Let $X$ be a countable set and $\mathcal M(X)$ be the set of **finitely additive** probability measures on $X$. If $\mu\in\mathcal M(X)$, I define the entropy of $\mu$ to be


$$
E(\mu)=\sup\left\{\left.-\sum\mu(A\_i)\log(\mu(A\_i))\\ \right|\\ X=\bigcup A\_i, A\_i \text{ pairwise disjoint}\right\}
$$


If $\mu$ is a Dirac measure, then $E(\mu)=0$. I am looking for non trivial examples for which this happens. 



> 
> **Question:** Does there exist a finitely additive (non countably additive) probability measure with zero entropy?
> 
> 
> 


P.s. using AC, one can ''construct'' an example taking the measure defined by a free ultrafilter on $X$. But... I would prefer to avoid the use of AC.


P.p.s. I could ask the same question also for $E(\mu)=\infty$, since the unique examples that I have in mind make use of the existence of an invariant measure on an amenable group, which follows by Hahn-Banach.


I have the feeling that the problem is deeper: it is difficult to prove the existence of a finitely additive non countably additive probability measure without using some version of AC. I am just asking if there is any particular case..



existence of finitely additive measures with zero entropy
Posted above as a comment, but re-posted here as an answer following suggestion of @Benoit Kloeckner so original q does not appear unanswered. 


The first question is equivalent to the existence of a non-principal ultrafilter: a measure has entropy 0 if and only if every set has measure 0 or 1. This question was asked (and answered) at [Non-principal ultrafilters on ω](https://mathoverflow.net/questions/15872/)


The second question is equivalent to the existence of a strictly finitely additive measure. Given any strictly finitely additive measure $\mu$, you can take its average with the countably additive measure $\nu(n)=C/(n\log n)^2$ to get a measure with infinite entropy.


