Boundary of open set always zero measure?

[jostpuur]

Is this true?

$$V\subset\mathbb{R}^n\;\textrm{open}\quad\implies\quad m_n(\partial V)=0$$

 

[mathman]

Assuming you mean ordinary Lebesgue measure, the answer is yes.

 

[Hurkyl]

Are you sure? Do you have a reference?

 

[quasar987]

I have a reference to the countrary! Spivak's calc on manifolds, page 56: "Problem 3-11 shows that even an open set C may not be Jordan-measurable, so that $\int_Cf$ is not necessarily defined even if C is open and f is continuous."

Jordan-measurable means that the boundary has Lebesgue measure zero. And the set of problem 3-11 is A subset of [0,1] given by a union of open intervals (a_i,b_i) such that each rational number in (0,1) is contained in some (a_i,b_i). Then bd(A) = [0,1]\A and if $\sum (b_i-a_i)<1$, bd(A) does not have measure zero.

 

[dvs]

Another family of examples can be obtained by letting V be the complement of a fat Cantor set in [0,1].

 

[jostpuur]

Thanks for pointing out the ugly fact

 

[Ja4Coltrane]

I was wondering this myself. I think I have an interesting example:
Let Q intersect [0,1] = {r_1, r_2, ...}. Then, find a countable sequence of intervals centered around each r_n, with the property that the total length of the intervals is less than 1. Then, let I be the union of these intervals.

This gives an open set whose boundary has positive lebesgue measure.

I think.