The Poisson integral of a finite complex boundary measure

Assume countable choice. Identify the torus $\mathbb{T}=\mathbb{R}\mathrm{/}\mathbb{Z}$ with the Euclidean unit circle through $\phi ([t])=e^{2\pi it}$ and write $m$ for the normalized Haar measure of $\mathbb{T}$ (The one-dimensional torus and its normalized Haar integral). For $z\in \mathbb{D}$ and $\zeta \in \mathbb{T}$ put $P(z,\zeta ):=P(z,\phi (\zeta ))=\frac{1-\mathrm{\mid }z{\mathrm{\mid }}^2}{\mathrm{\mid }\phi (\zeta )-z{\mathrm{\mid }}^2},$ where the second expression is the Poisson kernel of The Poisson kernel on the unit disc evaluated at the boundary point $\phi (\zeta )$ of the unit circle. For each fixed $z\in \mathbb{D}$ the function $\zeta \mapsto P(z,\zeta )$ is continuous and positive on the compact space $\mathbb{T}$, because $\phi$ is continuous and $\mathrm{\mid }\phi (\zeta )-z\mathrm{\mid }\ge 1-\mathrm{\mid }z\mathrm{\mid }>0$.

Integral against a finite complex measure. Let $\mu$ be a finite regular complex Borel measure on $\mathbb{T}$ (Regular complex Borel measures). Define the Poisson integral of $\mu$ by $P[\mu ](z):={\int }_{\mathbb{T}}P(z,\zeta )d\mu (\zeta )(z\in \mathbb{D}),$ the integral being the one against a signed or complex measure (Integration against a signed or complex measure, and the class L^1(nu) = L^1(|nu|)). This is well defined: for fixed $z$ the integrand is continuous on the compact space $\mathbb{T}$, hence bounded, so it belongs to $L^1(\mathrm{\mid }\mu \mathrm{\mid })$, and by Integrals against signed or complex measures are bounded by total variation $\mathrm{\mid }P[\mu ](z)\mathrm{\mid }\le ({\sup }_{\zeta \in \mathbb{T}}P(z,\zeta ))\mathrm{\mid }\mu \mathrm{\mid }(\mathbb{T})<+\mathrm{\infty }\mathrm{.}$ Linearity in $\mu$ is the linearity of the integral in the measure.

