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hello everybody, i'm studying the article of Bedford-Taylor "Fine topology silov boundary..." but i don't understand the proof of the following proposition.

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

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

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

Then since $O=\bigcap_{k}(u_k>v)$ and since $(dd^{c}\max(u_k,v))^{n}|O_k=(dd^{n}u_k)^{n}|O_k$ (this dont'understand),

where $O_k=(u_k>v)$, the statement holds.

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

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On $O_k$, $max\{u_k,v\}=u_k$. Or is it something else that is the problem? –  Margaret Friedland Oct 26 '12 at 1:02
yes i see it, but even on $O$ $u=\max(u,v)$ so i don't understand why he passes to an approximation sequence. –  user27561 Oct 26 '12 at 8:13
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). –  Vamsi Oct 26 '12 at 15:48
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). –  bonho Oct 26 '12 at 16:15
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