Local Riesz decomposition of a plane subharmonic function

Since $\bar{D}$ is compact and contained in $\mathrm{\Omega }$, [F4] and [F11] give ${\mu }_u(\bar{D})<+\mathrm{\infty }$; hence the restriction $M={\mu }_u{\mathrm{\mid }}_D$ is a finite positive Borel measure carried by $\bar{D}\subseteq \mathrm{\Omega }$, and in particular carried by $D$.

Let $D_0$ be a connected component of $D$, let $v$ be subharmonic on the complex domain $D_0$, and let $a\in D_0$. Then $v(a)\le A_r(v)(a)$ for every $r$ with $\overline{D(a,r)}\subseteq D_0$ by [F1]. For every real $N>v(a)$, upper semicontinuity gives $v\le N$ on $\overline{D(a,\delta )}\subseteq D_0$ for some $\delta >0$, hence $A_r(v)(a)\le N$ for $0<r\le \delta$. Thus ${\lim }_{r\downarrow 0}{A}_r(v)(a)=v(a)$, including $v(a)=-\mathrm{\infty }$, when every real $N$ works.

By [F3] Dependent Choice yields Countable Choice, so the choice hypotheses of the suppliers [F5], [F6], [F7], [F10] and [F13] are discharged for the whole argument.

Put $p:=p_M=\int \log \mathrm{\mid }z-w\mathrm{\mid }dM(w)$ with the diagonal value $-\mathrm{\infty }$; by [F5] (with $\rho =M$) the function $p$ is locally integrable and subharmonic on $\mathbb{C}$ and satisfies $\mathrm{\Delta }p=2\pi M$ distributionally, that is, $(2\pi {)}^{-1}\int p\mathrm{\Delta }\phi dA=\int \phi dM$ for every $\phi \in C_c^{\mathrm{\infty }}(\mathbb{C})$.

For uniqueness of the measure let $M^{\mathrm{\prime }}$ be a finite positive Borel measure carried by $D$ and $h^{\mathrm{\prime }}$ harmonic on $D$ with $u=h^{\mathrm{\prime }}+p_{M^{\mathrm{\prime }}}$ pointwise on $D$, and set $p^{\mathrm{\prime }}:=p_{M^{\mathrm{\prime }}}$, locally integrable and subharmonic with $\mathrm{\Delta }{p}^{\mathrm{\prime }}=2\pi M^{\mathrm{\prime }}$ by [F5]. For every $\phi \in C_c^{\mathrm{\infty }}(D)$, [F2], [F5], [F6] and the representation clause of [F4] give $2\pi \int \phi d{M}^{\mathrm{\prime }}=\int p^{\mathrm{\prime }}\mathrm{\Delta }\phi dA=⟨\mathrm{\Delta }{T}_{p^{\mathrm{\prime }}},\phi ⟩=⟨\mathrm{\Delta }{T}_{h^{\mathrm{\prime }}+p^{\mathrm{\prime }}},\phi ⟩=⟨\mathrm{\Delta }{T}_u,\phi ⟩=2\pi {\mu }_u(\phi )=2\pi \int \phi dM$, so $M$ and $M^{\mathrm{\prime }}$ define the same functional $\mathrm{\Phi }(\phi ):=\int \phi dM$ on $C_c^{\mathrm{\infty }}(D)$.

Let $T_u$ and $T_p$ be the regular distributions of the locally integrable functions $u$ and $p$. For every $\phi \in C_c^{\mathrm{\infty }}(D)$ one has $⟨\mathrm{\Delta }{T}_u,\phi ⟩=T_u(\mathrm{\Delta }\phi )={\int }_{\mathrm{\Omega }}u\mathrm{\Delta }\phi dA=2\pi {\mu }_u(\phi )=2\pi {\int }_{\mathrm{\Omega }}\phi d{\mu }_u=2\pi {\int }_{\mathrm{\Omega }}\phi dM={\int }_{\mathrm{\Omega }}p\mathrm{\Delta }\phi dA=⟨\mathrm{\Delta }{T}_p,\phi ⟩$, where the third equality is [F2], the fifth uses that $\phi$ is supported in $D$ and $M={\mu }_u{\mathrm{\mid }}_D$, and the sixth is step 2.1 and [F6]; hence $\mathrm{\Delta }(T_u-T_p)=0$ in ${\mathcal{D}}^{\mathrm{\prime }}(D)$.

