Proposition 4.2.4CAAD6BABA60C

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 $v(x)=0>-1/2$. $\square$