# Is there a finitely additive measure on R which is not sigma-additive?

Consider the usual measurable space of real number $$( \mathbb{R}, \mathcal{B}(\mathbb{R}))$$. My question is:

Is there an application $$\mu$$ on $$\mathcal{B}( \mathbb{R}) \rightarrow [0,+\infty]$$ such that :

i) $$\mu$$ is finitely additive

ii) $$\mu( \mathbb{R}) < \infty$$

iii) $$\exists (A_n) \subset \mathcal{B}( \mathbb{R})$$ that:

a) $$A_n \subset A_m \forall n

b) $$\lim_{n\rightarrow \infty} \mu(A_n) < \mu( \cup_{n} A_n)$$

## 1 Answer

Yes. Take a non-principal ultrafiler $$\omega$$ on $$\mathbb{N}$$, and define $$\mu$$ as follows:

$$\mu(A)=0$$ if $$A\cap\mathbb{N}$$ does not belong to $$\omega$$,

$$\mu(A)=1$$ if $$A\cap \mathbb{N}$$ belongs to $$\omega$$.

This measure if finitely additive, and if $$A_n=\{0,1,\ldots,n\}$$, then $$\mu(A_n)=0$$ for every $$n$$, while $$\mu(\bigcup A_n)=\mu(\mathbb{N})=1$$.

• On the side of plain existence, we may also argue: since $c_0$ is not reflexive, neither is its dual $\ell_1$, meaning there are bounded additive, non $\sigma$-additive measures on $\mathbb{N}$ (signed, but then also positive) Oct 19, 2018 at 12:14