Finitely additive function on an infinite set, s.t., $m(A)=0$ for any finite set and $m(X)=1$ (constructive approach) Other exercise which I found in Dudley's Analysis book: 

Show that there is a measure on a infinite set $X$, defined on $2^X$ s.t. is finitely additive, $m(A)=0$ for any finite set and $m(X)=1$.

The solution is very simple using the Frechet filter $\mathcal{F}:=\{A: X\setminus A \text{ finite}\}$ and defining the measure on the ultrafilter $\mathfrak{U}$ containing $\mathcal{F}$, as follows 
$$m(A)=\begin{cases} 1& A\in \mathfrak{U}\\
0&A\notin \mathfrak{U} \end{cases}$$
For the following

Lemma: Let $\mathfrak{U}$ be an ultrafilter of subsets of $X$ and let $m$ defined as above. Then $m$ is finitely additive on $2^X$.

PF: It's clear that $\varnothing\notin \mathfrak{U}$, so $m(\varnothing)=0$. Let $\{A_n\}_{n=1}^N\subset 2^{X}$ and disjoint, and let $A$ be their union. We consider two cases: If all are not in $ \mathfrak{U}$, i.e, $X\setminus A_n\in \mathfrak{U}$ for $n\le N$. Thus $X\setminus A=\bigcap_{n\le N}X\setminus A_n\in \mathfrak{U}$, so $A\notin \mathfrak{U}$. Hence $m(A)=\sum_{n\le N}m(A_n)=0$
Now suppose that at least one is in $\mathfrak{U}$. Let $A_1\in \mathfrak{U}$, so all the other elements are not in $\mathfrak{U}$ since otherwise $\varnothing=A\cap A_i\in \mathfrak{U}$ for $i\not=1$. Since $A_1\subset A$ and $A_1\in \mathfrak{U}$,  $A\in\mathfrak{U}$. Thus $m(A)=\sum_{n\le N}m(A_n)=1$. $\Box$
Does someone know if there is hope of a constructive approach? I believe the answer is negative...
 A: It is known that existence of a finitely additive measure defined on the whole powerset $2^{\mathbb N}$ which vanishes on finite sets cannot be proved in ZF. So we cannot hope in a constructive proof of existence of such measures.
Some posts where you can find further references:


*

*ZF and the Existence of Finitely additive measure on $\mathcal{P}(\mathbb{R})$

*How to construct a continuous finite additive measure on the natural numbers at MathOverflow

*Ref. request: Additive probability measure on $\mathcal P({\bf N})$ supplies subset of $\mathbf R$ without Baire property at MathOverflow


Some references where similar questions are studied:


*

*Howard, Rubin: Consequences of the Axiom of Choice, AMS, 1998. (See AMS website or Google Books.) This book is accompanied by a website where you can search for implications between various consequences of AC and search for models in which a particular consequence of AC holds (or does not hold). Existence of a nonprincipal measure on $2^{\mathbb N}$ is Form 222 in this book. Existence of non-principal measure is Form 221.

*E. Schechter: Handbook of Analysis and Its Foundations; It is shown in 29.37 that existence of such measures is equivalent (in ZF) to $\ell_1\subsetneq\ell_\infty^*$ and in 28.38 that it implies existence of a set without Baire property.

*David Pincus and Robert M. Solovay: Definability of measures and ultrafilters. J. Symbolic Logic 42 (1977), no. 2, 179–190.

*Eric K. van Douwen. Finitely additive measures on $\mathbb N$. Topology Appl., 47 (3), (1992), 223–268. 

