Compact capacity-zero sets and subharmonic minus-infinity loci

Dependent Choice is assumed in the statement, and by [F3] it yields Countable Choice, which is the selection principle used for the countable constructions below; Dependent Choice itself is used for the successive subsequences constructed later in this proof and through the Riesz suppliers [F8] and [F13].

The case $E=\mathrm{\varnothing }$: $\mathrm{cap}(\mathrm{\varnothing })=0$ by [F2], and the constant function $u\equiv 0$ is subharmonic on the complex domain $\mathbb{C}$ by [F11] with empty $-\mathrm{\infty }$ locus, so every empty compact set is the $-\mathrm{\infty }$ locus condition holds vacuously; conversely the condition $\mathrm{cap}(\mathrm{\varnothing })=0$ holds by convention. So the compact equivalence is true for $E=\mathrm{\varnothing }$, and below $E$ is assumed nonempty.

Now assume $E\mathrm{\ne }\mathrm{\varnothing }$ compact with $\mathrm{cap}(E)=0$. Choose $\rho >0$ with $E\subseteq D(0,\rho )$, put $X:=D(0,\rho )$ and $R:=2\rho +1$, so that $R>\mathrm{diam}(\bar{X})$ and $G(z,w):=\log \frac{R}{\mathrm{\mid }z-w\mathrm{\mid }}\in (0,+\mathrm{\infty }]$ is nonnegative on $X\times X$ and $+\mathrm{\infty }$ exactly on the diagonal. For every $\nu \in P(E)$ one has, by [F1] applied with this $R>\mathrm{diam}(E)$, $\int \int Gd\nu d\nu =I(\nu )+\log R$, and $I(\nu )=+\mathrm{\infty }$ by the characterization [F2] of $\mathrm{cap}(E)=0$; hence $\int \int Gd\nu d\nu =+\mathrm{\infty }$ for every Borel probability $\nu$ on $E$.

Let $E$ be nonempty compact and let $({\nu }_m{)}_m$ be a sequence of Borel probability measures on $E$. It will be shown that some subsequence converges weakly to a probability on $E$. By [F6] enumerate the rational boxes as $B_1,B_2,\dots$; applying Countable Choice of [F3] to the at-most-countable family $X_n:=E\cap B_n$ when this is nonempty and $X_n:=\{x_0\}$ for a fixed $x_0\in E$ otherwise gives points $x_n\in X_n$, and $D:=\{x_n:n\ge 1\}$ is countable. It is dense in $E$: if $U$ is open and $U\cap E\mathrm{\ne }\mathrm{\varnothing }$, [F6] gives a rational box $B_n$ with $x\in B_n\subseteq U$ for some $x\in U\cap E$, so $x_n\in E\cap B_n\subseteq U\cap E$.

With $G$ and $R$ as in step 1.3, for $n\ge 1$ put $G_n:=\min (G,n)$ and $I_n(\nu ):=\int \int G_{n}d\nu d\nu \in [0,n]$ for $\nu \in P(E)$. Each $G_n$ is continuous and bounded on $E\times E$, and $I_n$ is weakly continuous: the unital algebra of finite sums ${\sum }_l{f}_l(z)h_l(w)$ separates points of the compact metric space $E\times E$, so it is uniformly dense in $C(E\times E;\mathbb{R})$ by [F7]; for a product integrand the double integral factors into a product of single integrals, which converges along weakly convergent sequences by [F4]; uniform approximation handles the general integrand. Put $e_n:={\inf }_{\nu \in P(E)}{I}_n(\nu )$.

Let $V$ be the $\mathbb{Q}$-algebra generated inside $C(E;\mathbb{R})$ by the constant function $1$ and the functions $d(\cdot ,x)$, $x\in D$; being generated by countably many elements, $V$ is countable, so fix an enumeration $V=\{g_1,g_2,\dots \}$. $V$ is uniformly dense in $C(E;\mathbb{R})$: its closure $\bar{V}$ is a closed $\mathbb{R}$-subalgebra containing the generators, those generators separate points of $E$ (for $x\mathrm{\ne }y$ choose $a\in D$ with $d(x,a)<d(x,y)\mathrm{/}2$, then $d(y,a)\ge d(x,y)-d(x,a)>d(x,a)$), so $\bar{V}$ contains the unital algebra generated by the generators, which is all of $C(E;\mathbb{R})$ by [F7].

