Measure on Baire space

Question

Inspired by the first parenthetical sentence of Joel's answer to this question, I have the following question: is there any useful notion of measurability in the Baire space $\omega^\omega$?

Some initial thoughts:

• Any countably additive, translation-invariant measure has to give measure 0 to any set of the form $$\lbrace f: \sigma\prec f\rbrace.$$ The same holds if we ask for a finitely additive translation-invariant measure which gives finite measure to the whole space.

• There are several natural surjections $\omega^\omega\rightarrow 2^\omega$, and the latter space has a nice measure theory; so we could, for any such surjection $s$, define a "measure" given by $m_s(X)=m(s[X])$. However, it's unclear to me that any of these would be particularly interesting.

• Even short of a decent "measure" on $\omega^\omega$ (and indeed, I believe none exists), we might still be able to talk about subsets of $\omega^\omega$ being "measurable" or "non-measurable" in terms of what sorts of pathologies they admit. For example, call $X\subseteq\omega^\omega$ definably measurable if $L(\mathbb{R}, X)\models $ "Every set is measurable." This is, however, a terrible notion of measurability: there are measure zero (and hence measurable) sets $X\subseteq 2^\omega$ such that $L(\mathbb{R}, X)$ contains a non-measurable set, so this is too restrictive.

Thoughts?

Answer 1

First note that there is an extremely natural $\sigma$-algebra for this: the Borel $\sigma$-algebra generated by the open sets and in this case the clopen cones. We can later complete it for the Lebesgue $\sigma$-algebra.

Now note that if $X$ and $Y$ are uncountable Polish spaces then their Borel sets are isomorphic. Thus we can easily endow $\omega^\omega$ with a Borel measure (and a Lebesgue measure by completion).

The point of a measure is to allow us compare which sets are big and which sets are small. In this aspect the actual measure means nothing, only the $\sigma$-algebra is important.

Answer 2

There is a natural measure on Baire Space. Note that $\omega^\omega$ has basic open sets $[|\sigma|] = \{f \in \omega^\omega : \sigma \prec f\}$ where $\sigma \in \omega^{<\omega}$.

Then define $\mu([|\sigma|]) = \prod_{k = 0}^{n - 1}\frac{1}{2^{\sigma(k) + 1}}$ where $n = |\sigma|$.

Then a measure on $\omega^\omega$ is given by the usual extension of outer measure.