Now let $D_0$ be a connected component of the open set $D$; by [F12] it is open, hence a complex domain, contained in $\mathrm{\Omega }$ with $\overline{D_0}\subseteq \bar{D}\subseteq \mathrm{\Omega }$ compact, and $q:=p^{\mathrm{\prime }}{\mathrm{\mid }}_{D_0}$ is subharmonic on $D_0$ with Riesz functional $\phi \mapsto (2\pi {)}^{-1}{\int }_{D_0}q\mathrm{\Delta }\phi dA=\int \phi d{M}^{\mathrm{\prime }}$ for every $\phi \in C_c^{\mathrm{\infty }}(D_0)$ by [F5]. Both $M^{\mathrm{\prime }}{\mathrm{\mid }}_{D_0}$ and $M{\mathrm{\mid }}_{D_0}$ are finite Borel measures on the second-countable LCH space $D_0$, hence Radon by [F13] under the Countable Choice of step 1.3, and $M{\mathrm{\mid }}_{D_0}$ represents the same functional because $\int \phi d(M{\mathrm{\mid }}_{D_0})=\mathrm{\Phi }(\phi )=(2\pi {)}^{-1}{\int }_{D_0}q\mathrm{\Delta }\phi dA$ for every $\phi \in C_c^{\mathrm{\infty }}(D_0)$ by step 2.2; the uniqueness clause of [F4] applied on the domain $D_0$ therefore gives $M^{\mathrm{\prime }}{\mathrm{\mid }}_{D_0}=M{\mathrm{\mid }}_{D_0}$. Since $D$ is second countable and its components are pairwise disjoint nonempty open sets, each containing a member of a countable base, there are at most countably many components; they partition $D$, so countable additivity gives $M^{\mathrm{\prime }}=M$ on every Borel subset of $D$, and both measures are carried by $D$, hence $M^{\mathrm{\prime }}=M$ on $\mathcal{B}(\mathbb{C})$.

By [F7] applied to the distribution $T:=(T_u-T_p){\mathrm{\mid }}_D\in {\mathcal{D}}^{\mathrm{\prime }}(D)$ of step 3.1, there is a unique smooth harmonic $h$ on $D$ with $T=T_h$, that is, $⟨T_u-T_p,\phi ⟩={\int }_{D}h\phi dA$ for every $\phi \in C_c^{\mathrm{\infty }}(D)$.

The function $u-p$ is locally integrable on $D$ by [F9] and step 2.1, and $h$ is locally integrable; step 4.1 says that their regular distributions agree. By the injectivity of [F6], $u-p=h$ almost everywhere on $D$.

On each connected component $D_0$ of $D$, the function $g:=h+p$ is subharmonic: $h{\mathrm{\mid }}_{D_0}$ is harmonic and hence subharmonic by [F8], and $p{\mathrm{\mid }}_{D_0}$ inherits upper semicontinuity and the circle inequality from step 2.1 and cannot be identically $-\mathrm{\infty }$ because it is locally integrable. The sum is subharmonic by [F8]. Likewise $u{\mathrm{\mid }}_{D_0}$ is subharmonic by [F1] and its local integrability [F9]. Step 5.1 gives $u=g$ almost everywhere on $D$.

For fixed $a\in D$ choose $R>0$ with $\overline{D(a,R)}\subseteq D$. By step 2.1 and [F9], $u,g\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(D)$. Replace their values $-\mathrm{\infty }$ by $0$ to obtain finite Borel representatives $\tilde{u},\tilde{g}$; they agree with $u,g$ almost everywhere and satisfy $\tilde{u}=\tilde{g}$ almost everywhere by step 6.1. Thus $q:=\mathrm{\mid }\tilde{u}-\tilde{g}\mathrm{\mid }$ is a nonnegative Borel function with $q=0$ almost everywhere. Applying [F10] to $q+\mathrm{\mid }\tilde{u}\mathrm{\mid }+\mathrm{\mid }\tilde{g}\mathrm{\mid }$ on $B(a,R)$ shows that for almost every $r\in (0,R)$ the restrictions of $\tilde{u},\tilde{g}$ to the circle are integrable and agree almost everywhere in angle. Since the representatives differ from the subharmonic functions only on planar null sets, [F10] also makes those exceptional sets arclength-null for almost every $r$. Hence $A_r(u)(a)=A_r(g)(a)$ for almost every $r\in (0,R)$, where $A_r(v)(a):=(2\pi {)}^{-1}{\int }_0^{2\pi }v(a+r{e}^{it})dt$.

Let $S\subseteq (0,R)$ be the full-measure set of radii from step 7.1 for which the circle means agree. Each $S\cap (0,\min (R\mathrm{/}2,1\mathrm{/}(n+1)))$ is nonempty, so Countable Choice [F3] gives $r_n\in S\cap (0,\min (R\mathrm{/}2,1\mathrm{/}(n+1)))$ for every $n$; then $r_n\to 0$. By step 1.2, applied to the subharmonic functions $u$ and $g$ at $a$, $u(a)={\lim }_n{A}_{r_n}(u)(a)={\lim }_n{A}_{r_n}(g)(a)=g(a)$. Since $a\in D$ was arbitrary, $u=h+p$ everywhere on $D$, which is the asserted decomposition.

With $M^{\mathrm{\prime }}=M$ from step 3.2 we have $h^{\mathrm{\prime }}+p=h+p$ everywhere on $D$ by the decomposition of step 8.1, and $p$ is finite almost everywhere because $p\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(D)$ by step 2.1; hence $h^{\mathrm{\prime }}=h$ almost everywhere on $D$. The difference $h^{\mathrm{\prime }}-h$ is harmonic, hence continuous, on the open set $D$, and an almost-everywhere-vanishing continuous function on $D$ vanishes everywhere, since a set of full measure in a nonempty open set is dense; therefore $h^{\mathrm{\prime }}=h$, and the decomposition is unique in both entries.