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Let A be a borelian set with postivie measure. I was asking myself if it is possible to find an open set $B\subseteq A$ such that $B$ is an open set minus a set of null measure...

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do you mean:"... such that $B$ is $A$ minus a set of null measure? – Thomas Kragh Apr 6 '10 at 15:39
No, I mean that there exists an open set B' and a set N of null measure such that B=B'\N – Nicolò Apr 6 '10 at 15:43
Ok.. So $B$ is only open in the subspace topology of $A$, but not in the entire measure space? – Thomas Kragh Apr 6 '10 at 15:47
I mean that $B$ is equal almost everywhere to an open set... It can also be not open in $A$ topology, (if you add a point to $B$ the condition remains the same, and B can fail to be an open set in A topology) – Nicolò Apr 6 '10 at 16:07
Ok. Then remove the condition that $B$ IS open. – Thomas Kragh Apr 6 '10 at 16:15
up vote 11 down vote accepted

The Cantor set of positive measure is nowhere dense set. So it is an example.

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Isn't the Cantor set of null measure? – Nicolò Apr 6 '10 at 15:40
I mean the well-known construction (similar to the construction of Cantor set), when one delete middle intervals smaller then $1/3^n$. The resulting set has positive measure and homeomorphic to the standard Cantor set of measure zero. – Petya Apr 6 '10 at 15:43
ok, thanks! In wikipedia i've found such sets under the name of "cantor-smith-volterra sets" – Nicolò Apr 6 '10 at 15:45
I think it's also called a fat Cantor set? – Harald Hanche-Olsen Apr 6 '10 at 20:11
I'm not sure this answers the question, as I see it. Let $A$ be this fat Cantor set. We know it is nowhere dense, so contains no open set. But we actually want to know that $A \cup E$ contains no open set, where $E$ is any null set. Does this follow? Of course $A \cup E$ may no longer be nowhere dense. – Nate Eldredge Apr 6 '10 at 20:44

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