson May 19 '11 at 18:55
  • $\begingroup$ To see $W_2$ has property IM notice that we can use it to tile the integers minus the singleton zero. Then, using Alain's comment above, we see that the singleton is IM and has value $0$ so its complement is IM and has value $1$. You should be able to extend your argument for tiling the whole semigroup to tiling any IM subset. My claim should follow from that. $\endgroup$ – Ben Willson May 19 '11 at 18:58
  • $\begingroup$ It seems to me that $W_2$ would answer question 1 too. I'm going to edit the post in order to make it less trivial. $\endgroup$ – Valerio Capraro May 19 '11 at 22:29

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