The principle of descent and the logarithmic domination principle

Local Sobolev and strict-contact facts under DC

Write $dA$ for area measure and ${\mu }_f=(2\pi {)}^{-1}\mathrm{\Delta }f$ for the Riesz measure. Use the radial unit-mass mollifier $\rho (x)=C\exp \text{\hspace{0.17em}⁣}(-\frac{1}{1-\mathrm{\mid }x{\mathrm{\mid }}^2}){\mathbf{1}}_{\{\mathrm{\mid }x\mathrm{\mid }<1\}},{\rho }_{\delta }(x)={\delta }^{-2}\rho (x\mathrm{/}\delta ),$ with $C$ chosen so that $\int \rho dA=1$. The profile is smooth, radial, and nonincreasing. All mollifications below are taken on a safe interior of the relevant domain.

For a subharmonic $f$, let $A_{r}f(z)=\frac{1}{\pi r^2}{\int }_{B(z,r)}fdA\mathrm{.}$ Integrating its circle-mean inequality in the radius gives $A_{r}f(z)\ge f(z)$ when $f(z)$ is finite. If $f(z)=-\mathrm{\infty }$, upper semicontinuity bounds $f$ above by any prescribed real number on a sufficiently small disk. For finite $f(z)$, upper semicontinuity gives, for each $\eta >0$, an $r_0>0$ such that $A_{r}f(z)\le f(z)+\eta$ for every $0<r<r_0$. Thus $f(z)\le A_{r}f(z)\le f(z)+\eta$ for all such $r$, and $A_{r}f(z)\to f(z)$. The layer-cake formula for a radial decreasing density is $(f\ast {\rho }_{\delta })(z)={\int }_0^1\pi s^2(-{\rho }^{\mathrm{\prime }}(s))A_{\delta s}f(z)ds,{\int }_0^1\pi s^2(-{\rho }^{\mathrm{\prime }}(s))ds=1.$ If $f(z)$ is finite, choose $\delta <r_0$; every average in this weighted mean lies between $f(z)$ and $f(z)+\eta$, so the convolution does too. If $f(z)=-\mathrm{\infty }$, upper semicontinuity bounds $f$ by any real $N$ on a sufficiently small disk, so every average in the weighted mean is at most $N$. Consequently the actual upper-semicontinuous representative satisfies $(f\ast {\rho }_{\delta })(z)\to f(z)$ pointwise, including at $-\mathrm{\infty }$ values. This pointwise fact will be used against Riesz measures, including at points where a finite-energy potential equals $-\mathrm{\infty }$.

If $f$ is locally bounded above and below and subharmonic, then $f_{\delta }=f\ast {\rho }_{\delta }$ has $\mathrm{\Delta }{f}_{\delta }=2\pi {\mu }_f\ast {\rho }_{\delta }\ge 0$ by [F15]. For a smooth cutoff $0\le \eta \le 1$ supported in a fixed compact set, choose $C_0$ above $f_{\delta }$ on a fixed neighborhood of its support and let $B$ uniformly bound $C_0-f_{\delta }$ there. Integration by parts, with $C_0-f_{\delta }\ge 0$, and Young's inequality give $\int {\eta }^2\mathrm{\mid }\mathrm{\nabla }{f}_{\delta }{\mathrm{\mid }}^{2}dA\le 2B\int {\eta }^2\mathrm{\Delta }{f}_{\delta }dA+4B^2\int \mathrm{\mid }\mathrm{\nabla }\eta {\mathrm{\mid }}^{2}dA\mathrm{.}$ The first term is bounded by the Riesz mass on a fixed compact neighborhood, and local boundedness makes $B$ uniform. Hence $\mathrm{\nabla }{f}_{\delta }$ is bounded in local $L^2$. Pointwise convergence and local boundedness give $f_{\delta }\to f$ in local $L^2$. For a smooth test $\zeta$ on a smaller disk, the weak derivative functional is bounded by $\mid \int f{\mathrm{\partial }}_i\zeta dA\mid ={\lim }_{\delta \downarrow 0}\mid \int ({\mathrm{\partial }}_i{f}_{\delta })\zeta dA\mid \le C\mathrm{\parallel }\zeta {\mathrm{\parallel }}_2\mathrm{.}$ To use [F17] on that disk, exhaust it by concentric closed subdisks, extend each truncated $L^2$ function by zero, approximate by a compactly supported continuous function on ${\mathbb{R}}^2$, multiply by a smooth cutoff supported in the disk and equal to one on the truncation, and mollify with radius small enough to keep the support inside the disk. This proves density of smooth compactly supported tests in its $L^2$ space. The Hilbert-space representation in [F17] therefore gives an $L^2$ weak derivative. Thus $f\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}$.

