Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook