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# Is λ*(AUB) >= λ*(A)+λ*(B)? Watch

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1. 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?
2. (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.
3. (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.
4. (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.
5. (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.
6. (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
7. (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

by the interval property. But then this is

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

by the interval property. But then this is

by subadditivity, which then gives you the reverse inequality that you want.
I want to prove that
so I don't think I can't use the interval property.

I actually want to prove this property for just 2 disjoint intervals?
9. (Original post by cliveb2016)
I want to prove that
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

which was the original inequality you were seeking in order to prove the equality.
10. (Original post by cliveb2016)
I want to prove that
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 we can find a countable cover such that .

This follows from the infimum definition of the outer measure.

2. Since , we can split the cover into two disjoint covers for .

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

4. Note that

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Updated: October 11, 2016
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