It is well-known, and in fact not difficult to prove, that given a finitely additive measure $\mu$ of a family of sets $\{X_i\}_{i \in I}$, one can generate a filter $F_\mu = \{X \; | \; \mu(X) = 1\}$. My question is about the converse: given a filter on a family of sets $\{X_i\}_{i \in I}$, is it possible to generate a finitely additive measure $\mu_F$ such that $\mu_F(X_i) = 1$ iff $X_i \in F$? For instance, consider the Fréchet filter as defined on $\mathcal{P}(\mathbb{N})$; is there a finitely additive measure $\mu$ defined over $\mathcal{P}(\mathbb{N})$ such that $\mu(X)=1$ iff $X$ is co-finite?

(I tagged this as "set-theory" because it wouldn't surprise me if these issues relied on certain set-theoretical assumptions)

EDIT: Oh, I forgot to specify that I'm interested in probabilistic measures!


1 Answer 1


If $\mu$ is a finitely additive measure which vanishes on singletons defined on $\mathcal P(\Bbb N)$, then we can define the Bartle integral using $\mu$ which gives us a functional on $\ell^\infty$, say over the complex numbers:

For $x\in\ell^\infty$ and $\alpha\in\Bbb C$, let $x_\alpha=\{n\in\Bbb N\mid x_n=\alpha\}$.

$$\varphi_\mu(\vec x)=\int_\Bbb N\vec x\mathrm d\,\mu=\sum_{\alpha\in\Bbb C}\alpha\mu(x_\alpha)$$

You can check that this is in fact a continuous linear functional by the fact that $\mu$ is a finitely additive measure. But we can show that this linear functional is in fact not coming from a sequence in $\ell^1$, exactly because the measure vanishes on singletons.

And this means that $\ell^1$ is not reflexive, which implies the existence of irregular sets (e.g. without the Baire proprety and non-measurable).

So how does that help us? I mean, we know that these sets exists.

  1. This tells us that there is no robust way to match filters with finitely additive measures. If there was, then this matching would be definable and wouldn't rely on the axiom of choice in any kind of way, and in particular would exist in models where every set of reals has the Baire property, and no finitely additive measures exist (while the Fréchet filter always exists).

  2. This tells us that even assuming choice, not every filter corresponds to a measure. And in fact, the filters who do correspond to measures are kind of bizarre and arguably pathological. Certainly not the Fréchet filter.

  • $\begingroup$ This answer is not about probability measures, but it includes them... the only assumption is that $\mu(\Bbb N)$ is finite and that vanishes on singletons (so cofinite sets have "full measure" and then we can normalise). $\endgroup$
    – Asaf Karagila
    Commented Dec 16, 2019 at 22:12
  • $\begingroup$ You can find a lot more in Chapter 3 of "Zornian Functional Analysis: or How I Learned to Stop Worrying and Love the Axiom of Choice". $\endgroup$
    – Asaf Karagila
    Commented Dec 16, 2019 at 22:15
  • $\begingroup$ Thanks, especially for the reference for your paper. Theorem 31 there was particularly enlightening. $\endgroup$
    – Nagase
    Commented Dec 16, 2019 at 23:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .