Complex circle measures have finite regular total variation under countable choice

For every supplied disjoint Borel sequence $(D_j)$, ${\sum }_j\mathrm{\mid }\nu (D_j)\mathrm{\mid }$ is finite. To see this directly from [F1], split the real parts into their nonnegative and negative index groups. On each group, replace other cells by the empty set; [F1] says the complex series converges to the finite measure of that group's union. Its real part therefore has a finite sum of terms of one sign. Thus ${\sum }_j\mathrm{\mid }\mathrm{Re}\nu (D_j)\mathrm{\mid }<\mathrm{\infty }$. The two imaginary sign groups give ${\sum }_j\mathrm{\mid }\mathrm{Im}\nu (D_j)\mathrm{\mid }<\mathrm{\infty }$ as well. Since $\mathrm{\mid }z\mathrm{\mid }\le \mathrm{\mid }\mathrm{Re}z\mathrm{\mid }+\mathrm{\mid }\mathrm{Im}z\mathrm{\mid }$, the asserted absolute sum is finite. Empty groups cause zero sums.

Suppose $\mathrm{\mid }\nu \mathrm{\mid }(\mathbb{T})=\mathrm{\infty }$. For each $n\ge 0$ there is a finite ordered Borel partition ${\mathcal{P}}_n$ with sum of absolute measures greater than $2^n$: take a finite initial portion of a countable partition whose sum exceeds that threshold, and append its complement. CC supplies the sequence $({\mathcal{P}}_n)$. Let ${\mathcal{Q}}_n$ be the finite common refinement of ${\mathcal{P}}_0,\dots ,{\mathcal{P}}_n$, ordered lexicographically by their cell indices; empty cells may be retained. For Borel E put $S_n(E)={\sum }_{C\in {\mathcal{Q}}_n}\mathrm{\mid }\nu (E\cap C)\mathrm{\mid },V(E)={\sup }_n{S}_n(E)\mathrm{.}$ Refinement and the triangle inequality make $S_n(E)$ nondecreasing, and $V(\mathbb{T})=\mathrm{\infty }$. For any fixed m, refinement gives $S_n(E)={\sum }_{C\in {\mathcal{Q}}_m}{S}_n(E\cap C)$ for $n\ge m$; passing to the limit in this finite sum gives $V(E)={\sum }_{C\in {\mathcal{Q}}_m}V(E\cap C)\mathrm{.}$

Define a nested sequence deterministically, starting with $E_0=\mathbb{T}$ and index $m_0=-1$. Given $V(E_k)=\mathrm{\infty }$, take the least $n>m_k$ for which $S_n(E_k)>\mathrm{\mid }\nu (E_k)\mathrm{\mid }+2$. Among the finitely many cells of ${\mathcal{Q}}_n$ contained in $E_k$, choose the first C with $V(C)=\mathrm{\infty }$, possible by the finite-sum identity in step 2.1. Set $E_{k+1}=C$, $m_{k+1}=n$, and retain all other cells of the refinement inside $E_k$ as side cells $D_{k,j}$. Here $E_k$ is a cell of the previous refinement for $k>0$, so the new cells partition it. Write $s_k={\sum }_j\mathrm{\mid }\nu (D_{k,j})\mathrm{\mid }$. Finite additivity gives $\mathrm{\mid }\nu (C)\mathrm{\mid }\le \mathrm{\mid }\nu (E_k)\mathrm{\mid }+s_k$, while $S_n(E_k)=\mathrm{\mid }\nu (C)\mathrm{\mid }+s_k$, hence $s_k\ge \frac{S_n(E_k)-\mathrm{\mid }\nu (E_k)\mathrm{\mid }}{2}>1.$ Side cells from different stages are disjoint, since later parents lie in the retained nested child. Concatenating the prescribed finite ordered side lists is a disjoint countable Borel sequence with total absolute sum ${\sum }_k{s}_k=\mathrm{\infty }$, contradicting step 1.1. The recursion uses least natural numbers and first indices in supplied finite lists; it spends no dependent choice. Therefore $\mathrm{\mid }\nu \mathrm{\mid }(\mathbb{T})<\mathrm{\infty }$.

We also prove that variation is a measure directly. It has value zero on the empty set. Let $E={⨆}_j{E}_j$ be a supplied disjoint Borel union. Every piece has finite variation by step 3.1, since its partitions extend to partitions of the circle by appending the complement. For fixed N and epsilon, choose partitions of the first $N+1$ pieces within $\epsilon \mathrm{/}(N+1)$ of their variation suprema; [F2] supplies these finitely many choices. Concatenate their cells and append the remainder of E. The resulting partition gives $\mathrm{\mid }\nu \mathrm{\mid }(E)\ge {\sum }_{j=0}^N\mathrm{\mid }\nu \mathrm{\mid }(E_j)-\epsilon$. Let epsilon decrease to zero and then N increase to infinity. Conversely, for every partition $(B_l)$ of E, countable additivity gives $\mathrm{\mid }\nu (B_l)\mathrm{\mid }\le {\sum }_j\mathrm{\mid }\nu (B_l\cap E_j)\mathrm{\mid }$. Summing and interchanging the two nonnegative series yields ${\sum }_l\mathrm{\mid }\nu (B_l)\mathrm{\mid }\le {\sum }_j\mathrm{\mid }\nu \mathrm{\mid }(E_j)$. Taking the supremum over partitions proves the reverse bound. Thus $\mathrm{\mid }\nu \mathrm{\mid }$ is a finite positive Borel measure.

Step 4.1 and the finiteness from step 3.1 meet [F3], which gives regularity on the circle. For nu zero all sums are zero. CC was used only for the independent partition sequence in step 2.1 and the regularity theorem; the recursive refinement is deterministic and the measure proof uses only finite choice.