Finite complex circle measures are determined by Fourier coefficients and Poisson integrals

Set $\sigma =\mu -\nu$, a complex measure by countable additivity. If its Fourier coefficients vanish, its integral against every trigonometric polynomial is zero. Given a continuous u, [F2] supplies polynomials arbitrarily close to u uniformly; [F1] bounds $\mathrm{\mid }\int (u-p)d\sigma \mathrm{\mid }$ by $\mathrm{\parallel }u-p{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }\sigma \mathrm{\mid }(\mathbb{T})$. Therefore $\int ud\sigma =0$ for every continuous u.

For $z=r\zeta$, the geometric-series identity yields $P(r\zeta ,\eta )={\sum }_{k\in \mathbb{Z}}{r}^{\mathrm{\mid }k\mathrm{\mid }}{\zeta }^k{\eta }^{-k}\mathrm{.}$ At fixed r<1 this is uniformly absolutely convergent in both circle variables, with bound $1+2{\sum }_{k\ge 1}{r}^k$. Integrating first against $\mu$ is justified by [F1] and the uniform error bound, and produces the uniformly convergent circle series ${\sum }_k{r}^{\mathrm{\mid }k\mathrm{\mid }}\widehat{\mu }(k){\zeta }^k$, since $\mathrm{\mid }\widehat{\mu }(k)\mathrm{\mid }\le \mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})$. Integrating this series against ${\zeta }^{-n}dm$ and using [F2] gives the stated coefficient formula.

Fix a Borel E and $\epsilon >0$. Regularity in [F1] gives a compact $K\subseteq E$ and open $U\supseteq E$ with $\mathrm{\mid }\sigma \mathrm{\mid }(U\setminus K)<\epsilon$. There is a continuous $0\le u\le 1$ equal to one on K and zero outside U. Explicitly, if K is empty take u=0; if U is the circle take u=1; otherwise use $u(x)=\frac{d(x,\mathbb{T}\setminus U)}{d(x,\mathbb{T}\setminus U)+d(x,K)}\mathrm{.}$ The two sets are disjoint closed sets, so the denominator is positive at each x; distances are continuous because $\mathrm{\mid }d(x,A)-d(y,A)\mathrm{\mid }\le d(x,y)$ for a nonempty set A. Thus $\mathrm{\mid }{\mathbf{1}}_E-u\mathrm{\mid }\le {\mathbf{1}}_{U\setminus K}$. Step 1.1 and [F1] give $\mathrm{\mid }\sigma (E)\mathrm{\mid }=\mathrm{\mid }\int ({\mathbf{1}}_E-u)d\sigma \mathrm{\mid }<\epsilon$. Hence $\sigma (E)=0$ for every E, proving Fourier uniqueness.

If $P[\mu ]=P[\nu ]$, fix r=1/2 in step 1.2. Every factor $r^{\mathrm{\mid }n\mathrm{\mid }}$ is strictly positive, so all Fourier coefficients agree. Step 2.1 gives $\mu =\nu$. Zero measures and zero variation cause no division by a measure mass anywhere in the proof.