1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    Guess you didn't need TOO much help then.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook