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Outer measures are monotonic

  1. Sep 14, 2009 #1
    I have a sneaky suspicion that this is a fairly simple task, but I just can't seem to break through this particular proof.

    The problem statement, all variables and given/known data

    The (Lebesgue) outer measure of any set [itex]A\subseteq\mathbb{R}[/itex] is:

    [itex]m^*(A) = inf Z_A[/itex]

    where

    [itex]Z_A = \bigg\{\sum_{n=1}^\infty l(I_n)\;:\;I_n\;\text{are intervals},\;A\subseteq\bigcup_{n=1}^\infty I_n\bigg\}[/itex]

    My problem is to prove that [itex]m^*[/itex] is monotone, i.e

    If [itex]A\subset B[/itex] then [itex]m^*(A) \leq m^*(B)[/itex]

    The hints are to show that [itex]Z_B \subset Z_A[/itex] and then use the definition of the infimum to show that the larger set can't have an infimum greater than the smaller set.

    The attempt at a solution

    If [itex]I_n[/itex] covers B, then it will also cover A.
    [itex]A\subset B\subset \bigcup_n I_n[/itex]. Hence [itex]Z_B \subset Z_A[/itex].

    And now I am confused! :) I have two questions.

    1) I was under the impression that [itex]l(I_n)[/itex] was the length of a certain intervall. But that makes [itex]Z_A[/itex] and [itex]Z_B[/itex] numbers. How can one number be a subset of another number?

    2) My other question is regarding the statement about infimums. What does the size of a set have to do with what the infimum can be? For instance.
    [itex]A = \{2, 3\}[/itex] and [itex]B = \{1,2,3,4\}[/itex]. Here [itex]A\subset B[/itex] but [itex]\text{inf}\; B < \text{inf}\; A[/itex]

    Hopefully someone can shed some light on this!
     
  2. jcsd
  3. Sep 14, 2009 #2

    AKG

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    1) No, Z_A and Z_B are sets of numbers. For each open interval covering (I_n) of A, you get the number [itex]\sum l(I_n)[/itex]. As you range over all such coverings, you get a whole bunch of different numbers, and Z_A is the set of those numbers.

    2) The size of a set doesn't affect its inf, but if A is a subset of B, then B's inf is less than or equal to A's. Because inf(B) is the greatest lower bound of B, and if A is a subset of B, then every lower bound of B is a lower bound of A, and hence the greatest lower bound of B is a lower bound for A, and hence less than or equal to the greatest lower bound for A.
     
  4. Sep 14, 2009 #3
    Thank you!

    I will charge the proof with newfound courage and motivation. :)
     
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