For any $g\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}$, localization by a smooth cutoff, [F16], and Minkowski's inequality show that $g\ast {\rho }_{\delta }\to g$ strongly in local $W^{1,2}$. Weak derivatives commute with convolution on a safe interior by the direct test calculation ${\mathrm{\partial }}_i(g\ast {\rho }_{\delta })(x)=-\int g(y){\mathrm{\partial }}_{y_i}{\rho }_{\delta }(x-y)dA(y)=\int ({\mathrm{\partial }}_{i}g)(y){\rho }_{\delta }(x-y)dA(y)\mathrm{.}$ No full-Choice Sobolev chain-rule supplier is used.

For smooth $F:\mathbb{R}\to \mathbb{R}$ whose derivative is bounded and Lipschitz, approximate $g\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}$ strongly by smooth mollifications. Classical differentiation and the estimate $\mathrm{\parallel }{F}^{\mathrm{\prime }}(g_n)\mathrm{\nabla }{g}_n-F^{\mathrm{\prime }}(g)\mathrm{\nabla }g{\mathrm{\parallel }}_2\le \mathrm{\parallel }{F}^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }\mathrm{\nabla }{g}_n-\mathrm{\nabla }g{\mathrm{\parallel }}_2+\mathrm{\parallel }(F^{\mathrm{\prime }}(g_n)-F^{\mathrm{\prime }}(g))\mathrm{\nabla }g{\mathrm{\parallel }}_2$ prove $\mathrm{\nabla }F(g)=F^{\mathrm{\prime }}(g)\mathrm{\nabla }g$. For the second term, split into $\mathrm{\mid }{g}_n-g\mathrm{\mid }\le t$ and its complement, then use the Lipschitz bound on the first part and convergence in measure plus absolute continuity of $\int \mathrm{\mid }\mathrm{\nabla }g{\mathrm{\mid }}^2$ on the second. The same estimate proves strong local $W^{1,2}$ convergence under this composition. Choose smooth $\vartheta$ with $\vartheta =0$ on $(-\mathrm{\infty },0]$ and $\vartheta =1$ on $[1,\mathrm{\infty })$, put $Q(t)=t\vartheta (t)$ and $P_{\delta }(t)=\delta Q(t\mathrm{/}\delta )$. Then $\mathrm{\mid }{P}_{\delta }(t)-t^{+}\mathrm{\mid }\le \delta$, the derivatives are uniformly bounded, and $P_{\delta }^{\mathrm{\prime }}(t)\to 1_{\{t>0\}}$ (with value $0$ at $t=0$). Dominated convergence therefore gives $g^{+}\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2},\mathrm{\nabla }{g}^{+}=1_{\{g>0\}}\mathrm{\nabla }g\text{almost everywhere}\mathrm{.}$

