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The "support" of a measure

  1. Dec 27, 2007 #1
    [SOLVED] The "support" of a measure

    Posting a problem like this might help me get off my arse. This is #11 / chapter 2 of Rudin's Real and Complex Analysis.

    1. The problem statement, all variables and given/known data

    Let m be a regular Borel measure on a compact Hausdorff space X, assume m(X) = 1. Prove that there is a compact [tex]K \subseteq X[/tex] (the support of m) such that m(K) = 1 but m(H) < 1 for every proper compact subset H of K.

    Hint (given by Rudin): Let K be the intersection of all compact K_a such that m(K_a) = 1; show that every open set V which contains K also contains some K_a. Regularity of m is needed.

    2. Relevant equations

    A measure m is "regular" if the following two conditions hold for every measurable E:
    (1) m(E) = inf{m(V): V is an open set containing E}
    (2) m(E) = sup{m(K): K is a compact subset of E}

    3. The attempt at a solution

    ,,, work in progress

    Edit: "support of X" corrected to read "support of m"
     
    Last edited: Dec 28, 2007
  2. jcsd
  3. Dec 27, 2007 #2

    Dick

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    The rules of the forum are that you must show some attempt at solving the problem before anyone can help you. Just kidding :). But, seriously, without a start, we still don't how far you got before you ran into a problem. Off your arse yet?
     
    Last edited: Dec 27, 2007
  4. Dec 28, 2007 #3
    Kind of the opposite.. after posting I took a nap.

    The only things I can say so far is that given the hint is correct then it's not hard to show the constructed K satisfies the hypothesis...
    Because m(K) = inf{m(V): V open contains K}, and since each such V contains a K_a, we have m(V) >= m(K_a) = 1, hence m(K) = 1. Then if C is any proper compact subset of K, then K is not a subset of C, hence m(C) < 1.

    But as for proving the hint, I haven't figured anything out yet!
     
  5. Dec 28, 2007 #4
    Another point is that since m(X) = 1, whenever A,B are subsets such that m(A) = 1, m(B) = 1, then m(A [itex]\cap[/itex] B) = 1, because m(A^c [itex]\cup[/itex] B^c) = 0. It follows that every finite intersection of {K_a} is not empty (has measure 1), hence K is not empty. (But it does not follow directly that m(K) = 1, although that is what I'm trying to show...)
     
  6. Dec 28, 2007 #5
    Also sometimes it helps to look at an example... but the simple example of Lebesgue measure on X = [0,1] does not seem to help much.. If K_a is a compact subset of [0,1] such that m(K_a) = 1, then [0,1]-K_a is an open subset of measure 0: hence [0,1] - K_a is empty.

    In other words, [0,1] is the only compact subset of X with measure 1. So K = [0,1] in this case, so it is obvious. So I don't see any help in the general case here.
     
  7. Dec 28, 2007 #6
    OK I got it! This is a good example of how easy a problem looks after seeing it's solution...

    Again, let X be compact with m(X) = 1, let {K_a} be all the compacts such that m(K_a) = 1, and let K be the intersection, and let V be an open set containing K.
    Then [tex]\cap K_a \subseteq V \Rightarrow V^c \subseteq \cup K_a^c[/tex], hence [tex]\{ K_a^c \}[/tex] is an open cover of compact V^c, so has a finite subcover {K_1^c,...,K_n^c}. Thus [tex]K_1 \cap ... \cap K_n \subseteq V[/tex]. But as described above, a finite intersection of K_a's has measure 1 (and is compact), and that does it for proving the hint, and as described above, the problem.

    I didn't see the solution to this problem until I considered an equivalent formulation:

    Let {U_a} be the set of all open sets (in X above) such that m(U_a) = 0, and let U be the union of the U_a's. Then m(U) = 0. To prove this, show for any closed [tex]C \subseteq U[/tex], that you must have [tex]C \subseteq U_a[/tex] for some a. (That is where I realized you could use compactness...)

    Edit: one more remark, is that the "regularity is needed" particularly in applying the hint to proving the statement of the problem. And I discussed that in the first reply to Dick.
     
    Last edited: Dec 28, 2007
  8. Dec 28, 2007 #7

    Dick

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    Guess you didn't need TOO much help then.
     
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