1. Not finding help here? Sign up for a free 30min 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!

Proving Basic Measure Properties

  1. Aug 21, 2006 #1
    I want to prove that if E is a subset of F and both E and F are measurable, then m(E) </= m(F). (where </= is less than or equal to).

    Now I figured that I'd use one of the axioms for a measure to prove this, namely

    If A_i are measureable and disjoint A_n n A_m = {} for n not equal to m, then

    m(U A_i) = sum m(A_i)

    (for some reason tex is not working for me anymore.)

    But that requires me to form the following collection of sets:

    E*_1 = E
    E*_2 = F - E
    E*_3 = E*_4 = ... = emptyset

    to ensure that each E*_i are measureable and disjoint. Unfortunately I have no justification for doing so, only that I've seen it done before. In fact, Im not even sure of how to continue. Any help would be much appreciated.
     
  2. jcsd
  3. Aug 22, 2006 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, you said you were choosing your sets so you could invoke a theorem that said

    m(U A_i) = sum m(A_i)

    So, have you tried computing those?
     
  4. Aug 22, 2006 #3
    Well, I can see that

    m(U E*_i) = sum m(E*_i) = {}

    for i > 3. Is this the kind of thing you mean for me to do?
     
  5. Aug 22, 2006 #4
    Wait!! What about this:

    Let F = E U (F\E). Then m(F) = m(E) + m(F \ E) which implies that

    m(E) = m(F) - m(F \ E)

    => m(E) </= m(F)

    and m(E) = m(F) when m(F\E) = 0 <=> F = E.
     
    Last edited: Aug 22, 2006
  6. Aug 22, 2006 #5

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    There are non-empty sets of measure zero. Very important ones, like the rational numbers insided the reals.
     
  7. Aug 22, 2006 #6
    I have another interesting property of measure. If I have two (not necessarily disjoint) sets, say A and B, then what would m(A)
    + m(B) be? (where m is the the measure).

    Would it be:

    (1)...m(A) + m(B) = m(A u B)?

    But what if A and B overlap? Then the above formula may be too small! Perhaps then, in general,

    (2)...m(A) + m(B) = m(A u B) + m(A n B)

    But then this could be too large! Since then we would be counting points in the intersection twice! Perhaps it could be then

    (3)...m(A) + m(B) = m((A u B)\(A n B)) + m(A n B) = m(A u (B \ A)) + m(A n B)

    Unfortunately, I am told that the correct one, in general is (2), not (3) :(

    I can see that (1) would be correct if A and B are disjoint. But I dont see why (2) is the correct representation of the sum of measures. So I tried proving it. But I can only see it working if m(A n B) = 0, i.e. somehow the measure of two overlapping sets is zero.
     
    Last edited: Aug 22, 2006
  8. Aug 22, 2006 #7

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is why the measure of the intersection is subtracted, i.e., (2) rearranged is

    m(A u B) = m(A) + m(B) - m(A n B).

    (2) and (3) say the same thing, since

    A u (B\A) = A u (B n A^c) = (A u B) n (A u A^c) = A u B.

    I find that it helps to look at Venn diargrams.
     
  9. Aug 22, 2006 #8
    Well, my definition of measure says that the measure of the union of sets equals the sum of the measures of each individual set:

    m(A u B) = m(A) u m(B)

    nowhere does it require that you must subtract the intersection. Could you explain why this is?
     
  10. Aug 22, 2006 #9

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Let's back up a bit. (3) implies that

    m((A u B)\(A n B)) = m(A u (B \ A)),

    but this doesn't look right to me.
     
  11. Aug 22, 2006 #10

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Take A=B to see where your definition is wrong (assuming you actually meant a + on the RHS and not u which is just an easy to make oversight). (your definition works for disjoint sets. and it is easy to write the union of two sets as the union of disjoint sets as above.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Proving Basic Measure Properties
Loading...