Is a set whose boundary is measure zero a Borel set?

Does any given Borel set has a measure zero boundary?

I want to give my ideas first: If $E \subseteq R^n $ is some set whose boundary has Lebesgue measure zero, then we can safely do the Riemann integral $\int _E1dx$, which means we can calculate the "volumn" of $E$. Hence I think $E$ is a Borel set.

But how about the inverse arguement?


To your first question: Take the example of a non borel set from this page: http://en.m.wikipedia.org/wiki/Borel_set

Then use sterograpic representation to wrap it around the open unit disc. That set (this weird boundary union disc interior) has a measure zero boundary (closure minus interior) but is not borel, otherwise the original set would be.

Your second question:

http://en.wikipedia.org/wiki/Smith–Volterra–Cantor_set is a borel set with positive measure boundary.

So the answer to both your questions is "no".

My first example satisfies the theorem, but is not Borel. So what you would like to prove is false.

  • $\begingroup$ the cantor set is measure zero. and the cantor set is it's own boundary. $\endgroup$ – Stupid_Guy Oct 15 '14 at 18:53
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    $\begingroup$ For a non-constructive answer to the first question, note that there are $2^c$ many subsets of the boundary of the unit disk in the plane, all of which have (planar) measure zero, and only $c$ many of these can be Borel sets, so most subsets of the boundary of the unit disk have the property that its union with the open unit disk is not a Borel set (non-Borel union Borel is non-Borel), and all the boundaries of these non-Borel sets have measure zero. $\endgroup$ – Dave L. Renfro Oct 15 '14 at 19:59

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