Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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...

share|improve this question
    
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
show 4 more comments

1 Answer 1

up vote 11 down vote accepted

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

share|improve this answer
    
Isn't the Cantor set of null measure? –  Nicolò Apr 6 '10 at 15:40
1  
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
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.