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Recently,I am reading kolodziej's acta paper,there are some ditails that i do not know clearly.

In the top line of page 99,"it's no restriction to assume that for each s we have $\nu(\cup_{I\in{B_s}}\partial I)=0$",i do not know why.Here $\nu=(dd^{c}v)^n$ and $v\in PSH(\Omega)\cap L^{\infty}$,$\partial I$ is the boundary of some cube.

From Kolodziej's view,a poriori $\nu(\cup_{I\in{B_s}}\partial I)$ may not be zero.However,i think $\partial I$ can be seen as a part of a pluripolar set,then according to Bedford and Taylor's reasult,we know the above monge-ampere measure concentrates no mass on $\partial I$.So we get $\nu(\cup_{I\in{B_s}}\partial I)=0$,and it should not be a assumption!

I hope some expert in this field can help me.Thanks.

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closed as too localized by Franz Lemmermeyer, Angelo, Andrés Caicedo, Daniel Moskovich, Ryan Budney Dec 29 '10 at 6:26

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1 Answer 1

First of all, it would have been nice if you would have written the backgrounds for your questions, at least the assumptions and the claim which you were wondering. However, as I am currently reading things somewhat related to the paper you are referring to I looked it up.

One way to argue is this: We have $\nu(\Omega)<1$. (For example a $\sigma$-finite Borel measure would suffice.) If you fix the subdivisions $\beta_s$ of $I_0$, then for $\mathcal{H}^{2n}$-almost every $t \in C^n$ we have $\nu(\partial (I+t))=0$ for every $I \in \bigcup_s \beta_s$. Therefore you can move the grid slightly to have the claim satisfied.

Notice that I have not yet even read the whole paper up to that point (nor do I intend at this point), so I might be missing something.

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+1 for putting in more work in answering the question than the poster put in asking it. –  arsmath Dec 27 '10 at 16:58
Thanks. I should admit that I know so little about geometric measure theory. –  Unknown Dec 29 '10 at 10:40

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