# Invariant means on the integers

Let $A\subseteq\mathbb Z$, as usual we define the lower Beurling density $d^{-}(A)=\lim\inf_{n\rightarrow\infty}\frac{|A\cap[-n,n]|}{2n+1}$ and the upper Beurling density $d^+(A)=\lim\sup_{n\rightarrow\infty}\frac{|A\cap[-n,n]|}{2n+1}$. It seems to me I've found somewhere the following result, but I am not able to find the reference and I am not completely sure that I remember well:

for any $r$ in the interval $[d^{-}(A),d^+(A)]$, there is an invariant mean $m$ on $l^\infty(\mathbb Z)$ such that $m(\chi_A)=r$. Moreover, if $d^+(A)=d^{-}(A)$, then every invariant mean gives $\chi_A$ the same value and this is just the density $d(A)$.

Could anybody confirm that this result is true? reference?

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Could you just recall what a mean is in this context ? –  camomille Apr 7 '11 at 10:01
yep, a mean is a linear functional $m:l^\infty(\mathbb Z)\rightarrow\mathbb R$ such that $m(\chi_{\mathbb Z})=1$. Invariant mean is a mean such that $m(\chi_{n+A}=m(\chi_A)$ for all $A\subseteq\mathbb Z$ –  Valerio Capraro Apr 7 '11 at 10:11
and clearly for all $n\in\mathbb N$. By the way, how can I edit a comment? –  Valerio Capraro Apr 7 '11 at 10:13
Thanks for the definition. I don't know is one can edit one's comment but one can delete them (and copy/paste works on comments !). –  camomille Apr 7 '11 at 10:19
ops I've forgotten $m$ has to positive. It means that if $f$ is a positive and bounded function on $\mathbb Z$, then $m(f)\geq0$. –  Valerio Capraro Apr 7 '11 at 10:27

Consider an ultrafilter and take the ultralimit of the averages. Then you certainly get a mean in between the lower and the upper Beurling number. The question is, why you can obtain all the reals in between the two numbers. The reason is that for any $\epsilon>0$ there are infinitely many n's such that the averages in [-n,n] are greater than the upper Beurling number minus $\epsilon$ and there is an ultrafilter that contains this set. Hence if you choose this ultrafilter, the mean will be larger on the particular sequence than the upper Beurling number minus $\epsilon$. Similarly, you have a mean that is smaller on the sequence than the lower Beurling number plus $\epsilon$. Now, use the fact that the finite linear combination of means is a mean as well, and you can get a mean with the required property.

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Ultrafilter limits are not shift-invariant. –  Emil Jeřábek Apr 7 '11 at 11:50
Dear Gabor, it's a pleasure to hear from you! I've read a couple of your articles about sofic groups. Anyway, unfortunately I need invariant mean and ultrafilter limits are not. –  Valerio Capraro Apr 7 '11 at 13:09
It's not a big deal. Invariance is implicit in Gabor's answer (as he talks about ultralimits of averages). Essentially, he describes the same procedure as in my answer, the only difference being that he talks about ultrafilters, whereas I prefer more old-fashioned weak limit points. –  R W Apr 7 '11 at 13:28

When talking about means (which are highly non-constructive objects) it is very useful to use a much more constructive approach based on the Day-Reiter characterization of amenability. Namely, a sequence of probability measures $\lambda_n$ on a countable group $G$ (in your case it is just the group of integers $\mathbb Z$) is called (left) approximately invariant if $\|g\lambda_n-\lambda_n\|\to 0$ for any group element $g\in G$ (here $\|\cdot\|$ denotes the total variation). Any weak$^*$ limit point of an approximately invariant sequence is an invariant mean, and, conversely, any invariant mean can be obtained in this way (one can read more about it in the classical book of Greenleaf, which, at the end of Section 2.4, also contains a discussion pretty close to your question). The easiest example is provided by any sequence of uniform measures on intervals $I_n\subset\mathbb Z$ whose lengths go to infinity (in particular, any sequence of Cesaro averages, unilateral or bilateral, will do).

Now, back to your question. By the definition of the lower and upper densities, there are sequences of bilateral Cesaro averages whose limits realize these values. Therefore, by the above, there are invariant means which also realize the same values. By taking their convex combinations one can realize any intermediate value as well (one can also argue in a somewhat different way: for any given intermediate value there is a sequence of Cesaro averages whose limit realizes this value).

On the other hand, the converse is not true. Namely, it is well possible that an invariant mean gives to a set $A$ a value which is not sandwiched between its lower and upper densities. In particular, even if the lower and upper densities coincide, there may exist means with other values. The reason is very simple: one can modify any set $A\subset\mathbb Z$ in such a way that the new set $A'$ has the same upper and lower densities as $A$, but both $A'$ and its complement contain arbitrarily long intervals. Then (as it follows from the above example with intervals) there exist invariant means which give the set $A'$ values 0 and 1 (and therefore any intermediate value as well).

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You can look at $\mathbb N \subset \mathbb Z$. Then the Beurling densities conincide (and give $1/2$) whereas the invariant measure $$\mu(A) = \lim_{n \to \omega}\frac{|A \cap \{1,\dots,n\}|}n$$ gives it measure $1$. So, you have to be more careful.

What is true is that $$m_+(A) = \limsup_{k \to \infty} \ \ \sup_n \frac{|A \cap \lbrace n,\dots,n+k \rbrace|}k$$ and the corresponding $m_{-}(A)$ give the right endpoints of the interval of possible values. Again, as pointed out above, convex combinations allow for all possible values in this range.

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Thanks, Andreas, for giving a way to calculate the endpoints. This was indeed a very old post. Now I know that for any $s\in[0,1]$ there is invariant mean $m$ on $\mathbb Z$ such that $m(\chi_{\mathbb N})=s$ –  Valerio Capraro May 22 '11 at 11:28
Ok. I am sure you know this, but if you are not waiting for more answers, you could consider accepting an existing answer by clicking on the symbol below the voting count. –  Andreas Thom May 22 '11 at 12:51
ehm.. I did not know that actually! :D I'm little embarassed now.. which answer should I have to accept?! (I don't like to judge!) I have learnt something from each one. –  Valerio Capraro May 23 '11 at 21:46