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Is λ*(AUB) >= λ*(A)+λ*(B)?

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?
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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cliveb2016
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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.
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.
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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cliveb2016
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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
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

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

by the interval property. But then this is

λ∗((A⋃B)⋃C)≤λ∗(A⋃B)+λ∗(C)\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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cliveb2016
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Original post by Gregorius
The way I went about it was to note that

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

by the interval property. But then this is

λ∗((A⋃B)⋃C)≤λ∗(A⋃B)+λ∗(C)\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
λ∗(A⋃B)=λ∗(A)+λ∗(B)\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?
Original post by cliveb2016
I want to prove that
λ∗(A⋃B)=λ∗(A)+λ∗(B)\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

λ∗(A⋃B)≥λ∗(A)+λ∗(B)\lambda^{*}(A \bigcup B) \ge \lambda^{*}(A) + \lambda^{*}(B)

which was the original inequality you were seeking in order to prove the equality.
Original post by cliveb2016
I want to prove that
λ∗(A⋃B)=λ∗(A)+λ∗(B)\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 ϵ>0\epsilon >0 we can find a countable cover InI_n such that ∑n=0∞l(In)≤λ∗(A∪B)+ϵ \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∩B=∅A \cap B=\emptyset, we can split the InI_n cover into two disjoint covers for A,BA,B.

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

4. Note that ∀ϵ>0,x<ϵ⇒x=0 \forall \epsilon > 0, x < \epsilon \Rightarrow x = 0

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