5.15 Plurifine topologyDFFF533F1AE8

Proof. It suffices to show that the unit ball $\{ z\in \mathbb {C}^ n:|z|<1\} $ is $\mathcal{F}$-open. This follows from the observation that this set can be written as

This completes the proof. $\square$

Proof. By 5.15.5, there is an open neighborhood $U\subseteq \Omega $ of $x$ and $\psi \in \mathrm{PSH}(U)$ such that

Without loss of generality we may assume $x=0$ and that $U=B_ n(0,1)$ is the open unit ball.

Since $\psi $ is upper semicontinuous at $0$ with $\psi (0)=0$, we may choose a strictly decreasing sequence $(\delta _ j)_{j\geq 1}\subseteq (0,\mathrm{e}^{-1})$ such that

In particular $|\log \delta _ j|>1$ for every $j\geq 1$, so the series of constants

converges. We further set

which is also decreasing in $j$.

For each $j\geq 1$ define

On the boundary $|z|=\delta _ j$, the logarithmic term tends to $-2^{-j-1}$. By (5.15.7),

for every $z_0$ with $|z_0|=\delta _ j$. Hence the boundary condition in 5.14.16 is satisfied. We obtain that $\varphi _ j\in \mathrm{PSH}(\mathbb {C}^ n)$.

By construction, $\varphi _ j\leq 0$ on $U$, $\varphi _ j(0)=-2^{-j}$, and for $z\in E\setminus \{ 0\} $ with $|z|<\gamma _ j$ we have $\varphi _ j(z)\leq -1$, since

and, on $E\setminus \{ 0\} $, $\psi (z)-2^{-j}=-1-2^{-j}<-1$.

We then define

On the unit ball $U$ each $\varphi _ j\leq 0$, so the partial sums $S_ N\coloneqq \sum _{j=1}^ N\varphi _ j$ form a decreasing sequence of plurisubharmonic functions with

so $\varphi |_ U\in \mathrm{PSH}(U)$ by proposition 5.14.2. On $\mathbb {C}^ n\setminus B_ n(0,\delta _1)$ each $\varphi _ j$ equals $\frac{2^{-j-1}}{|\log \delta _ j|}\log |z|$, and (5.15.8) shows $S_ N$ converges uniformly on every compact subset to $\bigl (\sum _{j\geq 1}\tfrac {2^{-j-1}}{|\log \delta _ j|}\bigr )\log |z|$, a plurisubharmonic limit. Both formulas agree on $U\setminus B_ n(0,\delta _1)$, so altogether $\varphi \in \mathrm{PSH}(\mathbb {C}^ n)$.

We now verify the three claims.

Property 2. This is the equality $\varphi (0)=-\sum _{j\geq 1}2^{-j}=-1>-\infty $ obtained above.

Property 3. Fix $j_0\geq 1$. For $z\in E\setminus \{ 0\} $ with $|z|<\gamma _{j_0}$, the bound $\varphi _ j(z)\leq -1$ holds for every $j\leq j_0$ (since the $\gamma _ j$ are decreasing), while $\varphi _ j(z)\leq 0$ for $j>j_0$. Hence $\varphi (z)\leq -j_0$, and letting $j_0\to \infty $ yields

Property 1. Fix any $z_0\in \mathbb {C}^ n$ with $z_0\neq 0$ and choose $J\geq 1$ such that $\delta _ J<|z_0|/2$. On the open set $W=\{ w\in \mathbb {C}^ n:|w|>|z_0|/2\} $, all $\varphi _ j$ with $j\geq J$ lie in the outer regime, and (5.15.8) shows $\sum _{j\geq J}\varphi _ j(w) =\bigl (\sum _{j\geq J}\tfrac {2^{-j-1}}{|\log \delta _ j|}\bigr )\log |w|$ is locally bounded on $W$. The remaining finitely many terms $\varphi _1,\ldots ,\varphi _{J-1}$ are plurisubharmonic on $\mathbb {C}^ n$, hence locally bounded above on $W$, while $\varphi _ j(w)\geq \tfrac {2^{-j-1}}{|\log \delta _ j|}\log |w|$ keeps them locally bounded below as well. Therefore $\varphi $ is locally bounded on $W$. Since $z_0\neq 0$ was arbitrary, $\varphi $ is locally bounded on $\mathbb {C}^ n\setminus \{ 0\} $, which is property 1. $\square$

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

Then $\varphi _1+\varphi _2\in \mathrm{PSH}(U)$ and

In particular, $E_1\cup E_2$ is thin at $x$. $\square$

Proof. 2 $\implies $ 1. Assume 2. We may assume that $x\in \overline{\Omega \setminus E}$. Otherwise, our assertion follows from proposition 5.15.2.

By proposition 5.15.6, there is an open neighborhood $U$ of $x$ in $\Omega $ and $\varphi \in \mathrm{PSH}(\mathbb {C}^ n)$ such that

Let $F=\{ y\in \Omega : \varphi (y)>\lambda \} $. Then $x\in F$ and $F$ is $\mathcal{F}$-open by example 5.15.3. Moreover, $U\cap F\subseteq E$. By proposition 5.15.2, we conclude 1.

