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Boundary of open set always zero measure?

  1. Jan 19, 2009 #1
    Is this true?

    V\subset\mathbb{R}^n\;\textrm{open}\quad\implies\quad m_n(\partial V)=0
  2. jcsd
  3. Jan 19, 2009 #2


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    Assuming you mean ordinary Lebesgue measure, the answer is yes.
  4. Jan 19, 2009 #3


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    Are you sure? Do you have a reference?
  5. Jan 19, 2009 #4


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    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 [itex]\int_Cf[/itex] 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 [itex]\sum (b_i-a_i)<1[/itex], bd(A) does not have measure zero.
  6. Jan 19, 2009 #5


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    Another family of examples can be obtained by letting V be the complement of a fat Cantor set in [0,1].
  7. Jan 19, 2009 #6
    Thanks for pointing out the ugly fact :devil:
  8. Jul 8, 2009 #7
    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.
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