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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.


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You can try and take $\tau$ to be the counting measure on $\omega$ and then define $\tau_n = 2^{-n}\tau$, then take the product measure to be the product of all these. – Asaf Karagila Jun 17 '12 at 7:04
up vote 3 down vote accepted

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.

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I never got around to accepting this answer. Three years later, fixed! :) – Noah Schweber Jun 16 '15 at 2:14
Haha! Better late than sorry, I guess. :-) – Asaf Karagila Jun 16 '15 at 3:55

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.

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Could you explain what exactly is $\omega^{< \omega}$? is that subspace of elements with finite $supp$ (finite non-zero prefix)? – Stephen Dedalus Mar 5 '14 at 17:58
@Baranovskiy $\omega^{\omega}$ is the set of finite strings of natural numbers. – William Mar 11 '14 at 9:42

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