Successive subsequences are chosen by Dependent Choice. A state is a pair $(k,j)$ with $k\ge 0$ and $j:\mathbb{N}\to \mathbb{N}$ strictly increasing, and $(k,j)$ is related to $(k+1,j^{\mathrm{\prime }})$ when $j^{\mathrm{\prime }}=j\circ i$ for a strictly increasing $i:\mathbb{N}\to \mathbb{N}$ and the real sequence $m\mapsto \int g_{k+1}d{\nu }_{j^{\mathrm{\prime }}(m)}$ converges. The relation is entire: $m\mapsto \int g_{k+1}d{\nu }_{j(m)}$ is bounded by $\mathrm{\parallel }{g}_{k+1}{\mathrm{\parallel }}_{\mathrm{\infty }}$, so [F5] supplies a strictly increasing $i$ making it converge. Starting from $(0,\mathrm{i}\mathrm{d})$, Dependent Choice yields states $(k,j^{(k)}{)}_{k\ge 0}$ with $j^{(k+1)}$ a subsequence of $j^{(k)}$, and the diagonal $j_{\ast }(m):=j^{(m)}(m)$ is strictly increasing; for every $i$ the sequence $m\mapsto \int g_{i}d{\nu }_{j_{\ast }(m)}$ converges, since for $m\ge i$ it is a subsequence of the convergent sequence along $j^{(i)}$.

For $f\in C(E;\mathbb{R})$ and $\eta >0$ step 2.2 gives $i$ with $\mathrm{\parallel }f-g_i{\mathrm{\parallel }}_{\mathrm{\infty }}<\eta$, and then $\mid \int fd{\nu }_{j_{\ast }(m)}-\int fd{\nu }_{j_{\ast }(n)}\mid \le 2\eta +\mid \int g_{i}d{\nu }_{j_{\ast }(m)}-\int g_{i}d{\nu }_{j_{\ast }(n)}\mid$ shows that the $f$-integrals are Cauchy; define $L(f):={\lim }_m\int fd{\nu }_{j_{\ast }(m)}$. Limits of integrals against probability measures give that $L$ is linear, positive, $L(1)=1$ and $\mathrm{\mid }L(f)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$.

The compact metric space $E$ is LCH and $C_0(E;\mathbb{R})=C(E;\mathbb{R})$, so [F8], whose hypothesis is Dependent Choice, represents $L$ as integration against a unique Borel probability measure $\mu$ on $E$; by [F4] this says ${\nu }_{j_{\ast }(m)}\Rightarrow \mu$. This proves the claim of step 1.4.

The infimum $e_n$ is attained. By Countable Choice [F3] choose ${\nu }_{n,j}\in P(E)$ with $I_n({\nu }_{n,j})<e_n+1\mathrm{/}j$ for all $n,j\ge 1$. Fix $n$; step 5.1 applied to the sequence $({\nu }_{n,j}{)}_j$ give a weakly convergent subsequence with limit ${\mu }_n\in P(E)$, and weak continuity of $I_n$ from step 2.1 gives $I_n({\mu }_n)={\lim }_j{I}_n({\nu }_{n,j^{\mathrm{\prime }}})=e_n$. Hence the set of minimizers of $I_n$ is nonempty for every $n$, and Countable Choice selects one minimizer ${\mu }_n$ for each $n$.

The sequence $e_n$ is nondecreasing and $e_n\to +\mathrm{\infty }$. Monotonicity is immediate from $G_n\le G_{n+1}$. If $e_n\le L$ for all $n$, take minimizers ${\mu }_n$ from step 6.1 and apply step 5.1 to $({\mu }_n)$ to obtain a weakly convergent subsequence ${\mu }_{n_j}\Rightarrow \mu$. For fixed $m$, whenever $n_j\ge m$ one has $I_m({\mu }_{n_j})\le I_{n_j}({\mu }_{n_j})=e_{n_j}\le L$, so weak continuity of $I_m$ gives $I_m(\mu )\le L$; monotone convergence [F9] for $G_m\uparrow G$ then gives $\int \int Gd\mu d\mu ={\lim }_m{I}_m(\mu )\le L<+\mathrm{\infty }$, contradicting step 1.3.

