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I want to construct a nowhere dense set of positive measure by using the Smith-Volterra-Cantor method but subtracting closed sets intead of open sets.

Suppose we start with the open interval (0,1). Let $\alpha=1/2$. Now remove $U_1=[1/2-\alpha/4,1/2+\alpha/4]$. The measure $\mu(U_1)=\alpha/2$. For each of the two remaining open sets, subtract a closed from the center of width $\alpha/8$, thus a total measure of $\alpha/4$. Repeating this process results in removing a total measure of $\alpha/2+\alpha/4+\dots=\alpha$.

Does this give me a set of positive measure just as with subtracting closed intervals? I suspect not, since I have never heard of this, but what's the exact function of removing a closed set instead of an empty set?

Or, if I can do this, can I now consider this to be an open set that I can choose an admissible partition of unity for?

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You will end up with a set of measure $1-\alpha$ because the Lebesgue measure is $\sigma$-additive (countably additive). – Andrey Rekalo Dec 14 '10 at 7:48
Alternatively, the measure of the set will not depend on whether you subtract closed or open intervals since the difference is a countable sequence of endpoints, i.e. a measure zero set. – Andrey Rekalo Dec 14 '10 at 7:50

"Does this give me a set of positive measure just as with subtracting closed intervals?"

Yes, as follows from the basic property of continuity of measure: if $E_1 \supset E_2 \supset \ldots \supset E_n \supset \ldots$ is a nested sequence of subsets with $\mu(E_1) < \infty$, then if $E = \bigcap_{n=1}^{\infty} E_n$, we have $\mu(E) = \lim_{n \rightarrow \infty} \mu(E_n)$.

"[C]an I now consider this to be an open set...?" No. Here the variation you've made on the usual construction affects the answer: any intersection of closed sets is necessarily closed, whereas (even) a countably infinite intersection of open sets need not be open. I think it would be good for you to figure out for yourself exactly why the intersection $E$ is not open here. Think for instance about the fact that if you do the usual construction by removing open intervals you get a set which is nowhere dense and how your modified set $E$ compares to that closed, nowhere dense set.

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