Choice-free fair-coin content on open subsets of binary sequence space

The collection in the supremum is nonempty because it contains $\mathrm{\varnothing }$, and all its values lie in $[0,1]$ by [F1], so the real supremum exists. Monotonicity follows by inclusion of the collections being supremized. The empty and whole-space values follow directly from $p_0(\mathrm{\varnothing })=0$ and $p_0(\mathrm{\Omega })=1$. If $C\subseteq [\sigma ]$ is clopen, [F1] gives $p_0(C)\le p_0([\sigma ])$; taking $C=[\sigma ]$ gives the reverse bound and proves the cylinder formula.

Let $U={\bigcup }_i{U}_i$, and fix $C\in \mathcal{C}$ with $C\subseteq U$. The family $\{\mathrm{\Omega }\setminus C\}\cup \{U_i:i\ge 0\}$ is an open cover of $\mathrm{\Omega }$. By [F2] finitely many $U_i$, indexed by a finite set $F$, cover $C$. Consider all prefix cylinders contained in at least one $U_i$ with $i\in F$; they cover $C$ because prefix cylinders form a base. Together with $\mathrm{\Omega }\setminus C$ they cover $\mathrm{\Omega }$, so compactness gives finitely many such cylinders covering $C$. Assign each selected cylinder to the least eligible $i\in F$, and let $C_i$ be the finite union of cylinders assigned to $i$. Then $C\subseteq {\bigcup }_{i\in F}{C}_i$ and $C_i\subseteq U_i$. By [F1], $p_0(C)\le {\sum }_{i\in F}{p}_0(C_i)\le {\sum }_{i\in F}{\mu }_o(U_i)\le {\sum }_{i\ge 0}{\mu }_o(U_i)$. Taking the supremum over $C$ proves countable subadditivity. No choice of a cylinder for each point was made; the finite selections came from the stated compactness conclusion.

If $U,V$ are disjoint and open, clopen $C_U\subseteq U$ and $C_V\subseteq V$ are disjoint, so [F1] gives $p_0(C_U)+p_0(C_V)=p_0(C_U\cup C_V)\le {\mu }_o(U\cup V)$. Taking the two suprema yields ${\mu }_o(U)+{\mu }_o(V)\le {\mu }_o(U\cup V)$; step 2.1 gives the reverse inequality. Induction yields finite additivity on pairwise disjoint opens. For a disjoint sequence, monotonicity then gives ${\mu }_o({\bigcup }_i{U}_i)\ge {\sum }_{i<k}{\mu }_o(U_i)$ for every $k$; take the supremum in $k$ and combine with step 2.1 to obtain equality.

Now assume countable choice and take the Borel measure $\mu$ of [F3]. Enumerate all prefix cylinders contained in an open $U$ in the canonical length-then-binary-value order, and let $C_k$ be the union of those among the first $k$ strings that are contained in $U$. Each $C_k$ belongs to $\mathcal{C}$, the sequence increases, and ${\bigcup }_k{C}_k=U$ by [F2]. Continuity from below of $\mu$ gives $\mu (U)={\sup }_k\mu (C_k)={\sup }_k{p}_0(C_k)\le {\mu }_o(U)$. Conversely, if $C\in \mathcal{C}$ lies in $U$, then the open cover $\{C_k:k\ge 0\}\cup \{\mathrm{\Omega }\setminus C\}$ of compact $\mathrm{\Omega }$ has a finite subcover. Since the $C_k$ increase, $C\subseteq C_k$ for some $k$. Thus $p_0(C)\le p_0(C_k)\le \mu (U)$, and taking the supremum proves ${\mu }_o(U)\le \mu (U)$.