The fair-coin measure on Cantor space

For finite $F\subseteq \mathbb{N}$ and $a\in 2^F$ put $[a{]}_F:=\{x\in \mathcal{C}:x(i)=a(i)\text{ for every }i\in F\}$; each $[a{]}_F$ is open because the topology of $\mathcal{C}$ consists of unions of cylinders, and closed because $\mathcal{C}\setminus [a{]}_F=\bigcup \{[b{]}_F:b\in 2^F,b\mathrm{\ne }a\}$ is a finite union of cylinders and hence open; in particular $[\mathrm{\varnothing }{]}_{\mathrm{\varnothing }}=\mathcal{C}$. Consequently the family ${\mathcal{C}}_0$ of all finite unions of cylinders, the empty union included, contains $\mathrm{\varnothing }$ and $\mathcal{C}$ and is closed under complements, and it is closed under finite intersections because $[a{]}_F\cap [b{]}_G=[a\cup b{]}_{F\cup G}$ when $a$ and $b$ agree on $F\cap G$ and is $\mathrm{\varnothing }$ otherwise; by de Morgan it is therefore an algebra of subsets of $\mathcal{C}$.

If $F\subseteq G$ are finite and $a\in 2^F$, then the prescriptions $b\in 2^G$ with $b\upharpoonright F=a$ are in bijection with the functions $G\setminus F\to 2$, so there are $2^{\mid G\setminus F\mid }$ of them, and $[a{]}_F$ is their pairwise disjoint union: a point of $[a{]}_F$ agrees with exactly one such $b$ on $G$.

The fair-coin content is well defined on ${\mathcal{C}}_0$. Let $A\in {\mathcal{C}}_0$ and let $A=[a_0{]}_{F_0}\cup \cdots \cup [a_{\mathrm{\ell }}{]}_{F_{\mathrm{\ell }}}$ be a presentation; put $G:=F_0\cup \cdots \cup F_{\mathrm{\ell }}$, a finite set. By step 1.2 each $[a_i{]}_{F_i}$ is a disjoint union of cylinders $[b{]}_G$, and these lie inside $A$; since the $G$-cylinders partition $\mathcal{C}$, $A$ is the disjoint union of the $G$-cylinders contained in it, and we let $m(A,G)$ be their number and set $p_0(A):=m(A,G)\cdot 2^{-\mid G\mid }$. If $G\subseteq H$ are finite, step 1.2 splits each $G$-cylinder into $2^{\mid H\setminus G\mid }$ many $H$-cylinders, so $m(A,H)=m(A,G)\cdot 2^{\mid H\setminus G\mid }$ and $m(A,H)2^{-\mid H\mid }=m(A,G)2^{-\mid G\mid }$ because $\mid H\mid =\mid G\mid +\mid H\setminus G\mid$ and $2^{-\mid G\mid -\mid H\setminus G\mid }=2^{-\mid G\mid }{2}^{-\mid H\setminus G\mid }$; two finite sets containing all $F_i$ are compared through their union, which contains both, so the value does not depend on the presentation or on $G$.

From step 2.1, $p_0(\mathrm{\varnothing })=0$, $p_0(\mathcal{C})=1$ and $p_0([a{]}_F)=2^{-\mid F\mid }$; if $A,B\in {\mathcal{C}}_0$ are disjoint and $G$ is finite and contains the supports of presentations of both, then the $G$-cylinders inside $A\cup B$ are exactly those inside $A$ together with those inside $B$, because a $G$-cylinder meets the disjoint sets $A$ and $B$ in all of itself or in nothing, so $m(A\cup B,G)=m(A,G)+m(B,G)$ and $p_0(A\cup B)=p_0(A)+p_0(B)$; hence $p_0$ is monotone, and $0\le p_0(A)\le 1$ for every $A\in {\mathcal{C}}_0$.

Let $(A_k{)}_{k\in \mathbb{N}}$ be pairwise disjoint members of ${\mathcal{C}}_0$ whose union $A$ lies in ${\mathcal{C}}_0$. For every $m$ the set $A_0\cup \cdots \cup A_m$ is contained in $A$, so step 3.1 gives ${\sum }_{k\le m}{p}_0(A_k)=p_0(A_0\cup \cdots \cup A_m)\le p_0(A)$ and hence ${\sum }_{k\in \mathbb{N}}{p}_0(A_k)\le p_0(A)$. Conversely $A$ is a finite union of closed cylinders, hence closed by step 1.1 and [F3], so it is a compact subset of the compact space $\mathcal{C}$; the $A_k$ are open and cover $A$, so the finite-subcover property read in the ambient space through [F5] gives $m$ with $A\subseteq A_0\cup \cdots \cup A_m$, and then ${\sum }_{k\in \mathbb{N}}{p}_0(A_k)\ge {\sum }_{k\le m}{p}_0(A_k)=p_0(A_0\cup \cdots \cup A_m)\ge p_0(A)$ by step 3.1. Hence $p_0$ is a premeasure on the algebra ${\mathcal{C}}_0$.

Assume the Axiom of Countable Choice. By [F4] the outer set function ${\mu }^{\ast }$ induced by the premeasure $p_0$ has a restriction $\mu :={\mu }^{\ast }\upharpoonright \sigma ({\mathcal{C}}_0)$ that is a measure on the sigma-algebra generated by the cylinders, and $\mu (A)=p_0(A)$ for every $A\in {\mathcal{C}}_0$; in particular $\mu (\mathcal{C})=p_0(\mathcal{C})=1$, so $\mu$ is a probability measure.

The cylinders form a base of the topology of $\mathcal{C}$ and each is a union of open sets, so the sigma-algebra they generate is the Borel sigma-algebra: $\sigma ({\mathcal{C}}_0)=\mathcal{B}(\mathcal{C})$. Hence $\mu$ is a Borel probability measure and $\mu ([a{]}_F)=p_0([a{]}_F)=2^{-\mid F\mid }$ for every finite $F$ and $a\in 2^F$; for a finite binary word $s$ the cylinder $N_s$ is $[s{]}_F$ with $F=\{0,\dots ,\mid s\mid -1\}$, so $\mu (N_s)=2^{-\mid s\mid }$.

If $F\cap G=\mathrm{\varnothing }$, $a\in 2^F$ and $b\in 2^G$, then $[a{]}_F\cap [b{]}_G=[a\cup b{]}_{F\cup G}$ by step 1.1, so $\mu ([a{]}_F\cap [b{]}_G)=2^{-\mid F\cup G\mid }=2^{-\mid F\mid -\mid G\mid }=\mu ([a{]}_F)\mu ([b{]}_G)$, the middle equality using $\mid F\cup G\mid =\mid F\mid +\mid G\mid$ and the power laws.

Steps 6.1 and 6.2 are the two claims: $2^{\mathbb{N}}$ carries the Borel probability measure $\mu$ with $\mu (N_s)=2^{-\mid s\mid }$, obtained as the Carathéodory extension of the fair-coin content, and $\mu$ is multiplicative on cylinders with disjoint coordinate sets. ∎