Finite positive circle measures admit a Lebesgue decomposition under countable choice

Set $\nu =\mu +m$, a finite positive measure on the circle Borel sigma-algebra. The functional $\mathrm{\Lambda }(f)=\int fdm$ on real $L^2(\nu )$ is well-defined: a $\nu$-null set is $m$-null since $m\le \nu$, and Cauchy-Schwarz gives $\int \mathrm{\mid }f\mathrm{\mid }dm\le (\int \mathrm{\mid }f{\mathrm{\mid }}^{2}dm{)}^{1\mathrm{/}2}\le \mathrm{\parallel }f{\mathrm{\parallel }}_{L^2(\nu )}$. By [F1] there is a real Borel representative $h\in L^2(\nu )$ with $\mathrm{\Lambda }(f)=\int fhd\nu$. Every Borel indicator lies in this $L^2$, so $m(E)={\int }_{E}hd\nu (E\text{ Borel})\mathrm{.}$

This identity forces $0\le h\le 1$ $\nu$-almost everywhere. On $E_k=\{h<-1\mathrm{/}k\}$ the integral is at most $-\nu (E_k)\mathrm{/}k$ but equals the nonnegative $m(E_k)$, hence $\nu (E_k)=0$. On $D_k=\{h>1+1\mathrm{/}k\}$ it is at least $(1+1\mathrm{/}k)\nu (D_k)$ but $m(D_k)\le \nu (D_k)$, so $\nu (D_k)=0$. Clip $h$ into $[0,1]$ on the countable union of these null Borel sets. The identity remains true. By linearity, $\mu (E)=\nu (E)-m(E)={\int }_E(1-h)d\nu \mathrm{.}$

Put $N=\{h=0\}$, and define $w=(1-h)\mathrm{/}h$ on $\mathbb{T}\setminus N$ and $w=0$ on $N$. Step 1.1 gives $m(N)=0$. Its indicator identity implies $\int vdm=\int vhd\nu$ for every nonnegative Borel $v$: first for simple functions by linearity, then for arbitrary nonnegative functions by increasing simple approximation and [F2]. Applying it to $v=w{\mathbf{1}}_E$ gives ${\int }_{E}wdm={\int }_{E\setminus N}(1-h)d\nu =\mu (E\setminus N)\mathrm{.}$ In particular $w$ is integrable, with integral at most $\mu (\mathbb{T})$. Set ${\mu }_s=\nu {\mathrm{\mid }}_N=\mu {\mathrm{\mid }}_N$ by step 2.1. It is positive, finite and carried by the $m$-null Borel set $N$, and the displayed identity proves $\mu =wm+{\mu }_s$.

For uniqueness, suppose also $\mu =vm+{\rho }_s$ with the stated properties. Choose null Borel carriers for ${\mu }_s$ and ${\rho }_s$ and let $A$ be their union. For every Borel $E\subseteq \mathbb{T}\setminus A$, equality of the two measures gives ${\int }_E(w-v)dm=0$. Testing the sets where $w-v>1\mathrm{/}k$ or $v-w>1\mathrm{/}k$ proves $w=v$ almost everywhere outside $A$, hence everywhere almost surely. The density measures are equal, and subtraction then gives ${\mu }_s={\rho }_s$. For $\mu =0$, positivity forces both parts zero. Countable choice was used only for the Hilbert-space interface [F1]; all other constructions use explicit measurable formulas.