Every integer-base circle map is strongly mixing

The b-adic intervals at all depths, together with the empty set, form a pi-system: two such intervals are nested or disjoint, and depth zero is X. They generate the circle Borel sets. Indeed every ordinary open interval in (0,1) is the union of the b-adic cells whose closures lie within it; the cell containing any specified interior point has arbitrarily small diameter. The endpoint zero is the intersection of $[0,b^{-r})$ over r, so relative interval-open sets are generated as well. Conversely each cell is Borel. The countable family of cells permits these unions in the generated sigma-algebra.

Let $I=I_{r,k}$ and $J=I_{s,l}$. For $n\ge r$, the interval I contains exactly $b^{n-r}$ depth-n cells. On each such cell, the inverse image under $D_b^n$ of J is a half-open interval of length $b^{-(n+s)}$. Hence $\lambda (I\cap D_b^{-n}J)=b^{n-r}{b}^{-(n+s)}=b^{-r}{b}^{-s}=\lambda (I)\lambda (J)$. The same equality holds when either set is empty. The generator theorem proves Borel strong mixing.

For completed A,B take Borel A_0,B_0 and Borel null sets Z_A,Z_B containing their respective symmetric differences. For every n, $(A\cap D_b^{-n}B)\mathrm{△}(A_0\cap D_b^{-n}{B}_0)\subseteq Z_A\cup D_b^{-n}{Z}_B$, a completed null set by preservation. Thus the correlation and both marginal measures agree with those for A_0,B_0. The Borel limit proves the completed limit. Countable choice is inherited from the measure and completion suppliers; only finitely many representatives are selected here.