We now prove the bounded plane strict-contact identity. Let $x,y$ be locally bounded above and below subharmonic functions, put $w=\max (x,y)$ and $f=x-y$. The preceding bound gives $x,y,w\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}$, and the positive-part formula gives $\mathrm{\nabla }(w-y)=\mathrm{\nabla }f$ almost everywhere on $\{f>0\}$. Fix $\phi \in C_c^{\mathrm{\infty }}$ and choose smooth ${\chi }_n$ with $0\le {\chi }_n\le 1$, ${\chi }_n(t)=0$ for $t\le 1\mathrm{/}(2n)$, and ${\chi }_n(t)=1$ for $t\ge 1\mathrm{/}n$. Put ${\psi }_n=\phi {\chi }_n(f)$. Its gradient vanishes where $f\le 1\mathrm{/}(2n)$; where $f>1\mathrm{/}(2n)$ the displayed gradients agree. Thus $\mathrm{\nabla }(w-y)\cdot \mathrm{\nabla }{\psi }_n=\mathrm{\nabla }(x-y)\cdot \mathrm{\nabla }{\psi }_n\text{almost everywhere}\mathrm{.}$ For $\delta >0$ use smooth tests ${\psi }_{n,\delta }=\phi {\chi }_n(x\ast {\rho }_{\delta }-y\ast {\rho }_{\delta })$. Strong local $W^{1,2}$ convergence and the scalar chain rule give ${\psi }_{n,\delta }\to {\psi }_n$ strongly in $W^{1,2}$; pointwise convergence of the radial mollifications gives pointwise convergence to the actual $f$, and $\mathrm{\mid }{\psi }_{n,\delta }\mathrm{\mid }\le \mathrm{\mid }\phi \mathrm{\mid }$. For each $p\in \{x,y,w\}$ the distributional Riesz identity and integration by parts give $\int {\psi }_{n,\delta }d{\mu }_p=-\frac{1}{2\pi }\int \mathrm{\nabla }p\cdot \mathrm{\nabla }{\psi }_{n,\delta }dA\mathrm{.}$ Strong convergence on the right and dominated convergence against the locally finite Radon measure on the left pass this identity to ${\psi }_n$. Subtract the $p=y$ identity from those for $p=w$ and $p=x$, use the gradient equality, and then let $n\to \mathrm{\infty }$. Since ${\chi }_n(t)\to 1_{\{t>0\}}$, dominated convergence gives, as Borel measures, $1_{\{x>y\}}{\mu }_x=1_{\{x>y\}}{\mu }_{\max (x,y)}\mathrm{.}$ To pass from smooth tests to $C_c$, extend any $C_c$ function by zero and convolve it with ${\rho }_{\delta }$; uniform continuity gives uniform convergence, and small $\delta$ keeps the supports in a fixed compact subset of the domain. For a Borel contact set $E$, let $\lambda =1_E{\mu }_p$. It is finite on compact sets because $\lambda \le {\mu }_p$. It is inner regular on every Borel set $B$: take a compact exhaustion $K_n$ of the plane domain. For fixed $n$ and $\eta >0$, outer regularity of ${\mu }_p$ gives an open $O_n\supset K_n\setminus (E\cap B)$ with ${\mu }_p(O_n)\le {\mu }_p(K_n\setminus (E\cap B))+\eta$. Then $F_n=K_n\setminus O_n$ is compact in $E\cap B$, and $\lambda ((B\cap E\cap K_n)\setminus F_n)\le \eta$. Letting $\eta \downarrow 0$ and then exhausting the domain proves inner regularity on $B$; the countable selections are supplied by DC$\Rightarrow$CC. To get outer regularity, suppose $\lambda (B)<\mathrm{\infty }$ and choose a relatively compact open exhaustion $U_n$ of the domain. By the inner regularity just proved, choose compact $F_n\subset U_n\setminus B$ with $\lambda ((U_n\setminus B)\setminus F_n)<\eta 2^{-n}$. The open set $V={\bigcup }_n(U_n\setminus F_n)$ contains $B$ and satisfies $\lambda (V\setminus B)<\eta$. If $\lambda (B)=\mathrm{\infty }$, outer regularity is automatic. Thus $\lambda$ is Radon. Smooth compactly supported functions are uniformly dense in $C_c$ by zero extension and convolution with ${\rho }_{\delta }$ on a safe interior. Therefore [F19] identifies the restricted measures from their $C_c$ integrals. This proves the displayed identity as an exact Borel measure identity. The proof uses the actual upper-semicontinuous representatives and no quasicontinuous replacement.

