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<m $

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


1 Answer 1


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 $\mu(\bigcup A_n)=\mu(\mathbb{N})=1$.

  • 1
    $\begingroup$ 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) $\endgroup$ Oct 19, 2018 at 12:14

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