## monge-ampere operator

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

-
 On $O_k$, $max\{u_k,v\}=u_k$. Or is it something else that is the problem? – Margaret Friedland Oct 26 at 1:02 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 at 15:48

yes i see it, but even on $O$ $u=\max(u,v)$ so i don't understand why he passes to an approximation sequence.