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Measurability and Distance

  1. Apr 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Let X, Y be subsets of R and defined d(X, Y) = inf {|x - y| : x in X and y in Y}.

    (i) Prove that if d(X, Y) > 0, then m*(X cup Y) = m*(X) + m*(Y)

    (ii) Prove, without using any facts about measurability, that if X, Y are disjoint and compact, then m*(X cup Y) = m*(X) + m*(Y).


    2. Relevant equations
    m* is Lebesgue outer measure.


    3. The attempt at a solution
    Concerning (i), d(X, Y) > 0 implies that X and Y are disjoint right? But being disjoint is not enough to conclude that m*(X cup Y) = m*(X) + m*(Y).

    Concerning (ii), since X and Y are disjoint, d(X, Y) > 0 and so part (i) applies and we're done. Since we're not supposed to use measurability, I imagine that (i) uses measurability so this isn't allowed. I guess I'm forced to work with the definition of outer measure. By monotonicity of m*, m*(X cup Y) ≤ m*(X) + m*(Y) so I only need to prove the reverse inequality. Let {U_n} and {V_n} be coverings of X and Y by open intervals. Then {W_n} = {U_n} cup {V_n} is an covering of X cup Y. By compactness, we can shrink each of {U_n}, {V_n} and {W_n} so that they are finite. Now {W_n} subseteq {U_n} cup {V_n}, so that sum L(W_n) ≤ sum L(U_n) + sum L(V_n), where L() returns the length of the interval. This is all I can think of. Any tips?
     
  2. jcsd
  3. Apr 1, 2009 #2
    for (i) let d = d(X,Y), and [tex]O = \bigcup_{x\in X}(x - d,x+d)[/tex] so [tex]O\bigcap Y = empty set, [/tex] and [tex] X \subseteq O[/tex] then since O is measurable we get:

    [tex]m(X\cup Y) = m((X\cup Y)\cap O) + m((X\cup Y)\cap O^{c}) = m(X) + m(Y). [/tex]
     
    Last edited: Apr 1, 2009
  4. Apr 2, 2009 #3
    I thought of that exact same argument. Thanks. Any ideas on (ii)?
     
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