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Prove a set consisting of a single point is measurable and has zero area

  1. Feb 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that a set consisting of a single point is measurable and has zero area.


    2. Relevant equations
    Area Axioms: [tex] M [/tex] is a class of measurable sets.
    (a) Every rectangle [tex]R \in M [/tex]. If the edges of R have lengths h and k, then the area [tex] a(R) = hk [/tex].

    Additionally, a rectangle can be represented as [tex] R=((x,y)|0≤x≤h,0≤y≤k) [/tex]

    (b) If a set [tex] S \in M [/tex] and if [tex] S [/tex] is congruent to [tex] T [/tex], then [tex] T \in M [/tex] and the areas [tex] a(S) = a(T) [/tex]


    3. The attempt at a solution

    Let [tex] (x_{0},y_{0}) \in ((x_{0},y_{0})) [/tex] be an arbitrary point on the plane. Since [tex] ((x_{0},y_{0})) = ((x,y)|x_{0}≤x≤x_{0},y_{0}≤y≤y_{0}) [/tex] (a) which is a rectangle [tex] R [/tex] and so [tex] ((x_{0},y_{0})) \in M [/tex] (measurable) by (b). Since the side lenghts of [tex] ((x_{0},y_{0})) [/tex] are zero, then [tex] a(((x_{0},y_{0}))) = 0 [/tex] .

    Is this right? I apologize for my tex.
     
  2. jcsd
  3. Feb 3, 2012 #2

    HallsofIvy

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    Staff Emeritus
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    Yes, that is correct. You might want to specify that the side lengths of the "rectangle" are [itex]x_0- x_0[/itex] and [itex]y_0- y_0[/itex] which are, as you say, 0.
     
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