# Unicity of additive, $(-1)$-homogeneous, and shift invariant probability measures on $\mathbf N^+$

Let $\mathcal D$ be the set of all (finitely) additive probability measures $\mu^\ast: \mathcal P(\mathbf N^+) \to [0,\infty[$ such that $\mu^\ast(k \cdot X + h) = \frac{1}{k} \mu^\ast(X)$ for all $X \subseteq \mathbf N^+$ and $h,k \in \mathbf N^+$. It is proved in [1, Section 10] that $\mathcal D$ is nonempty, provided that we lay out the foundations of our mathematical believes, say, in ZFC (see here for some discussion on this point). However, [1] doesn't imply anything, insofar as I can tell, about the size of $\mathcal D$ (again, in ZFC). So my question is:

Do you know of any place in the literature where it's proved, say, that $\mathcal D$ has more than one element? And what about $\mathcal D$ being infinite?

I've also read through [2], but it's not clear to me whether the answer is there or not.

Bibliography.

[1] R. P. Agnew and A. P. Morse, Extensions of linear functionals, with applications to limits, integrals, measures, and densities, Ann. of Math. (2) 39 (1938), No. 1, 20-30.

[2] E. K. van Douwen, Finitely additive measures on $\mathbb N$, Topology Appl. 47 (1992), No. 3, 223-268.

This is basically Second construction 5.3 from Eric K. van Douwen. Finitely additive measures on $\mathbb N$. Topology Appl., 47 (3), (1992), 223–268. MR1192311 (94c:28004). This issue of Topology and Its Applications was a special issue dedicated to Eric K. van Douwen.

EDIT: Only after making the construction I posted below I returned to the paper and noticed that it contains Appendix 5C: The number of elastic measures. It is shown here that the set of elastic measures has cardinality $2^{\mathfrak c}$. Every elastic measure has properties required in the above question.

So this gives a better result than I obtained below. I have decided to keep the original version of my answer here. Just in case some of the information given there might be interesting for some people reading this question.


For any filter $\FF$ we can define the $\FF$-limit of a sequence $(x_n)$ as $$\Flim x_n=L \qquad\Leftrightarrow\qquad (\forall \varepsilon>0) \{n; |x_n-L|<\varepsilon\}\in\FF.$$ If $\FF$ is a free ultrafilter, then $\Flim x_n$ exists for each bounded sequence. Moreover, we know that for a free ultrafilter $\FF$ we have

• If $(x_n)$ is convergent, then $\Flim x_n=\lim x_n$ (=extends limit);
• $\Flim (x_n+y_n) = \Flim x_n + \Flim y_n$ and $\Flim (cx_n)=c\Flim x_n$ (=linearity);
• $x_n\le y_n$ $\Rightarrow$ $\Flim x_n\le\Flim y_n$ (=is positive);
• For a given sequence $(x_n)$ and any cluster point $c$ of $(x_n)$ there exists a free ultrafilter such that $\Flim x_n=c$.

Some references for $\FF$-limits can be found here or in the comments to this post. (There is also a Wikipedia article on ultralimit. However, this article discusses ultralimits of sequences only very briefly.)

For $A\subseteq\mathbb N$ we define $$\mu_n(A) = \frac{\sum_{j\le n}\frac1j \chi_A(j)}{\ln n}.$$

Notice that $\overline\delta(A)=\limsup\limits_{n\to\infty} \mu_n(A)$ is precisely the upper logarithmic density. For limit inferior we get the lower logarithmic density $\underline\delta(A)=\liminf\limits_{n\to\infty} \mu_n(A)$.

Let $\FF$ be a free ultrafilter. We define $$\mu(A)=\Flim \mu_n(A).$$ (Maybe $\mu_F$ would be a better notation, since it depends on the choice of $\FF$, but I will use $\mu$ for brevity.)

It is relatively easy to see that $\mu(A)$ is a finitely additive measure on $\mathbb N$.

We will show below that $\mu(kA+h)=\frac1k\mu(A)$.

Now if we fix some set $A$ such that $\underline\delta(A)=0$ and $\overline\delta(A)=1$, then every point in the interval $[0,1]$ is a cluster point of the sequence $(\mu_n(A))$. (Note that the set of all cluster points of this sequence is connected.)

So there are at least $\mathfrak c$ such measures, since $\mu(A)$ can attain all values between $0$ and $1$ (for various choices of the free ultrafilter $\FF$).

So it remains to show that the measures of the form described above fulfill the condition $$\mu(kA+h)=\frac1k\mu(A).$$ In fact, this is shown van Douwen's paper. But since I changed notation a little bit I will include sketch of the proof.

$\mu$ is shift-invariant, i.e., $\mu(A+1)=\mu(A)$.

Let $B=A+1$. We have $$|\mu_n(A)-\mu_n(B)| \le \frac{\frac1n+\sum_{k<n} \left(\frac1k-\frac1{k+1}\right)}{\ln n}\le\frac1{\ln n}.$$ This implies that $\lim\limits_{n\to\infty} (\mu_n(A)-\mu_n(B)) =0$, hence $\Flim (\mu_n(A)-\mu_n(B))=0$ which means that $\Flim\mu_n(A)=\Flim\mu_n(B)$ and $$\mu(B)=\mu(A).$$

$\mu$ is $\mathbb N$-scale invariant, i.e., $\mu(kA)=\frac1k\mu(A)$ for $k\in\mathbb N$.

Let $B=kA$. Now we have $$\mu_n(B)=\frac{\sum_{j\le n}\frac1j \chi_B(j)}{\ln n} = \frac{\sum_{j\le n/k}\frac1{kj} \chi_A(j)}{\ln n} = \frac1k \frac{\sum_{j\le n/k}\frac1j \chi_A(j)}{\ln n}.$$ This yields $$|\frac1k\mu_n(A)-\mu_n(B)| \le \frac1k \frac{\sum_{n/k<j\le n} \frac1j}{\ln n} \sim \frac1k \cdot \frac{\ln k}{\ln n}.$$ Again $\lim\limits_{n\to\infty} (\frac1k\mu_n(A)-\mu_n(B)) =0$ and $\mu(B)=\frac1k\mu(A)$.