Finite-energy potentials are locally Sobolev

Let $\sigma$ be finite positive with compact support $K$, mass $m>0$ and $I(\sigma )<\mathrm{\infty }$, and write $p(z)=\int \log \mathrm{\mid }z-w\mathrm{\mid }d\sigma (w)$. For a compact plane set $Q$, Tonelli and the local integrability of $\mathrm{\mid }\log \mathrm{\mid }z-w\mathrm{\mid }{\mathrm{\mid }}^2$ uniformly for $w\in K$ show that ${\int }_K\mathrm{\mid }\log \mathrm{\mid }z-w\mathrm{\mid }{\mathrm{\mid }}^{2}d\sigma (w)<\mathrm{\infty }$ for area-a.e. $z\in Q$. Jensen [F18] then gives ${\int }_Q\mathrm{\mid }p(z){\mathrm{\mid }}^{2}dA(z)\le m{\int }_K{\int }_Q\mathrm{\mid }\log \mathrm{\mid }z-w\mathrm{\mid }{\mathrm{\mid }}^{2}dA(z)d\sigma (w)<\mathrm{\infty }\mathrm{.}$ Thus $p\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^2$. Put ${\sigma }_{\delta }=\sigma \ast {\rho }_{\delta }$ and $p_{\delta }=p\ast {\rho }_{\delta }=p_{{\sigma }_{\delta }}$; Fubini justifies the last identity because the logarithm is locally area-integrable on the compact sets involved. The convolution ${\kappa }_{\delta }={\rho }_{\delta }\ast {\rho }_{\delta }$ is radial and has mass one. The circle-mean calculation $\frac{1}{2\pi }{\int }_0^{2\pi }\log \mathrm{\mid }a+r{e}^{it}\mathrm{\mid }dt=\log \max (\mathrm{\mid }a\mathrm{\mid },r)\ge \log \mathrm{\mid }a\mathrm{\mid }$ follows by factoring out the larger radius and averaging the logarithm series, with the boundary case obtained by its integrable limit. Hence $k_R\ast {\kappa }_{\delta }\le k_R$ for $k_R(a)=\log (R\mathrm{/}\mathrm{\mid }a\mathrm{\mid })$. Choose $R>1+\mathrm{diam}K+2$ and $0<\delta <1$, so the shifted kernel is nonnegative on all smoothed supports. Tonelli gives $I({\sigma }_{\delta })+m^2\log R=\iint (k_R\ast {\kappa }_{\delta })(z-w)d\sigma (z)d\sigma (w)\le \iint k_R(z-w)d\sigma (z)d\sigma (w)=I(\sigma )+m^2\log R\mathrm{.}$ On $\mathrm{supp}{\sigma }_{\delta }$, $p_{\delta }\le m\log R$, so $\int p_{\delta }^{-}d{\sigma }_{\delta }=I({\sigma }_{\delta })+\int p_{\delta }^{+}d{\sigma }_{\delta }\le I(\sigma )+m^2\log R\mathrm{.}$ For a fixed smooth cutoff $\eta$, integration by parts and $\mathrm{\Delta }{p}_{\delta }=2\pi {\sigma }_{\delta }$ give $\int {\eta }^2\mathrm{\mid }\mathrm{\nabla }{p}_{\delta }{\mathrm{\mid }}^{2}dA\le 4\pi \int p_{\delta }^{-}d{\sigma }_{\delta }+4\int p_{\delta }^2\mathrm{\mid }\mathrm{\nabla }\eta {\mathrm{\mid }}^{2}dA\mathrm{.}$ The last term is uniformly bounded by Jensen's inequality applied to $p\ast {\rho }_{\delta }$ and $p\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^2$ on a slightly larger compact set. Thus $\mathrm{\nabla }{p}_{\delta }$ is uniformly bounded in local $L^2$. Approximate identity convergence from [F16] gives $p_{\delta }\to p$ in local $L^2$; the bounded-functional and Hilbert-space argument above then produces the weak derivatives of $p$ in local $L^2$. Consequently every finite-energy logarithmic potential belongs to $W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}$ under DC.