First variation at an exact minimizer. Let $n\ge 1$, let $x\in E$ and let ${\mu }_n$ be a minimizer of $I_n$. For $0<t\le 1$ the measure ${\nu }_t:=(1-t){\mu }_n+t{\delta }_x$ is a probability on $E$, and expanding the double integral gives $I_n({\nu }_t)=(1-t{)}^2{I}_n({\mu }_n)+2t(1-t)G_n{\mu }_n(x)+t^2{G}_n(x,x)$, where $G_n{\mu }_n(x):=\int G_n(x,w)d{\mu }_n(w)$ and $G_n(x,x)=n$. Since $I_n({\nu }_t)\ge e_n=I_n({\mu }_n)$, dividing the inequality $I_n({\nu }_t)-I_n({\mu }_n)\ge 0$ by $t$ and letting $t\downarrow 0$ gives $2(G_n{\mu }_n(x)-e_n)\ge 0$, that is $G_n{\mu }_n(x)\ge e_n$ for every $x\in E$.

For $k\ge 1$ let $n_k:=\min \{n:e_n\ge k^3\}$, finite by step 7.1, and by Countable Choice choose minimizers ${\mu }_{n_k}$; the measure $\sigma :={\sum }_{k\ge 1}{k}^{-2}{\mu }_{n_k}$ is a finite positive Borel measure on $E$ with $\sigma (E)={\sum }_{k\ge 1}{k}^{-2}<+\mathrm{\infty }$ by [F15]. For $z\in E$ the potential of $\sigma$ against the shifted kernel satisfies $G\sigma (z)=\int G(z,w)d\sigma (w)={\sum }_{k\ge 1}{k}^{-2}G{\mu }_{n_k}(z)\ge {\sum }_{k\ge 1}{k}^{-2}{G}_{n_k}{\mu }_{n_k}(z)\ge {\sum }_{k\ge 1}{k}^{-2}{e}_{n_k}\ge {\sum }_{k\ge 1}k=+\mathrm{\infty }$.

For $z\in E$, $p_{\sigma }(z)=\int \log \mathrm{\mid }z-w\mathrm{\mid }d\sigma (w)=\sigma (E)\log R-G\sigma (z)=-\mathrm{\infty }$, because $\log \mathrm{\mid }z-w\mathrm{\mid }=\log R-G(z,w)$ holds for $z\mathrm{\ne }w$ and both sides are $-\mathrm{\infty }$ at $z=w$. The measure $\sigma$ is finite with compact support $E$, so [F16], whose hypothesis Countable Choice is available by step 1.1, makes $p_{\sigma }$ locally integrable and subharmonic on the complex domain $\mathbb{C}$; in particular $p_{\sigma }$ is not identically $-\mathrm{\infty }$. This proves the forward direction $\mathrm{cap}(E)=0\Rightarrow$ witness for nonempty compact $E$.

