Understanding the range of a strange finitely additive measure

Question

I came across the following example.

Let $\mu_0$ be a finitely additive probability measure defined on every subset of $\mathbb N$ such that $\mu_0(A)=0$ if $A$ is finite. For $n \geq 1$, let $\mu_n$ be point mass at $n$, i.e. the probability measure defined for every subset of $\mathbb N$ by $\mu_n(\{n\})=1$. Let

$$\mu = \sum_{n=0}^\infty \frac{1}{2^{n+1}} \mu_n$$

Claim $\mu$ does not take the value $1/2$.

Here's what I see. If $A$ is finite, then $\mu_0(A)=0$ and $\sum_{n \geq 1} 1/2^{n+1} \mu_n(A) < 1/2$ because all but finitely many terms in the series are $0$, so $\mu(A) < 1/2$. If $A$ is co-finite, then $\mu_0(A)=1$ and $\sum_{n \geq 1} 1/2^{n+1} \mu_n(A) > 0$ because infinitely many terms in the series are positive, so $\mu(A) > 1/2$.

How to conclude for infinite $A$ with infinite complement? I guess I'm failing to grasp something about the possible values that sub-series of $\sum_{n \geq 1} 1/2^{n+1}$ can take.

Added. The reasoning above goes through for the case where $\mu_0(A)=0$, whether or not $A$ is finite, and the case where $\mu_0(A)=1$, whether or not $A$ is co-finite. So, really, my question is how to conclude when $\mu_0(A) \in (0,1)$.

Answer 1

A generalised limit will be defined as an extension of the classical limit functional $(a_n)_n \mapsto \lim_{n \to \infty} a_n$ from the space of convergent sequences to the whole of $\ell^\infty$ with operator norm $1$.
You can refer to this forum post by Terence Tao for more information on generalised limits, and Chapter VII of Sequences and Series in Banach Spaces by Diestel for an account on the (linear isometric surjective) correspondence between finitely additive measures and the dual of $\ell^\infty$, of the form: $$\mu \leftrightarrow L$$ $$\|\mu\|_{TV} = \|L\|_{(\ell^\infty)^*}$$ $$\mu(E) = L(1_E)$$ $$L(u) = \int_{\mathbb{N}} u\mathrm{d}\mu$$

where $1_E$ is the indicator sequence of $E$ and $\|\cdot\|_{TV}$ the total variation norm.

One can prove two facts:

• The aforementioned correspondence induces another correspondence between finitely additive, positive, probability measures vanishing on finite sets and generalised limits (this is based on a careful study of said correspondence);
• For any divergent sequence $u$ and any $\alpha \in [\liminf u, \limsup u]$, one can define a generalised limit $L$ such that $L(u) = \alpha$ (this is an application of the Hahn-Banach theorem).

It then suffices to take any set such that $1_E$ diverges, $E := 2\mathbb{N}$ for example, and choose a generalised limit $L_0$ such that: $$L_0(1_E) = 1 - \sum_{n \geq 1} \frac{\mu_n(E)}{2^n} \in [0,1] = [\liminf 1_E, \limsup 1_E]$$

It is then evident that, if $\mu_0$ is the measure corresponding to $L_0$, we have: $$\mu(E) = \frac{1}{2} \mu_0(E) + \frac{1}{2}\sum_{n \geq 1} \frac{\mu_n(E)}{2^n} = \frac{1}{2} \left(1 - \sum_{n \geq 1} \frac{\mu_n(E)}{2^n} + \sum_{n \geq 1} \frac{\mu_n(E)}{2^n}\right) = \frac{1}{2}$$ thus disproving your claim as it stands.