the outer measure of an interval is its length

[[concept]]

[!themes] Topics

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Theorem

If I is an interval, then

m(I)=(I)
Proof

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

m(I)(aϵ,b+ϵ)=ba+2ϵ

Thus we have m(I)ba.

Now we show bam(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 ak=0mJk, there exists k1 such that aJk1. Rearranging intervals, I let k1=1 and aJ1=(a1,b1). If b1<b, then b1[a,b]. So ther eis some k2 such that b1Jk2. By rearranging again, I assume k2=2. So b1J2=(a2,b2).

Continuing in the same manner, we conclude that there exists a K,1Km such that b<bK. And for all k{1,2,,K1} we have bkb and ak+1bk<bk+1.

But then we have

n(In)k=1m(Jk)k=1K(Jk)=(bKaK)+(bK1aK1)++(b1a1)=bK+(bK1aK)+(bK2aK1)++(b1a2)a1bKa1ba=(I)

Thus we have m(I)(I)=ba

Thus we conclude m(I)=(I)

References

References

See Also

Mentions

Mentions

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