4.2.1 Fine neighborhoods and thinnessE15D6A501F63

Proof. Suppose first that $A$ is thin at $x$. If $x$ does not belong to the Euclidean closure of $A\setminus \{ x\} $, some Euclidean neighborhood of $x$ does not meet $A\setminus \{ x\} $. Otherwise let $u$ be a subharmonic function on a Euclidean open neighborhood $V$ of $x$ as in definition 4.2.3, and choose $a\in \mathbb R$ strictly between the upper limit in that definition and $u(x)$. By the definition of the upper limit, after shrinking $V$ we have $u<a$ on $(A\setminus \{ x\} )\cap V$. Hence $\{ y\in V:u(y)>a\} $ is a fine neighborhood of $x$ that does not meet $A\setminus \{ x\} $.

Conversely, let $N$ be a fine neighborhood of $x$. By the description of the basis above, $N$ contains a set

where $V$ is a Euclidean open neighborhood of $x$; adding the function $0$ with the constant $-1$ if necessary, we may assume that $m\geq 1$. Each $u_ i(x)$ is finite, since $u_ i(x)>a_ i$. Put $b_ i=u_ i(x)-a_ i>0$ and

Then $v(x)=0$. We claim that

There is nothing to prove if $x$ is not an accumulation point of $V\setminus N_0$. Otherwise let $y_ k\to x$ be a sequence in $V\setminus N_0$ along which $v(y_ k)$ converges to the upper limit. For each $k$ some index $i$ satisfies $u_ i(y_ k)\leq a_ i$, that is, $(u_ i(y_ k)-u_ i(x))/b_ i\leq -1$; passing to a subsequence, we may assume that this index $i$ is the same for all $k$. For every other index $j$, upper semicontinuity of $u_ j$ at $x$ gives $\varlimsup _ k(u_ j(y_ k)-u_ j(x))/b_ j\leq 0$. Hence $\lim _ kv(y_ k)\leq -1$, which proves the claim.

If $N$ does not meet $A\setminus \{ x\} $, then $(A\setminus \{ x\} )\cap V\subset V\setminus N_0$, and the claim shows that either $x$ is not in the Euclidean closure of $A\setminus \{ x\} $ or the upper limit of $v$ along $A\setminus \{ x\} $ is at most $-1<v(x)$. In both cases $A$ is thin at $x$. For the last assertion, the claim provides a Euclidean open neighborhood $V'\subset V$ of $x$ such that $v<-1/2$ on $V'\setminus N_0$. Thus $\{ y\in V':v(y)>-1/2\} \subset N_0\subset N$, and $v(x)=0>-1/2$. $\square$

Proof. By proposition 4.2.4, $N$ contains a set $\{ y\in V:u(y)>a\} $ with $u\in \mathrm{SH}(V)$ and $u(x)>a$. It suffices to consider balls $B(x,r)$ with $\overline{B(x,r)}\subset V$. The set $E=\{ y\in B(x,r):u(y)>a\} $ is a Borel subset of $N$, because $u$ is Borel measurable. Let $M$ be a finite upper bound for $u$ on $\overline{B(x,r)}$; then $M\geq u(x)>a$. Since $u\leq M$ on $E$ and $u\leq a$ on $B(x,r)\setminus E$, the ball submean inequality of theorem 3.4.5 gives

Hence $|E|\geq |B(x,r)|\, (u(x)-a)/(M-a)>0$. As a singleton has Lebesgue measure zero, it is not a fine neighborhood of its point. $\square$

Proof. We construct a subharmonic function on $\mathbb R^ d$ that is finite at $x$ and equals $-\infty $ on $A\setminus \{ x\} $ near $x$. Fix $x\in \Omega $ and $\rho >0$ with $\overline{B(x,\rho )}\subset \Omega $. For $k\geq 0$ put $r_ k=2^{-k}\rho $, let $S_ k=\{ y:r_{k+1}\leq |y-x|\leq r_ k\} $ be the closed annulus, and let $A_ k=A\cap S_ k$; these sets cover $A\cap \overline{B(x,\rho )}\setminus \{ x\} $. Each $A_ k$ is polar and has compact closure, so proposition 4.1.11 provides a finite positive measure $\nu _ k'$ of compact support with $E_ d*\nu _ k'=-\infty $ on $A_ k$. We restrict it to the open set $U_ k=\{ y:r_{k+2}<|y-x|<2\rho \} $, which contains $S_ k$, and put $\nu _ k=\nu _ k'|_{U_ k}$. For $y\in A_ k$ and $z\in \operatorname{Supp}\nu _ k'\setminus U_ k$ we have $|y-z|\geq \operatorname {dist}(S_ k,\mathbb R^ d\setminus U_ k)>0$, so $E_ d(y-z)$ is bounded on this set of pairs and

Moreover, $\nu _ k$ is supported in $\overline{U_ k}$, which does not meet $B(x,r_{k+2})$, so $p_ k=E_ d*\nu _ k$ is finite at $x$.

