PomeLemma 5.15.9.D3F2BF3D82BD Let $E_1,E_2\subseteq \Omega $. Assume that $E_1$, $E_2$ are both thin at $x\in \Omega $. Then so is $E_1\cup E_2$.
Lemma 5.15.9D3F2BF3D82BD
Proof. We may assume that $x\in \overline{E_1}\cap \overline{E_2}$. Take an open neighborhood $U\subseteq \Omega $ of $x$ and $\varphi _1,\varphi _2\in \mathrm{PSH}(U)$ such that
\[ \varlimsup _{y\to x,y\in E_ i\setminus \{ x\} } \varphi _ i(y)<\varphi _ i(x),\quad i=1,2. \]
Then $\varphi _1+\varphi _2\in \mathrm{PSH}(U)$ and
\[ \varlimsup _{y\to x,y\in (E_1\cup E_2)\setminus \{ x\} } (\varphi _1+\varphi _2)(y)<\varphi _1(x)+\varphi _2(x). \]
In particular, $E_1\cup E_2$ is thin at $x$. $\square$
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