Dyadic coding supplies coin measure and its completed Lebesgue transfer

Put $I_{\mathrm{\varnothing }}=[0,1)$. Recursively split $I_s=[a,a+2^{-n})$, $\mathrm{\mid }s\mathrm{\mid }=n$, into $I_{s^{⌢}0}=[a,a+2^{-n-1})$ and $I_{s^{⌢}1}=[a+2^{-n-1},a+2^{-n})$. The two halves partition I_s, including the midpoint in the right half only. For each $x\in [0,1)$ the unique half containing x at each stage determines b(x) by F2. Thus $b^{-1}[N_s]=I_s$ at every n, including the root. If b(x)=b(y), both points lie in one interval of length $2^{-n}$ for every n, so $\mathrm{\mid }x-y\mathrm{\mid }<2^{-n}$. Since $2^n\ge n+1$, F3 implies these bounds tend to zero, giving x=y.

Assume A1's DC. Given any sequence of nonempty sets $(X_n)$, let S be the set of finite selections on initial segments, including the empty selection. Every selection of length n has an extension of length n+1, because X_n is nonempty. The relation of one-coordinate extension is entire on this nonempty set. DC with starting value empty gives a chain whose nth term has length n. Its union selects one member of every X_n, exactly countable choice. This licenses the countable-choice hypotheses of F5 and F6, without assuming AC.

Every open subset of $\mathcal{C}$ is the union of those cylinders it contains, an explicitly countably coded family by F1. Its b-preimage is the union of the corresponding I_s, hence Borel in $[0,1)$ and in $\mathbb{R}$, since these half-open intervals are Borel. The family of subsets D of $\mathcal{C}$ for which $b^{-1}[D]$ is real Borel is a sigma-algebra: preimages commute with countable unions and relative complements, the latter taken inside the Borel set [0,1). F4's leastness therefore proves b-preimages of all Borel D are Borel.

Independently in ZF define $\pi :\mathcal{C}\to [0,1]$ by the unique point of ${\bigcap }_n\overline{I_{z\upharpoonright n}}$. These are nested nonempty bounded closed intervals with lengths $2^{-n}$ tending to zero, so F3 applies. If z,w share their first n bits, the two images belong to the same closed interval and differ by at most $2^{-n}$; hence $\pi$ is continuous by F8 and F1. Since x belongs to every interval chosen by b(x), uniqueness gives $\pi (b(x))=x$. Thus $\pi$ is injective on b[A] for every A.

Define $\nu (D)=\lambda (b^{-1}[D])$ on Borel D. Step 2.1 and F5 make the expression defined. Preimages of disjoint sequences are disjoint, so F5 and F7 give $\nu ({\bigcup }_n{D}_n)={\sum }_n\nu (D_n)$ and $\nu (\mathrm{\varnothing })=0$. By F6 and step 1.1, $\nu (N_s)=\lambda (I_s)=2^{-\mathrm{\mid }s\mathrm{\mid }}$, in particular $\nu (\mathcal{C})=1$. Thus it is a probability measure. F7's continuity from below applies to any increasing Borel sequence; continuity from above applies to any decreasing one because its first measure is at most one.

The inner supremum and outer infimum are over nonempty bounded sets of values: K empty and O whole are admissible. If $K\subseteq E\subseteq O$, monotonicity gives $0\le \nu (K)\le \nu (O)\le 1$, proving the four inequalities. Complementation bijects closed $K\subseteq \mathcal{C}\setminus E$ with open $O\supseteq E$. Finite additivity gives $\nu (O)=1-\nu (K)$, so taking the infimum on one side and supremum on the other proves the complement identity.

If both envelope values equal t, their supremum and infimum definitions give, for each n, a closed $K_n\subseteq E$ and open $O_n\supseteq E$ with $\nu (O_n)-\nu (K_n)<2^{-n}$. Indeed choose each value within $2^{-n-1}$ of t; if t=0 the empty K suffices, and if t=1 the whole O suffices. Step 1.2 selects these pairs simultaneously. Put $K={\bigcup }_n{K}_n$, $O={\bigcap }_n{O}_n$. They are Borel with $K\subseteq E\subseteq O$. For each n, $O\setminus K\subseteq O_n\setminus K_n$, whose measure is the displayed difference; hence $\nu (O\setminus K)=0$ by step 4.1 and the shrinking bound. Their b-preimages are Borel by step 2.1, and the difference is Lebesgue null by definition of $\nu$. Completeness F5 makes every subset of that difference measurable, so $b^{-1}[E]$, lying between those two Borel sets, is measurable.

Let L be a compact Cantor copy in b[A]. Then $\pi {\mathrm{\mid }}_L$ is a continuous injection into A by step 2.2. Its image is compact: pull an open cover back and use F8's finite-subcover condition. A compact set in a metric space is closed, since for an exterior point x the balls $B(y,d(x,y)\mathrm{/}3)$ about compact-set points have a finite subcover, and a ball about x smaller than all the corresponding radii avoids the compact set. Closed subsets of L are compact by adjoining the open complement to a cover; their images are therefore closed by this same separation argument. Consequently the inverse of $\pi {\mathrm{\mid }}_L$ is continuous, and the image is a compact Cantor copy in A. No unique binary expansion at dyadic endpoints was required; only the one-sided inverse equation for the fixed half-open coding was used. QED.