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How to show a set has measure zero?

  1. Nov 6, 2009 #1
    I'm not completely familiar with measures yet, but am trying to be. I'm trying to show a that a few sets in Rn have measure zero and am having difficulty showing this using countable covers with a total volume less than [tex]\epsilon[/tex]. Is there an easier way to show that a set has measure zero? If it's needed, I could give the problems that I'm working with. Any help would be appreciated.
  2. jcsd
  3. Nov 6, 2009 #2
    The method you mention is really the only general one. If you give the specific sets you're working with I may be able to give a simpler method in that specific situation.
  4. Nov 6, 2009 #3
    Show that Rn-1 x {0} has measure zero in Rn.

    This is one of the ones I'm working on.
  5. Nov 6, 2009 #4
    This one is easy using epsilon neighborhoods.
  6. Nov 6, 2009 #5
    I'll expand on wofsy's comment a bit. You do use the open cover formulation on this one. Let [tex]\{ q_k \}_{ k = 1 }^\infty[/tex] be an enumeration of [tex]\mathbb{Q} \times \{ 0 \}[/tex]. Show that if [tex]\varepsilon > 0[/tex], [tex]\left \{ ( q_k - \frac{ 1 }{ 2 }, q_k + \frac{ 1 }{ 2 } ) \times ( -\varepsilon\ 2^{ - ( k + 1 ) }, \varepsilon\ 2^{ - ( k + 1 ) } ) \right \}_{ k = 1 }^\infty [/tex] is an open cover of [tex]\mathbb{ R } \times \{ 0 \}[/tex] with total measure less than or equal to [tex]\varepsilon[/tex].

    EDIT: This proof works for showing that [tex]\mathbb{ R } \times \{ 0 \} \subset \mathbb{ R }^2[/tex] has measure zero. The [tex]n > 2[/tex] case is a straightforward generalization.
  7. Nov 7, 2009 #6
    The way I did it earlier used
    [tex]\bigcup[/tex]n=1[tex]\infty[/tex] [-n,n]n-1x{0} which is a countable cover of Rn-1x{0}. Then using the fact that a countable union of sets of measure zero also has measure zero, you can show that each set has measure zero and the rest follows. After reading a bit on it today, I feel like I have a much better understanding of it now.
  8. Nov 7, 2009 #7
    That works too.
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