The same strict-contact test proof also applies to a finite-energy subharmonic $x$ and a smooth finite-valued subharmonic obstacle $y$: then $x,y,w=\max (x,y)\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}$, and at points where $x=-\mathrm{\infty }$ the radial mollifications tend to $-\mathrm{\infty }$ while $y$ stays finite. Define ${\chi }_n(-\mathrm{\infty })=0$; the tests converge pointwise there as well. The DCT argument therefore proves the exact Borel restriction identity on the strict contact set also in this one-obstacle case.

Full-mass truncation and the unbounded contact identity

Let ${\mathbb{P}}^1=\mathbb{C}\cup \{\mathrm{\infty }\}$, ${\rho }_{FS}(z)=\frac{1}{2}\log (1+\mathrm{\mid }z{\mathrm{\mid }}^2),\omega =d{d}^c{\rho }_{FS},d{d}^c=(2\pi {)}^{-1}\mathrm{\Delta }dA,{\int }_{{\mathbb{P}}^1}\omega =1.$ Indeed $\mathrm{\Delta }{\rho }_{FS}=2(1+\mathrm{\mid }z{\mathrm{\mid }}^2{)}^{-2}$, so polar integration gives $(2\pi {)}^{-1}{\int }_{\mathbb{C}}\mathrm{\Delta }{\rho }_{FS}dA=1$; the form extends smoothly over infinity using the coordinate $t=1\mathrm{/}z$. Here ${\rho }_{FS}$ is a local potential on the affine chart. An $\omega$-psh function $q$ is an upper-semicontinuous locally integrable function on ${\mathbb{P}}^1$ such that $q+{\rho }_U$ is subharmonic in every chart with $d{d}^c{\rho }_U=\omega$. Its ordinary current $T_q=\omega +d{d}^{c}q$ is a positive Radon measure: the local Riesz measures agree on overlaps. Its total mass is one, since on the compact surface $⟨d{d}^{c}q,1⟩=⟨q,d{d}^{c}1⟩=0$ and $\int \omega =1$. These statements use the local Riesz-measure interface [F4] and the finite-atlas gluing just described.

For bounded $\omega$-psh $x,y$, adding a chart potential turns them into locally bounded subharmonic functions. The bounded plane identity above then gives globally on ${\mathbb{P}}^1$ $1_{\{x>y\}}{T}_x=1_{\{x>y\}}{T}_{\max (x,y)}\mathrm{.}$ The strict set is Borel: for extended-valued upper-semicontinuous functions, $\{x>y\}={\bigcup }_{r\in \mathbb{Q}}(\{x>r\}\cap \{y<r\})$.

For any $\omega$-psh $q$, put $q_j=\max (q,-j)$, $T_j=T_{q_j}$ and $F_j=\{q>-j\}$. Each $T_j$ is positive and has mass one. For $j\ge k$, apply the bounded identity to $(q_j,-k)$; since $\{q_j>-k\}=F_k$ and $\max (q_j,-k)=q_k$, it gives $T_j{\mathrm{\mid }}_{F_k}=T_k{\mathrm{\mid }}_{F_k}\mathrm{.}$ Thus $R_j=1_{F_j}{T}_j$ increases. Define the nonpluripolar truncation limit $T_q^{np}={\lim }_j{R}_j$ and the full-mass class $\mathcal{E}=\{q:T_q^{np}({\mathbb{P}}^1)=1\}\mathrm{.}$ Since $T_j({\mathbb{P}}^1)=1$, this definition is equivalent to

If $q\in \mathcal{E}$, stabilization gives $T_j{\mathrm{\mid }}_{F_j}=T_q^{np}{\mathrm{\mid }}_{F_j}$; the complementary tails of both measures have mass $1-R_j({\mathbb{P}}^1)$. Consequently

