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](https://mathoverflow.net/questions/209067/ref-request-additive-probability-measure-on-mathcal-p-bf-n-supplies-sub) for some discussion on this point). However, [[1](https://mathoverflow.net/questions/209067/ref-request-additive-probability-measure-on-mathcal-p-bf-n-supplies-sub)] 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](https://mathoverflow.net/questions/209067/ref-request-additive-probability-measure-on-mathcal-p-bf-n-supplies-sub)] 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.



Unicity of additive, $(-1)$-homogeneous, and shift invariant probability measures on $\mathbf N^+$
This is basically Second construction 5.3 from Eric K. van Douwen. *[Finitely additive measures on $\mathbb N$](http://dx.doi.org/10.1016/0166-8641(92)90032-U)*. Topology Appl., **47 (3)**, (1992), 223–268. [MR1192311 (94c:28004)](http://www.ams.org/mathscinet-getitem?mr=1192311). This issue of Topology and Its Applications was a special issue [dedicated to Eric K. van Douwen](http://dx.doi.org/10.1016/0166-8641(92)90025-U). 



> 
> **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.
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> 
> 




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$\newcommand{\Flim}{\operatorname{\mathcal F-\lim}}\newcommand{\FF}{\mathcal F}$
The author uses diffuse mean in this construction. I will use $\FF$-limit, which is also a diffuse mean. (See that paper for terminology.) I will summarize which properties of limit along an ultrafilter I need. (Of course, you can use some other functional fulfilling these properties instead of $x\mapsto \Flim x\_n$. I believe that such functionals can be obtained, for example, [using Hahn-Banach theorem](https://math.stackexchange.com/questions/55651/nonnegative-linear-functionals-over-l-infty).)




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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](https://math.stackexchange.com/questions/51476/basic-facts-about-ultrafilters-and-convergence-of-a-sequence-along-an-ultrafilte) or in the comments to [this post](https://math.stackexchange.com/questions/1045827/references-on-filter-quantifiers). (There is also a [Wikipedia article on ultralimit](https://en.wikipedia.org/wiki/Ultralimit). However, this article discusses ultralimits of sequences only very briefly.)




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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](https://math.stackexchange.com/questions/195646/if-a-sequence-satisfies-lim-limits-n-to-inftya-n1-a-n-0-then-the-set).)


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 $\FF$).




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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 $\mu(B)=\frac1k\mu(A)$. 


