A boundary atom gives an h1 function without an L1 density

Integrating bounded Borel functions against ${\delta }_{{\zeta }_0}$. Since ${\delta }_{{\zeta }_0}$ is a probability measure, $\mathrm{\mid }{\delta }_{{\zeta }_0}\mathrm{\mid }={\delta }_{{\zeta }_0}$ and ${\delta }_{{\zeta }_0}(\mathbb{T})=1$ by [L6] and [L5]. Let $h$ be a bounded Borel function on $\mathbb{T}$ and let $s:=h({\zeta }_0){\mathbf{1}}_{\mathbb{T}}$, a complex simple function with simple integral ${\int }_{\mathbb{T}}sd{\delta }_{{\zeta }_0}=h({\zeta }_0){\delta }_{{\zeta }_0}(\mathbb{T})=h({\zeta }_0)$. The constant sequence $(s)$ is admissible in the definition of ${\int }_{\mathbb{T}}hd{\delta }_{{\zeta }_0}$, because $\mathrm{\mid }h-s\mathrm{\mid }=\mathrm{\mid }h-h({\zeta }_0)\mathrm{\mid }$ is a nonnegative measurable function with ${\int }_{\mathbb{T}}\mathrm{\mid }h-s\mathrm{\mid }d\mathrm{\mid }{\delta }_{{\zeta }_0}\mathrm{\mid }=0$: the integrand vanishes at ${\zeta }_0$ and is supported on $\mathbb{T}\setminus \{{\zeta }_0\}$, which is ${\delta }_{{\zeta }_0}$-null by [L5], so its integral vanishes by A nonnegative integral over a null set vanishes. Hence ${\int }_{\mathbb{T}}hd{\delta }_{{\zeta }_0}=h({\zeta }_0)$.

The function $u$ is the translate of the kernel. Fix $z\in \mathbb{D}$. The function $\zeta \mapsto P(z,\zeta )$ is continuous on $\mathbb{T}$ by [L2], hence bounded Borel, so step 1.1 and the definition of the Poisson integral give $u(z)={\int }_{\mathbb{T}}P(z,\zeta )d{\delta }_{{\zeta }_0}(\zeta )=P(z,{\zeta }_0)=(1-\mathrm{\mid }z{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }\phi ({\zeta }_0)-z{\mathrm{\mid }}^2$, and this is strictly positive because $\mathrm{\mid }\phi ({\zeta }_0)-z\mathrm{\mid }\ge 1-\mathrm{\mid }z\mathrm{\mid }>0$.

Membership and norm. By [L7] the Dirac measure is a finite regular complex Borel measure, so the converse clause of [L3] shows that $u=P[{\delta }_{{\zeta }_0}]$ lies in $h^1(\mathbb{D})$ with $\mathrm{\parallel }u{\mathrm{\parallel }}_{h^1}=\mathrm{\mid }{\delta }_{{\zeta }_0}\mathrm{\mid }(\mathbb{T})={\delta }_{{\zeta }_0}(\mathbb{T})=1$, and $u$ is harmonic by [L4].

Limits at the other boundary points. Let $\zeta \in \mathbb{T}\setminus \{{\zeta }_0\}$ and let $z_m\to \zeta$ with $z_m\in \mathbb{D}$. Then step 2.1 gives $u(z_m)=(1-\mathrm{\mid }{z}_m{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }\phi ({\zeta }_0)-z_m{\mathrm{\mid }}^2$; here $1-\mathrm{\mid }{z}_m{\mathrm{\mid }}^2\to 0$, while by injectivity of $\phi$ and $\zeta \mathrm{\ne }{\zeta }_0$ one has $\phi ({\zeta }_0)\mathrm{\ne }\zeta$, so $\mathrm{\mid }\phi ({\zeta }_0)-z_m{\mathrm{\mid }}^2\to \mathrm{\mid }\phi ({\zeta }_0)-\zeta {\mathrm{\mid }}^2>0$. Hence $u(z_m)\to 0$: the limit is $0$ at every boundary point other than ${\zeta }_0$, along arbitrary sequences inside the disc.

The radial blow-up at ${\zeta }_0$. For $0\le r<1$ one has $\phi ({\zeta }_0)-r\phi ({\zeta }_0)=(1-r)\phi ({\zeta }_0)$ with $\mathrm{\mid }\phi ({\zeta }_0)\mathrm{\mid }=1$, so step 2.1 gives $u(r{\zeta }_0)=\frac{1-r^2}{(1-r{)}^2}=\frac{1+r}{1-r}$. As $r\uparrow 1$ the numerator tends to $2$ and the denominator to $0$ through positive values, so $u(r{\zeta }_0)\to +\mathrm{\infty }$; in particular $u$ is unbounded on $\mathbb{D}$.

No $L^1$ density. Suppose $f\in L^1(\mathbb{T},m)$ satisfied $P[f]=u$. Then $P[fm]=P[f]=u=P[{\delta }_{{\zeta }_0}]$, and by [L7] the density measure $fm$ is a finite complex measure with $\mathrm{\mid }fm\mathrm{\mid }(\mathbb{T})=\mathrm{\parallel }f{\mathrm{\parallel }}_1<\mathrm{\infty }$, hence a finite regular complex Borel measure; being finite and regular it is one of the measures to which the uniqueness clause of [L3] applies, so $fm={\delta }_{{\zeta }_0}$. Evaluating both sides at the singleton $\{{\zeta }_0\}$ gives $0={\int }_{\{{\zeta }_0\}}fdm=(fm)(\{{\zeta }_0\})={\delta }_{{\zeta }_0}(\{{\zeta }_0\})=1$: the middle integral vanishes because $m(\{{\zeta }_0\})=0$ and $f\in L^1(\mathbb{T},m)$, while the last value is $1$ by [L5]. This contradiction shows that no $f\in L^1(\mathbb{T},m)$ represents $u$.

Assembly. Steps 2.1, 3.1, 4.1, 3.2 and 3.3 establish, respectively, the identification $u(z)=P(z,{\zeta }_0)>0$, the membership $u\in h^1(\mathbb{D})$ with norm $1$ and harmonicity, the absence of a representing $L^1$ density, the boundary limit $0$ at every point other than ${\zeta }_0$, and the radial formula $u(r{\zeta }_0)=(1+r)\mathrm{/}(1-r)\to +\mathrm{\infty }$. So the boundary atom ${\delta }_{{\zeta }_0}$ produces an unbounded positive $h^1$ function whose boundary measure is singular with respect to $m$ and which therefore has no $L^1$ density; the Axiom of Choice is used only through the representation theorem [L3] and the countable-choice regularity corollary [L7]. ∎