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Is λ*(AUB) >= λ*(A)+λ*(B)?

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    Let λ* be the Lebesgue outer measure. Suppose A and B are disjoint intervals. And they are subsets of R.

    Is it true that λ*(AUB) >= λ*(A)+λ*(B)?

    I want to show λ*(AUB) = λ*(A)+λ*(B) but so far have only managed to show λ*(AUB) <= λ*(A)+λ*(B) using countable subadditivity.

    Any ideas how to get the inequality I need to finish this off?

    The only tools I have are the properties of λ* (countable subadditivity, monotonicity)

    Any help?
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    (Original post by cliveb2016)
    Let λ* be the Lebesgue outer measure. Suppose A and B are disjoint intervals. And they are subsets of R.

    λ*(AUB) = λ*(A)+λ*(B)
    Isn't this false as stated, for general subsets of R? You need both A and B to be Lebesgue measurable to get to work, I think, unless I'm misreading you. For the case when they are Lebesgue measurable, the proof that I have seen proceeds from definition of the Caratheodory criterion.
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    (Original post by atsruser)
    Isn't this false as stated, for general subsets of R? You need both A and B to be Lebesgue measurable to get to work, I think, unless I'm misreading you. For the case when they are Lebesgue measurable, the proof that I have seen proceeds from definition of the Caratheodory criterion.
    Well A and B are both intervals rather than general subsets so they are Lebesgue measurable.
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    (Original post by cliveb2016)
    Well A and B are both intervals rather than general subsets so they are Lebesgue measurable.
    Fair enough - I missed this. Then maybe my comment re: the Caratheodory criterion will help - I'm afraid I can't give a better hint without looking up the proof myself, I'm afraid.
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    (Original post by cliveb2016)
    Let λ* be the Lebesgue outer measure. Suppose A and B are disjoint intervals. And they are subsets of R.

    Is it true that λ*(AUB) >= λ*(A)+λ*(B)?

    I want to show λ*(AUB) = λ*(A)+λ*(B) but so far have only managed to show λ*(AUB) <= λ*(A)+λ*(B) using countable subadditivity.

    Any ideas how to get the inequality I need to finish this off?

    The only tools I have are the properties of λ* (countable subadditivity, monotonicity)

    Any help?
    The quickest way i can think of for this is to go through the standard proof that the outer lebesgue measure of an interval is its length, and then think about the interval A U B U C where C is the bit between the intervals A and B.
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    (Original post by Gregorius)
    The quickest way i can think of for this is to go through the standard proof that the outer lebesgue measure of an interval is its length, and then think about the interval A U B U C where C is the bit between the intervals A and B.
    Suppose the Lebesgue outer measure of an interval (a,b) (can be open closed or half and half) is |b-a|.

    Suppose WLOG that the intervals are positive and can be split as A=[a,b], B=[c,d] and C=(b,c) where a<b<c<d.

    Suppose that the intervals are also finite (since if not inequality holds trivially)

    Then λ*(AUBUC)-λ*(C) = λ*(AUB) = (d-a)-(c-b) >= (b-a)+(d-c)=λ*(A)+λ*(B)

    Does that make any sense or have I missed your point completely? I'm not sure about λ*(AUBUC)-λ*(C) = λ*(AUB) .

    Thanks
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    (Original post by cliveb2016)
    Suppose the Lebesgue outer measure of an interval (a,b) (can be open closed or half and half) is |b-a|.

    Suppose WLOG that the intervals are positive and can be split as A=[a,b], B=[c,d] and C=(b,c) where a<b<c<d.

    Suppose that the intervals are also finite (since if not inequality holds trivially)

    Then λ*(AUBUC)-λ*(C) = λ*(AUB) = (d-a)-(c-b) >= (b-a)+(d-c)=λ*(A)+λ*(B)

    Does that make any sense or have I missed your point completely? I'm not sure about λ*(AUBUC)-λ*(C) = λ*(AUB) .

    Thanks
    The way I went about it was to note that

    \lambda^{*}(A \bigcup B \bigcup C) = \lambda^{*}(A) + \lambda^{*}(B) + \lambda^{*}(C)

    by the interval property. But then this is

    \lambda^{*}((A \bigcup B) \bigcup C) \le \lambda^{*}(A \bigcup B) + \lambda^{*}(C)

    by subadditivity, which then gives you the reverse inequality that you want.
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    (Original post by Gregorius)
    The way I went about it was to note that

    \lambda^{*}(A \bigcup B \bigcup C) = \lambda^{*}(A) + \lambda^{*}(B) + \lambda^{*}(C)

    by the interval property. But then this is

    \lambda^{*}((A \bigcup B) \bigcup C) \le \lambda^{*}(A \bigcup B) + \lambda^{*}(C)

    by subadditivity, which then gives you the reverse inequality that you want.
    I want to prove that
    \lambda^{*}(A \bigcup B) = \lambda^{*}(A) + \lambda^{*}(B) so I don't think I can't use the interval property.

    I actually want to prove this property for just 2 disjoint intervals?
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    (Original post by cliveb2016)
    I want to prove that
    \lambda^{*}(A \bigcup B) = \lambda^{*}(A) + \lambda^{*}(B) so I don't think I can't use the interval property.

    I actually want to prove this property for just 2 disjoint intervals?
    My previous post gives you

    \lambda^{*}(A \bigcup B) \ge \lambda^{*}(A) + \lambda^{*}(B)

    which was the original inequality you were seeking in order to prove the equality.
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    (Original post by cliveb2016)
    I want to prove that
    \lambda^{*}(A \bigcup B) = \lambda^{*}(A) + \lambda^{*}(B) so I don't think I can't use the interval property.

    I actually want to prove this property for just 2 disjoint intervals?
    There is a proof that goes something like this:

    1. Given \epsilon &gt;0 we can find a countable cover I_n such that  \sum_{n=0}^\infty l(I_n) \le \lambda^*(A \cup B) + \epsilon .

    This follows from the infimum definition of the outer measure.

    2. Since A \cap B=\emptyset, we can split the I_n cover into two disjoint covers for A,B.

    3. We can apply countable subadditivity of outer measure to complete the inequality with \epsilon in place.

    4. Note that  \forall \epsilon &gt; 0, x  &lt; \epsilon \Rightarrow x = 0
 
 
 
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