For $q\in \mathcal{E}$, this total-variation limit is Radon: given a Borel set, approximate it from outside by an open set for a sufficiently close Radon $T_j$; the total-variation error transfers the outer-regularity estimate to $T_q^{np}$. For an open set, approximate it from inside by a compact set for $T_j$ and transfer the estimate in the same way. The ordinary current is this same measure. Indeed $q_j\downarrow q$ and $\mathrm{\mid }{q}_j\mathrm{\mid }\le \mathrm{\mid }q\mathrm{\mid }$ almost everywhere, so dominated convergence gives $q_j\to q$ in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ on every chart. Hence $T_j=\omega +d{d}^c{q}_j\to T_q=\omega +d{d}^{c}q$ distributionally. By (2), the same sequence converges in total variation to $T_q^{np}$. The two limits therefore agree on smooth tests, and uniqueness of the local Riesz measure in [F4] gives

This identifies the truncation limit with the ordinary current instead of leaving two measures unnamed as though they were equal.

The integer truncation levels are cofinal among real levels: for $s\le t$, the bounded contact identity applied to $(q_t,-s)$ gives stabilization on $\{q>-s\}$. Thus changing every level by a fixed constant leaves the truncation limit, condition (1), and membership in $\mathcal{E}$ unchanged.

We need that $\mathcal{E}$ is upward closed. First derive bounded comparison. For bounded $\omega$-psh $\alpha ,\beta$, put $r=\max (\alpha ,\beta )$. On $\{\alpha <\beta \}$ one has $T_r=T_{\beta }$, and on $\{\alpha >\beta \}$ one has $T_r=T_{\alpha }$. Since $T_r$ has mass one, $T_{\beta }(\alpha <\beta )=1-T_r(\alpha \ge \beta )\le 1-T_{\alpha }(\alpha >\beta )=T_{\alpha }(\alpha \le \beta )\mathrm{.}$ Apply this to $(\alpha ,\beta -\delta )$ and let $\delta \downarrow 0$; the strict sublevel sets increase to $\{\alpha <\beta \}$, giving

For $\alpha ,\beta \in \mathcal{E}$, apply (4) to ${\alpha }_j=\max (\alpha ,-j)$ and ${\beta }_k=\max (\beta ,-k)$. Let $k\to \mathrm{\infty }$ and then $j\to \mathrm{\infty }$. The indicators converge pointwise, and (2)-(3) give total-variation convergence of both truncated currents, so

Now suppose $\theta \in \mathcal{E}$ and $\theta \le q$ for an $\omega$-psh $q$. By the constant-shift observation, shift both down so $q\le -2$. Put $v=q\mathrm{/}2$, $v_j=\max (v,-j)=q_{2j}\mathrm{/}2$ and ${\theta }_{2j}=\max (\theta ,-2j)$. Since $v\le -1$, the Borel sets $S_j=\{{\theta }_{2j}<v_j-j+1\}$ satisfy $\{v\le -j\}\subseteq S_j\subseteq \{\theta \le -j\}$: on the first set ${\theta }_{2j}\le q_{2j}=2v_j=-2j<v_j-j+1$, and the second inclusion follows from $v_j\le -1$. Apply (5) to the bounded pair $({\theta }_{2j},v_j-j+1)$; adding a constant leaves the current unchanged. Then $T_{v_j}(v\le -j)\le T_{v_j}(S_j)\le T_{{\theta }_{2j}}(S_j)\le T_{{\theta }_{2j}}(\theta \le -j)=T_{{\theta }_j}(\theta \le -j)⟶0.$ For the equality, $T_{{\theta }_{2j}}$ and $T_{{\theta }_j}$ agree on $\{\theta >-j\}$ by stabilization and both have mass one. Hence $v\in \mathcal{E}$ by (1). Since $T_{v_j}=\frac{1}{2}{T}_{q_{2j}}+\frac{1}{2}\omega$, $T_{q_{2j}}(q\le -2j)\le 2T_{v_j}(v\le -j)⟶0.$ These are the even-index tails in (1); the tail masses are decreasing, so all tails tend to zero and $q\in \mathcal{E}$. This proves upward closure.

