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Measure of unbounded set

  1. Oct 13, 2013 #1
    Suppose A is not a bounded set and m(A∩B)≤(3/4)m(B) for every B. what is m(A)??

    here, m is Lebesgue Outer Measure

    My attemption is :

    Let An=A∩[-n,n], then m(A)=lim m(An)= lim m(An∩[-n,n]) ≤ lim (3/4)m([-n,n]) = infinite.

    is my solution right? I am confusing m(A) < infinite , it doest make sence for me. Could someone help me???
  2. jcsd
  3. Oct 14, 2013 #2


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    What you wrote is correct as far as it goes, but ##m(A) \leq \infty## doesn't tell you anything new: this is of course true of the outer measure of any set.

    Certainly ##m(A) = 0## is possible: consider ##A = \mathbb{Q}##, for example.

    Is ##m(A) > 0## possible? Hint: consider ##A = B##.
  4. Oct 14, 2013 #3
    There are three possible cases worth thinking about.
    - [itex]m(A)=0[/itex], which jbunniii showed is possible.
    - [itex]0<m(A)<\infty[/itex], for which jbunniii provided a very useful hint.
    - [itex]m(A)=\infty[/itex]... Is this possible? Consider the sets [itex]A_n=A\cap[-n,n][/itex] you defined. If we have to have [itex]0<m(A_n)<\infty[/itex] for some [itex]n\in \mathbb N[/itex] (Is this true?), then maybe the same trick as above can be reused.

    It's worth noting that the answer to this question depends on a special property of the Lebesgue measure on [itex]\mathbb R[/itex], which fails for some other infinite measures. Namely, we're using the property that the whole space is a countable union of finite-measure sets.
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