the outer measure of an interval is its length

[[concept]]

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Theorem

If I is an interval, then

m∗(I)=ℓ(I)
Proof

Suppose I=[a,b]. Then I⊂(a−ϵ,b+ϵ) for all ϵ>0. But then we have

m∗(I)≤ℓ(a−ϵ,b+ϵ)=b−a+2ϵ

Thus we have m∗(I)≤b−a.

Now we show b−a≤m∗(I). Let {In}n be a collection of open intervals such that [a,b]⊂⋃nIn.

Now, since [a,b] is compact (by the Heine-Borel theorem), there exists a finite subcover {Jk}k=0m⊂{In}n such that [a,b]⊂⋃k=0mJk

Since a∈⋃k=0mJk, there exists k1 such that a∈Jk1. Rearranging intervals, I let k1=1 and a∈J1=(a1,b1). If b1<b, then b1∈[a,b]. So ther eis some k2 such that b1∈Jk2. By rearranging again, I assume k2=2. So b1∈J2=(a2,b2).

Continuing in the same manner, we conclude that there exists a K,1≤K≤m such that b<bK. And for all k∈{1,2,…,K−1} we have bk≤b and ak+1≤bk<bk+1.

But then we have

∑nℓ(In)≥∑k=1mℓ(Jk)≥∑k=1Kℓ(Jk)=(bK−aK)+(bK−1−aK−1)+⋯+(b1−a1)=bK+(bK−1−aK)+(bK−2−aK−1)+⋯+(b1−a2)−a1≥bK−a1≥b−a=ℓ(I)

Thus we have m∗(I)≥ℓ(I)=b−a

Thus we conclude m∗(I)=ℓ(I)

◼

References

References

See Also

Mentions

Mentions

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