Probability and the first uncountable ordinal number

Question

Let's suppose we can put a probability measure on the set of countable ordinals. (which is the same as the first uncountable ordinal). Let's now play a game. I pick a countable ordinal, say $\alpha$. Now you pick one. (Clearly our choices are independent). But $\alpha$ has countably many ordinals less than it and for you the number of choices greater than $\alpha$ are uncountable. So with a large probability, possibly one, your choice is bigger than mine. But the choices were independent! How to explain? Of course, you could take this as a proof that no such measure exists but intuitively it seems to make sense that such a game could exist.

Answer 1

A probability measure is on the set of measurable subsets of the set of outcomes, so which subsets do you want to consider measurable? If it's a discrete probability measure (i.e. all point masses), then only countably many outcomes can have positive measure (that is well known and easy to prove). But let's say you have a one-to-one correspondence between $\mathbb R$ and the set of countable ordinals (that can be done if the continuum hypothesis and the axiom of choice hold). Then say you consider all sets corresponding to Lebesgue-measurable sets to be measurable. In that case, the set of all ordinals less than any particular outcome has measure $0.$

Answer 2

This is indeed a nonexistence proof in disguise: it shows that there is no "nice" measure on $\omega_1$ according to which the set $\{\langle \alpha,\beta\rangle:\alpha<\beta\}$ is also "nice" as a subset of $(\omega_1)^2$ with respect to the product measure $\mu^2$.

Specifically, once we strip away the game aspect of the question, what you're really doing is pushing back against Fubini's theorem. We partition $(\omega_1)^2$ into three pieces: the diagonal $\{\langle\alpha,\beta\rangle: \alpha=\beta\}$, and the two "triangles" $T_1=\{\langle \alpha,\beta\rangle: \alpha<\beta\}$ and $T_2=\{\langle \alpha,\beta\rangle: \alpha>\beta\}$. As long as $\mu$ is "reasonable," the product measure will satisfy $\mu^2(D)=0$ and $\mu^2(T_1)=\mu^2(T_2)$. And of course we'll have $\mu^2((\omega_1)^2)=1$.

You now want to argue that this gives a contradiction: again as long as $\mu$ is "reasonable" we'll have $\mu(\{x: x<\eta\})=0$ for all $\eta<\omega_1$, but by Fubini's theorem this should give $\mu^2(T_1)=\mu^2(T_2)=0$ (a special case of Fubini is: "Any measurable subset of the square of the space, each of whose 'fibers' has measure zero, itself has measure zero."). But in order to apply Fubini here, we need that $T_1$ and $T_2$ are in fact $\mu^2$-measurable.

So really what this line of thought shows is that there is no measure on $\omega_1$ satisfying a couple basic properties and making the ordering relation on $\omega_1$ well-behaved.

• For example, suppose $\mathsf{CH}$ holds. Then we can "port over" the usual Lebesgue measure $\lambda$ on $[0,1]$ to a measure $\mu$ on $\omega_1$ via some bijection $f:[0,1]\rightarrow\omega_1$. This $\mu$ satisfies all the relevant reasonability conditions, but $\{\langle a,b\rangle: f(a)<f(b)\}$ is a non-Lebesgue-measurable subset of $[0,1]^2$.

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If you want to keep the "sequential" flavor of the question, we can think in terms of order of integration (and to be fair this is how Fubini is usually presented). Let $\chi:(\omega_1)^2\rightarrow\{0,1\}$ be the characteristic function of the set $\{\langle \alpha,\beta\rangle: \alpha<\beta\}$. Then as long as we have a "reasonable" notion of measure, the maps $$\alpha\mapsto \int_{\omega_1} \chi(\alpha,x)dx\quad\mbox{and}\quad\beta\mapsto\int_{\omega_1} \chi(x,\beta)dx$$ are each the constant-zero function. However, this does not mean that $$\iint_{\omega_1\times\omega_1}\chi(x,y)d(x,y)=0:$$ for that equality to hold we'd need that the whole function $\chi$ is integrable with respect to the product measure, and just because each of its "sections" is doesn't mean it itself is. In fact, the non-integrability of $\chi$ with respect to any "reasonable" measure is exactly what your argument shows.