Conversely, let $E\mathrm{\ne }\mathrm{\varnothing }$ be compact, let $\mathrm{\Omega }$ be a complex domain with $E\subseteq \mathrm{\Omega }$, and let $u$ be subharmonic on $\mathrm{\Omega }$ with $u(z)=-\mathrm{\infty }$ for every $z\in E$. Suppose $\mathrm{cap}(E)>0$. Then $V_E<+\mathrm{\infty }$ by [F2], so there is $\mu \in P(E)$ with $I(\mu )<+\mathrm{\infty }$. Fix $R>1+\mathrm{diam}(E)$ and put $G(z,w):=\log \frac{R}{\mathrm{\mid }z-w\mathrm{\mid }}$ and ${\tilde{U}}^{\lambda }:=U^{\lambda }+\lambda (\mathbb{C})\log R=\int G(\cdot ,w)d\lambda (w)$ for finite $\lambda$. Then $\int {\tilde{U}}^{\mu }d\mu =I(\mu )+\log R<+\mathrm{\infty }$. For $M>0$ put $G_M:=\min (M,G)$ on $E\times E$. This is continuous and bounded, and monotone convergence gives ${\tilde{U}}^{\mu }(z)={\lim }_{M\to \mathrm{\infty }}\int G_M(z,w)d\mu (w)$ for $z\in E$. Each truncated integral is continuous on $E$, so ${\tilde{U}}^{\mu }$ is lower semicontinuous there. Hence for every real $b$ the set $F_b:=\{z\in E:{\tilde{U}}^{\mu }(z)\le b\}$ is closed in $E$, and for $b>I(\mu )+\log R$ it has positive measure: $\mu (E\setminus F_b)\le \frac{1}{b}\int {\tilde{U}}^{\mu }d\mu <1=\mu (E)$, since ${\tilde{U}}^{\mu }\ge 0$ on $E$ and ${\tilde{U}}^{\mu }>b$ on $E\setminus F_b$. Fix such a $b$ and put $\lambda :=\mu {\mathrm{\mid }}_{F_b}$, the restriction of $\mu$ to the measurable set $F_b$ (Restriction of a measure to a measurable set), a nonzero finite positive measure with $\mathrm{supp}\lambda \subseteq F_b$ because $F_b$ is closed. For $z\in F_b$, ${\tilde{U}}^{\lambda }(z)={\int }_{F_b}G(z,w)d\mu (w)\le {\tilde{U}}^{\mu }(z)\le b$, since $G\ge 0$ on $E\times E$. Cover $E$ by finitely many open discs $D_1,\dots ,D_m$ whose closures lie in $\mathrm{\Omega }$ and whose radii are less than $1\mathrm{/}2$: such discs cover $E$ because $\mathrm{\Omega }$ is open and $R>1$, so compactness gives a finite subcover. Then $\mathrm{diam}{\bar{D}}_i<1<R$, and since $\lambda (E)>0$ some $i$ satisfies $\lambda (D_i)>0$. Put ${\lambda }_i:=\lambda {\mathrm{\mid }}_{D_i}$, a nonzero finite positive measure with $\mathrm{supp}{\lambda }_i\subseteq F_b\cap {\bar{D}}_i$; for $z\in F_b$ one has ${\tilde{U}}^{{\lambda }_i}(z)\le {\tilde{U}}^{\lambda }(z)\le b$ because ${\lambda }_i\le \lambda$ and $G\ge 0$. Thus $U^{{\lambda }_i}\le b-{\lambda }_i(\mathbb{C})\log R$ on $\mathrm{supp}{\lambda }_i$, and the maximum principle [F14] (applied to the nonzero measure ${\lambda }_i$) gives $U^{{\lambda }_i}\le b-{\lambda }_i(\mathbb{C})\log R$ on all of $\mathbb{C}$, that is, ${\tilde{U}}^{{\lambda }_i}\le b$ everywhere. On the disc $D_i$ the Riesz decomposition [F13] gives a finite positive measure $M_i$ carried by $D_i$ and a harmonic $h_i$ on $D_i$ with $u=h_i-U^{M_i}$ there. Since $u=-\mathrm{\infty }$ on $E\cap D_i$ and $h_i$ is finite there, $U^{M_i}(z)=+\mathrm{\infty }$ for every $z\in E\cap D_i$, hence on $F_b\cap D_i$, which has ${\lambda }_i$-measure ${\lambda }_i(D_i)>0$; therefore $\int U^{M_i}d{\lambda }_i=+\mathrm{\infty }$, and since ${\tilde{U}}^{M_i}=U^{M_i}+M_i(\mathbb{C})\log R\ge U^{M_i}$ also $\int {\tilde{U}}^{M_i}d{\lambda }_i=+\mathrm{\infty }$. Tonelli [F10] computes the same product integral in the other order: $\int {\tilde{U}}^{M_i}d{\lambda }_i=\int \int Gd{M}_{i}d{\lambda }_i=\int \int Gd{\lambda }_{i}d{M}_i=\int {\tilde{U}}^{{\lambda }_i}d{M}_i\le b{M}_i(\mathbb{C})<+\mathrm{\infty }$, because ${\tilde{U}}^{{\lambda }_i}\le b$ everywhere and $M_i$ is finite. This contradiction gives $\mathrm{cap}(E)=0$, the converse implication.