Choose $c_ k>0$ with

and put $\nu =\sum _ kc_ k\nu _ k$, a finite measure supported in $\overline{B(x,2\rho )}$, and $p=E_ d*\nu $. Let $C$ be an upper bound for $E_ d(y-z)$ when $|y-z|\leq 3\rho $. For $y\in \overline{B(x,\rho )}$, monotone convergence applied to the nonnegative functions $C-E_ d(y-\cdot )$ gives

where every term on the right is nonpositive. Hence $p(x)$ is finite, and $p=-\infty $ on $A\cap \overline{B(x,\rho )}\setminus \{ x\} $, because there one term equals $-\infty $. Since $p$ is subharmonic on $\mathbb R^ d$ (proposition 3.4.11), the set

is a fine neighborhood of $x$, and $N\cap A\subset \{ x\} $. If $x\notin A$, this shows that $\Omega \setminus A$ is a fine neighborhood of each of its points, so $A$ is finely closed. If $x\in A$, it shows that $\{ x\} $ is relatively open in $A$ for the fine topology. $\square$

Proof. The function $G$ is upper semicontinuous, because $\{ G<a\} =\{ g<a\} \times \mathbb R^{d-k}$ for every real $a$. Since subharmonicity is a local property, it suffices to show that $G$ is subharmonic on $D\times \mathbb R^{d-k}$ for every open ball $D=B(z_0,s)\subset \mathbb R^ k$ with $\overline{B(z_0,2s)}\subset V$. If $g$ is identically $-\infty $ on $B(z_0,2s)$, then so is $G$ on $B(z_0,2s)\times \mathbb R^{d-k}$, and there is nothing to prove. Otherwise $g$ is locally integrable on the ball $B(z_0,2s)$ by proposition 3.4.8.

For $0<\varepsilon <s$, the $k$-dimensional mollification $g_\varepsilon =g*\rho _\varepsilon $ is smooth on $D$ and satisfies $\Delta _ zg_\varepsilon =(\Delta g)*\rho _\varepsilon \geq 0$ there, by part (1) of theorem 3.4.12 and parts (1) and (3) of proposition 2.5.3; moreover, $g_\varepsilon $ decreases to $g$ on $D$ as $\varepsilon \downarrow 0$, by (3.4.13). The function $G_\varepsilon (z,y)=g_\varepsilon (z)$ is smooth on $D\times \mathbb R^{d-k}$, with $\Delta _ xG_\varepsilon (z,y)=\Delta _ zg_\varepsilon (z)\geq 0$. By (3.1.6), its sphere means in $\mathbb R^ d$ increase with the radius, and they tend to its value at the center as the radius tends to zero, because $G_\varepsilon $ is continuous. Hence $G_\varepsilon $ satisfies the submean inequality and is subharmonic on $D\times \mathbb R^{d-k}$. Along the sequence $\varepsilon _ j=s/(j+1)$, the functions $G_{\varepsilon _ j}$ decrease to $G$, which is therefore subharmonic on $D\times \mathbb R^{d-k}$ by theorem 3.4.9. $\square$

Proof. A similarity $x\mapsto a+tx$ with $t>0$ preserves upper semicontinuity and sphere means, and hence maps polar sets to polar sets. It therefore suffices to find one uncountable compact polar subset of $\mathbb R^ d$. For $d=2$ we take the Cantor set $K$ of proposition 4.1.12. It is compact and polar, and it is uncountable because it carries an atom-free probability measure.

For $d>2$, write $x=(z,y)\in \mathbb R^2\times \mathbb R^{d-2}$. By lemma 4.2.7, applied with $k=2$ to the subharmonic function $p$ of proposition 4.1.12, the function $P(x)=p(z)$ is subharmonic on $\mathbb R^ d$. It is finite when $z\notin K$ and equals $-\infty $ on $K\times \mathbb R^{d-2}$. Hence $K\times \{ 0\} $ is an uncountable compact polar subset of $\mathbb R^ d$.

Now let $\Omega \neq \varnothing $, and let $K'\subset \Omega $ be an uncountable compact polar set. By proposition 4.2.6, $K'$ is finely closed, and its induced fine topology is discrete. A closed subspace $F$ of a Lindelöf space $X$ is Lindelöf: extend the members of an open cover of $F$ to open subsets of $X$, add $X\setminus F$, and pass to a countable subcover. On the other hand, the cover of an uncountable discrete space by its singletons has no countable subcover. Hence the fine topology on $\Omega $ is not Lindelöf. Nor is it second countable, because a second countable space is Lindelöf: given an open cover and a countable basis, choose, for each basis element contained in some member of the cover, one such member; these countably many members cover the space. $\square$