Finally let $\theta \in \mathcal{E}$ and let $\xi$ be any $\omega$-psh function, with no lower bound and no finiteness assumption on its current. Upward closure gives $q=\max (\theta ,\xi )\in \mathcal{E}$. Define ${\theta }_j=\max (\theta ,-j),{\xi }_{j+1}=\max (\xi ,-j-1),q_j=\max (q,-j),$ and $E=\{\theta >\xi \}$, $E_j=\{{\theta }_j>{\xi }_{j+1}\}$. Then $E\subseteq E_j$ and $\max ({\theta }_j,{\xi }_{j+1})=q_j$. Bounded contact gives $T_{{\theta }_j}{\mathrm{\mid }}_{E_j}=T_{q_j}{\mathrm{\mid }}_{E_j}$, hence for every Borel $B$, $T_{{\theta }_j}(B\cap E)=T_{q_j}(B\cap E)\mathrm{.}$ By (2)-(3), $T_{{\theta }_j}\to T_{\theta }$ and $T_{q_j}\to T_q$ in total variation. Passing to the limit proves the exact Borel measure identity

The proof uses bounded strict-contact only as established above; it does not assume a quasicontinuous-representative theorem or any external unbounded contact identity. It permits $\xi =-\mathrm{\infty }$ on arbitrary Borel sets.

Let $P=\{z:p_{\mu }(z)=-\mathrm{\infty }\}$. This is Borel by upper semicontinuity. For $R>\mathrm{diam}(\mathrm{supp}\mu )$ with $R>1$, put $\tilde{U}(z)=\int k_R(z,w)d\mu (w)$. Then $\tilde{U}=U^{\mu }+M\log R$ and, by finite energy and Tonelli, $\int \tilde{U}d\mu =I(\mu )+M^2\log R<\mathrm{\infty }\mathrm{.}$ Since $\tilde{U}=+\mathrm{\infty }$ on $P$, it follows that $\mu (P)=0$, so $T_{\varphi }(P)=0$.

For $E_j=\{\varphi >-j\}$ and ${\varphi }_j=\max (\varphi ,-j)$, write $T_j=\omega +d{d}^c{\varphi }_j$. On $\mathbb{C}$, ${\varphi }_j+{\rho }_{FS}=\max (u,{\rho }_{FS}-j)\mathrm{.}$ The finite-energy estimate above gives $u\in W_{loc}^{1,2}$; the one-obstacle strict-contact proof therefore gives $1_{E_j}{T}_j=1_{E_j}{T}_{\varphi }$ on $\mathbb{C}$. For all sufficiently large $j$, $E_j$ contains a neighborhood of infinity and ${\varphi }_j=\varphi$ there, so this equality holds on all of ${\mathbb{P}}^1$. The truncation measure ${\lim }_j{1}_{E_j}{T}_j$ consequently equals $T_{\varphi }{\mathrm{\mid }}_{{\mathbb{P}}^1\setminus P}$ and has mass one. Thus $\varphi \in \mathcal{E}$. The constant-shift property proved above gives $\varphi +\epsilon \in \mathcal{E}$ for every $\epsilon >0$. [F1, F3, F4, F5, F6, F8, F9, F10, F13]

Apply (6) with $\theta =\varphi +\epsilon$ and $\xi =\psi$. On $\mathbb{C}$, the strict set is $A_{\epsilon }$ and $\max (\varphi +\epsilon ,\psi )+{\rho }_{FS}=w_{\epsilon }$. Restricting (6) to $\mathbb{C}$ yields the exact Borel measure identity $1_{A_{\epsilon }}(\mu \mathrm{/}M)=1_{A_{\epsilon }}{\lambda }_{\epsilon },{\lambda }_{\epsilon }:={\mu }_{w_{\epsilon }}\mathrm{.}$ [F15, F16, F17, F18, F19]