For the $F_{\sigma }$ extension, let $(E_j{)}_{j\ge 1}$ be a specified sequence of compact subsets of $\mathbb{C}$ with $\mathrm{cap}(E_j)=0$ for every $j$, and put $E:={\bigcup }_{j\ge 1}{E}_j$. For each $j$ with $E_j\mathrm{\ne }\mathrm{\varnothing }$, step 9.1 apply to $E_j$ and yield a finite positive measure ${\sigma }_j$ carried by $E_j$ with $p_{{\sigma }_j}(z)=-\mathrm{\infty }$ for every $z\in E_j$; for $E_j=\mathrm{\varnothing }$ put ${\sigma }_j:=0$. Countable Choice selects the family $({\sigma }_j{)}_{j\ge 1}$.

With $m_j:={\sigma }_j(\mathbb{C})$ and $r_j:={\max }_{w\in E_j}\mathrm{\mid }w\mathrm{\mid }$ for $E_j\mathrm{\ne }\mathrm{\varnothing }$, and $m_j=r_j:=0$ otherwise, set $a_j:=2^{-j}/(1+m_j(1+\log (1+r_j)))>0$ and $\sigma :={\sum }_{j\ge 1}{a}_j{\sigma }_j$. Then $\sigma$ is a finite positive Borel measure with $\sigma (\mathbb{C})\le {\sum }_{j\ge 1}{2}^{-j}=1$ and $\int \log (1+\mathrm{\mid }w\mathrm{\mid })d\sigma (w)\le {\sum }_{j\ge 1}{a}_j{m}_j\log (1+r_j)\le {\sum }_{j\ge 1}{2}^{-j}=1$; in particular the logarithmic moment of $\sigma$ is finite.

Put $u(z):=p_{\sigma }(z)=\int \log \mathrm{\mid }z-w\mathrm{\mid }d\sigma (w)$. If $z\in E_{j_0}$ for some $j_0$ with $E_{j_0}\mathrm{\ne }\mathrm{\varnothing }$, then $\int {\log }^{+}\mathrm{\mid }z-w\mathrm{\mid }d\sigma (w)\le \log (1+\mathrm{\mid }z\mathrm{\mid })+\int \log (1+\mathrm{\mid }w\mathrm{\mid })d\sigma (w)<+\mathrm{\infty }$ because $\mathrm{\mid }z-w\mathrm{\mid }\le (1+\mathrm{\mid }z\mathrm{\mid })(1+\mathrm{\mid }w\mathrm{\mid })$, while the term $a_{j_0}{\sigma }_{j_0}$ contributes $a_{j_0}\int {\log }^{-}\mathrm{\mid }z-w\mathrm{\mid }d{\sigma }_{j_0}(w)=+\mathrm{\infty }$: indeed $p_{{\sigma }_{j_0}}(z)=-\mathrm{\infty }$ by step 10.2 and $\int {\log }^{+}\mathrm{\mid }z-w\mathrm{\mid }d{\sigma }_{j_0}(w)<+\mathrm{\infty }$ since ${\sigma }_{j_0}$ is finite with compact support. Hence $\int {\log }^{-}\mathrm{\mid }z-w\mathrm{\mid }d\sigma (w)=+\mathrm{\infty }$ and $u(z)=\int {\log }^{+}-\int {\log }^{-}=-\mathrm{\infty }$; that is, $u=-\mathrm{\infty }$ on $E$.

Local integrability: there is for every compact $Q\subseteq \mathbb{C}$ a constant $C_Q<\mathrm{\infty }$ with ${\int }_Q\mathrm{\mid }\log \mathrm{\mid }z-w\mathrm{\mid }\mathrm{\mid }dA(z)\le C_Q(1+\log (1+\mathrm{\mid }w\mathrm{\mid }))$ for every $w\in \mathbb{C}$. Indeed, if $\mathrm{\mid }w\mathrm{\mid }\le 2+2{\sup }_{z\in Q}\mathrm{\mid }z\mathrm{\mid }$ then $z-w$ ranges over a fixed bounded region and the integral is bounded by ${\int }_{B(0,K)}\mathrm{\mid }\log \mathrm{\mid }v\mathrm{\mid }\mathrm{\mid }dA(v)<+\mathrm{\infty }$ for a suitable ball $B(0,K)$ by [F12]; if $\mathrm{\mid }w\mathrm{\mid }$ is larger then $\mathrm{\mid }z-w\mathrm{\mid }\ge \mathrm{\mid }w\mathrm{\mid }\mathrm{/}2>1$ on $Q$, so $\mathrm{\mid }\log \mathrm{\mid }z-w\mathrm{\mid }\mathrm{\mid }\le \log (2\mathrm{\mid }w\mathrm{\mid })\le 1+\log (1+\mathrm{\mid }w\mathrm{\mid })$ and the bound follows with $\mathrm{area}(Q)$. Tonelli [F10] with the nonnegative integrand $\mathrm{\mid }\log \mathrm{\mid }z-w\mathrm{\mid }\mathrm{\mid }$ and the finite measure $\sigma$ therefore gives ${\int }_Q{\int }_{\mathbb{C}}\mathrm{\mid }\log \mathrm{\mid }z-w\mathrm{\mid }\mathrm{\mid }d\sigma (w)dA(z)\le C_Q(1+\int \log (1+\mathrm{\mid }w\mathrm{\mid })d\sigma (w))<+\mathrm{\infty }$, so $u\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathbb{C})$; in particular $u$ is finite Lebesgue-a.e. and is not identically $-\mathrm{\infty }$ on the domain $\mathbb{C}$.