1 $\implies $ 2. Assume 1. We may always replace $E$ by smaller $\mathcal{F}$-neighborhoods of $x$. In particular, we may assume that $E$ has the following form

where $U\subseteq \Omega $ is an open neighborhood of $x$, and $\varphi _1,\ldots ,\varphi _ m$ are $\mathbb {R}$-valued psh functions on $\Omega $, and $\lambda _1,\ldots ,\lambda _ m\in \mathbb {R}$. Since $x\in U$, the set $\Omega \setminus U$ is thin at $x$. Moreover,

By 5.15.9, it is enough to prove that each set $\{ y\in U:\varphi _ i(y)\leq \lambda _ i\} $ is thin at $x$. Thus we may assume $m=1$. In this case, $\Omega \setminus E$ is clearly thin at $x$. $\square$

Proof. Observe that sets of the form (5.15.12) are $\mathcal{F}$-open.² By theorem 5.15.10, it suffices to show its complement in $\Omega $ is thin at each point of (5.15.12), which is obvious.

Now consider $x\in \Omega $ and an $\mathcal{F}$-open neighborhood $V\subseteq \Omega $ of $x$. We want to find a set of the form (5.15.12) contained in $V$ and containing $x$.

Write $E=\Omega \setminus V$. In case $x\in \operatorname{Int}V$, there is nothing to prove. So we may assume that $x\in \overline{E}$. By theorem 5.15.10, $E$ is thin at $x$. By definition, there is an open neighborhood $U\subseteq \Omega $ of $x$ and $\varphi \in \mathrm{PSH}(U)$ such that

We may assume that $\varphi |_{E\cap U}\leq 0<\varphi (x)$. Then the set $\{ y\in U:\varphi (y)>0\} $ suffices for our purpose. $\square$

Proof. By theorem 5.15.11, it suffices to prove the assertion after replacing $\Omega $ by a smaller $\mathcal F$-open neighborhood of $x$. Thus we may assume that

for some open set $U\subseteq \mathbb C^ n$ and some $\varphi \in \mathrm{PSH}(U)$.

Take a compact neighborhood $K$ of $x$ in $U$. Now $\{ y\in K: \varphi (y)\geq \varphi (x)/2\} $ is a compact $\mathcal{F}$-neighborhood of $x$ contained in $\Omega $. $\square$

Proof. It suffices to show that the inverse image $F^{-1}(U)$ of each $\mathcal{F}$-open subset $U\subseteq \Omega $ is $\mathcal{F}$-open. By theorem 5.15.11, after possibly shrinking $\Omega $ and $\Omega '$, we may assume that $U$ has the form $\{ x\in \Omega :\psi (x)>0\} $, where $\psi \in \mathrm{PSH}(\Omega )$. Clearly, we may assume that $F^{-1}(U)$ is non-empty. Then $F^*\psi \in \mathrm{PSH}(\Omega ')$ by proposition 5.13.17, we find that

is $\mathcal{F}$-open. $\square$

Proof. It suffices to show that $\varphi |_{\{ \varphi \neq -\infty \} }$ is $\mathcal{F}$-continuous, since $\{ \varphi =-\infty \} $ is $\mathcal{F}$-closed.

The problem is local, so we may assume that $X\subseteq \mathbb {C}^ n$ is a domain and $\varphi =\psi +g$, where $\psi \in \mathrm{PSH}(X)$ and $g\in C^{\infty }(X)$ and $|g|\leq C$ for some $C>0$. For an open interval $(a,b)\subseteq \mathbb {R}$, it suffices to show that

is $\mathcal{F}$-open. Take $x\in U$. We can find an open neighborhood $V$ of $x$ in $U$ such that

Therefore,

is an $\mathcal{F}$-open neighborhood of $x$ in $U$. We conclude that $U$ is $\mathcal{F}$-open. $\square$

Proof. It suffices to show that for any $x\in X$ such that $\varphi (x)>\psi (x)$, the same holds on an $\mathcal{F}$-neighborhood $U$ of $x$. Observe that $\varphi (x)\neq -\infty $. If $\psi (x)\neq -\infty $, then it suffices to apply proposition 5.15.18. Otherwise, take

This is an $\mathcal{F}$-open neighborhood of $x$ contained in $\{ \varphi >\psi \} $. $\square$

Proof. The problem is local, so we may assume that $X$ is a domain in $\mathbb {C}^ n$ and that there is $\varphi \in \mathrm{PSH}(X)$ with $Z\subseteq \{ \varphi =-\infty \} $. It suffices to show that $\{ \varphi >-\infty \} $ is $\mathcal{F}$-dense.

Let $x\in X$ be a point with $\varphi (x)=-\infty $ and $U\subseteq X$ be an $\mathcal{F}$-open neighborhood of $x$ in $X$. We need to show that $U\cap \{ \varphi >-\infty \} \neq \varnothing $.

Thanks to theorem 5.15.11, after shrinking $U$, we may assume that there is $\psi \in \mathrm{PSH}(X)$ such that $U=\{ \psi >0\} $. Observe that $U$ is not a pluripolar set: Otherwise, $\psi \leq 0$ almost everywhere by proposition 5.14.4, and hence everywhere by proposition 5.14.6. So $\varphi |_ U\not\equiv -\infty $. We conclude that $U\cap \{ \varphi >-\infty \} \neq \varnothing $, as desired. $\square$

Proof. This is an immediate consequence of 5.15.20 and proposition 5.15.18. $\square$