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I'm studying the article of Bedford–Taylor "Fine topology, Šilov boundary…" but I don't understand the proof of the following proposition.

Let $u$, $v$ be plurisubharmonic functions defined on $\Omega$, an open subset of $\mathbb{C}^n$, such that $u,v\in L^{\infty}_{\text{loc}}(\Omega)$.

Let $O$ be a fine-open subset of $\Omega$ where $u>v$. Then $(dd^{c}\max(u,v))^{n}\rvert O=(dd^{n}u)^{n}\rvert O$.

They say that if $O$ is open then it is obvious, while if $O$ is just fine open then it is shown using a decreasing approximation sequence $u_k$ of smooth functions for $u$.

Then since $O=\bigcap_{k}(u_k>v)$ and since $(dd^{c}\max(u_k,v))^{n}\rvert O_k=(dd^{n}u_k)^{n}\rvert O_k$ (this I don't understand), where $O_k=(u_k>v)$, the statement holds.

Maybe it is trivial but I'm not able to see it.

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  • $\begingroup$ On $O_k$, $max\{u_k,v\}=u_k$. Or is it something else that is the problem? $\endgroup$ Oct 26, 2012 at 1:02
  • $\begingroup$ yes i see it, but even on $O$ $u=\max(u,v)$ so i don't understand why he passes to an approximation sequence. $\endgroup$
    – user27561
    Oct 26, 2012 at 8:13
  • $\begingroup$ This isn't a complete answer, but may help. For smooth u_k, u_k >v is an open set. For non-smooth u, this is not necessarily the case. It may have a boundary. Just outside the boundary max(u,v) = v and inside max(u,v) = u. So, whilst testing the current against test functions whose support "ends" at the boundary, it isn't obvious (to me) that dd^c(max(u,v)) = dd^c (u). $\endgroup$
    – Vamsi
    Oct 26, 2012 at 15:48
  • $\begingroup$ Hi digital: I can not give comments, hence give answer here for your answer. If O is not open, then u=max(u,v) on O does not neccessarily imply that dd^cu=dd^cmax (u,v) on a neighborhood of O, so that when restricting to O you get the equality. (If you want to do derivatives, you need to do it on an open set). $\endgroup$
    – bonho
    Oct 26, 2012 at 16:15

1 Answer 1

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As a psh function, v is usc, so the set O_k is open, hence one can apply the usual locality property of Bedford--Taylor product.

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