Fix $N\ge 1$ and split $\sigma ={\sigma }_N+{\sigma }^N$ on $D(0,N)$, where ${\sigma }_N$ is its restriction to $\{\mathrm{\mid }w\mathrm{\mid }\le 2N+1\}$ and ${\sigma }^N$ its restriction to $\{\mathrm{\mid }w\mathrm{\mid }>2N+1\}$. The measure ${\sigma }_N$ is finite with compact support, so $p_{{\sigma }_N}$ is subharmonic by [F16]. For $z\in D(0,N)$ and $w$ in the tail one has $\mathrm{\mid }z-w\mathrm{\mid }>N+1>1$; the logarithmic moment from step 11.1 makes $\log \mathrm{\mid }z-w\mathrm{\mid }$ integrable against ${\sigma }^N$, and its first and second derivatives in $z$ are bounded on this disc by constants because the distance is bounded below by $N+1$. The kernel and all its derivatives are Borel in $w$ and smooth in $z$ away from $w$. The bounds on derivatives of orders one and two are integrable constants because ${\sigma }^N$ is finite. Applying [F18] along each coordinate interval inside the disc, first to the kernel and then to its first derivatives, permits differentiation under the integral twice; dominated convergence in [F18] makes those derivatives continuous. By [F19], and $\mathrm{\Delta }{p}_{{\sigma }^N}(z)=\int {\mathrm{\Delta }}_z\log \mathrm{\mid }z-w\mathrm{\mid }d{\sigma }^N(w)=0$. Hence $p_{{\sigma }^N}$ is harmonic on $D(0,N)$ and $u=p_{{\sigma }_N}+p_{{\sigma }^N}$ is subharmonic there by [F19]. Each closed disc is contained in one of these discs, which exhaust $\mathbb{C}$, so $u$ is upper semicontinuous and satisfies the circle mean inequality locally on $\mathbb{C}$; it is not identically $-\mathrm{\infty }$ by step 12.2. Thus $u$ is subharmonic on $\mathbb{C}$ by [F11] and is $-\mathrm{\infty }$ on $E$ by step 12.1.

Conversely to the forward direction of steps 10.2–13.1, if $u$ is subharmonic on a complex domain $\mathrm{\Omega }$ with $E\subseteq \mathrm{\Omega }$ and $u=-\mathrm{\infty }$ on $E$, then every $E_j$ is a nonempty or empty compact subset of the domain $\mathrm{\Omega }$; for nonempty $E_j$, step 10.1 applied to $E_j$ give $\mathrm{cap}(E_j)=0$, and empty pieces have capacity zero by [F2].

Assembling: step 1.2 and step 9.1 give the compact forward direction, with the witness $\sigma$ of step 8.1 and $U^{\sigma }=+\mathrm{\infty }$ on $E$ by step 9.1; step 10.1 gives the compact converse; steps 10.2–13.1 give the witness $\sigma$ of step 11.1 for a specified $F_{\sigma }$ union of compact capacity-zero sets, with $U^{\sigma }=+\mathrm{\infty }$ on $E$ by steps 12.1 and 13.1; and step 14.1 gives the converse for such unions. The potential-form clauses of the statement are thus the witnesses actually constructed, so both assertions and their Evans-measure refinements are proved.