Integral against an $L^1$ density. For $f\in L^1(\mathbb{T},m)$ (The class $L^1(\mu )$ of integrable functions) let $fm$ denote the finite complex measure $(fm)(E):={\int }_{E}fdm(E\subseteq \mathbb{T}\text{ Borel}),$ which is a complex measure with $\mathrm{\mid }fm\mathrm{\mid }(E)={\int }_E\mathrm{\mid }f\mathrm{\mid }dm$ (A complex L^1 density defines a complex measure whose total variation is |h| dmu). Since $\mathbb{T}$ is compact Hausdorff and second-countable, the finite positive Borel measure $\mathrm{\mid }fm\mathrm{\mid }$ is regular by Locally finite Borel measures on second-countable LCH spaces are regular under the assumed countable choice. Thus $fm$ is a finite regular complex measure. The displayed measurable-set formula supplies the density $f$ directly, uniquely up to $m$-almost-everywhere equality by Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree. For a simple measurable $g={\sum }_{j=1}^k{c}_j{\mathbf{1}}_{E_j}$ in canonical disjoint form, each $f{\mathbf{1}}_{E_j}$ is integrable since $\mathrm{\mid }f{\mathbf{1}}_{E_j}\mathrm{\mid }\le \mathrm{\mid }f\mathrm{\mid }$, and $\mathrm{\mid }fm\mathrm{\mid }(E_j)\le \mathrm{\parallel }f{\mathrm{\parallel }}_1<\mathrm{\infty }$. Thus The simple integral against a signed or complex measure and The Lebesgue integral is linear on $L^1(\mu )$ give ${\int }_{\mathbb{T}}gd(fm)={\sum }_{j=1}^k{c}_j(fm)(E_j)={\sum }_{j=1}^k{c}_j{\int }_{E_j}fdm={\int }_{\mathbb{T}}gfdm\mathrm{.}$ For bounded measurable $g$, the same identity follows as follows. Set $s_n:=2^{-n}(\lfloor 2^n\mathrm{Re}g\rfloor +i\lfloor 2^n\mathrm{Im}g\rfloor )\mathrm{.}$ Each $s_n$ is a measurable simple function with finite range, and $\mathrm{\mid }{s}_n-g\mathrm{\mid }\le \sqrt{2}{2}^{-n}$. Hence ${\int }_{\mathbb{T}}\mathrm{\mid }{s}_n-g\mathrm{\mid }d\mathrm{\mid }fm\mathrm{\mid }\le \sqrt{2}{2}^{-n}\mathrm{\parallel }f{\mathrm{\parallel }}_1⟶0,$ so the definition of integration against a complex measure gives $\int s_{n}d(fm)\to \int gd(fm)$. Also, by The Lebesgue integral is linear on $L^1(\mu )$ and The modulus of an integral is bounded by the integral of the modulus, $\mid {\int }_{\mathbb{T}}{s}_{n}fdm-{\int }_{\mathbb{T}}gfdm\mid \le \sqrt{2}{2}^{-n}\mathrm{\parallel }f{\mathrm{\parallel }}_1⟶0.$ Passing to the limit in the simple-function identity proves it for every bounded measurable $g$. The density and approximants are supplied explicitly, so this argument uses no Radon-Nikodym existence theorem or additional choice assumption. In particular, for $z\in \mathbb{D}$ with $g=P(z,\cdot )$, $P[f](z):=P[fm](z)={\int }_{\mathbb{T}}P(z,\zeta )f(\zeta )dm(\zeta )\mathrm{.}$ If $f=f^{\mathrm{\prime }}$ $m$-almost everywhere, then $(fm)(E)={\int }_{E}fdm={\int }_E{f}^{\mathrm{\prime }}dm$ for every Borel $E$ (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree), so $fm=f^{\mathrm{\prime }}m$ and $P[f]=P[f^{\mathrm{\prime }}]$: the Poisson integral of an $L^1$ class is independent of the chosen representative.

Radial functions. For $0\le r<1$ and $f\in L^1(\mathbb{T},m)$ write $P_r\ast f:\mathbb{T}\to \mathbb{C},(P_r\ast f)(\zeta ):={\int }_{\mathbb{T}}P(r\zeta ,\eta )f(\eta )dm(\eta ),$ so that $(P_r\ast f)(\zeta )=P[f](r\zeta )$ by the preceding display. Equivalently, writing $P_r(u):=(1-r^2)\mathrm{/}(1-2r\cos (2\pi u)+r^2)$ for $u\in \mathbb{T}$, one has $(P_r\ast f)(\zeta )={\int }_{\mathbb{T}}{P}_r(\zeta -\eta )f(\eta )dm(\eta )$, the torus convolution of $f$ with the kernel $P_r$, in agreement with the second display of The Poisson kernel on the unit disc under the identification $\zeta =e^{2\pi i\theta }$.

Agreement with the published continuous-data integral. Let $\psi :\mathrm{\partial }\mathbb{D}\to \mathbb{R}$ be continuous and let $F:=\psi \circ \phi :\mathbb{T}\to \mathbb{R}$, a continuous hence bounded function. By the definition of the torus integral, ${\int }_{\mathbb{T}}P(z,\zeta )F(\zeta )dm(\zeta )={\int }_{[0,1)}P(z,e^{2\pi it})\psi \left(e^{2\pi it}\right)dt=\frac{1}{2\pi }{\int }_0^{2\pi }P(z,e^{is})\psi \left(e^{is}\right)ds,$ the last step by the linear change of variables $s=2\pi t$ on $(0,1)$ (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions). The right-hand side is exactly the Poisson integral of $\psi$ from The Poisson integral on the unit disc, so $P[\psi \circ \phi ]=P[\psi ]$ on $\mathbb{D}$. The definition above therefore extends, and does not conflict with, the published real continuous-data definition.