proof that the outer (Lebesgue)
measure of an interval is its length∗
Simone†
2013-03-21 18:14:39
We begin with the case in which we have a closed bounded interval, say [a, b].
Since the open interval (a − ε, b + ε) contains [a, b] for each positive number ε,
we have m∗ [a, b] ≤ b − a + 2ε. But since this is true for each positive ε, we must
have m∗ [a, b] ≤ b − a. Thus we only have to show that m∗ [a, b] ≥ b − a; for this
it suffices to show that if {In } is a countable open cover by intervals of [a, b],
then
X
l(In ) ≥ b − a.
By the Heine-Borel theorem, any collection of open intervals covering [a, b] contains a finite subcollection that also cover [a, b] and since the sum of the lengths
of the finite subcollection is no greater than the sum of the original one, it suffices to prove the
S inequality for finite collections {In } that cover [a, b]. Since a
is contained in In , there must be one of the In ’s that contains a. Let this be
the interval (a1 , b1 ). We then have a1 < a < b1 . If b1 ≤ b, then b1 ∈ [a, b], and
since b1 6∈ (a1 , b1 ), there must be an interval (a2 , b2 ) in the collection {In } such
that b1 ∈ (a2 , b2 ), that is a2 < b1 < b2 . Continuing in this fashion, we obtain a
sequence (a1 , b1 ), . . . , (ak , bk ) from the collection {In } such that ai < bi−1 < bi .
Since {In } is a finite collection our process must terminate with some interval
(ak , bk ). But it terminates only if b ∈ (ak , bk ), that is if ak < b < bk . Thus
X
X
l(In ) ≥
l(ai , bi )
= (bk − ak ) + (bk−1 − ak−1 ) + · · · + (b1 − a1 )
= bk − (ak − bk−1 ) − (ak−1 − bk−2 ) − · · · − (a2 − b1 ) − a1
> bk − a1 ,
since
P ai < bi−1 . But bk > b and a1 <∗ a and so we have bk − a1 > b − a, whence
l(In ) > b − a. This shows that m [a, b] = b − a.
∗ hProofThatTheOuterLebesgueMeasureOfAnIntervalIsItsLengthi
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If I is any finite interval, then given ε > 0, there is a closed intervalJ ⊂ I
such that l(J) > l(I) − ε. Hence
l(I) − ε < l(J) = m∗ J ≤ m∗ I ≤ m∗ I = l(I) = l(I),
where by I we mean the topological closure of I. Thus for each ε > 0, we have
l(I) − ε < m∗ I ≤ l(I), and so m∗ I = l(I).
If now I is an unbounded interval, then given any real number ∆, there is
a closed interval J ⊂ I with l(J) = ∆. Hence m∗ I ≥ m∗ J = l(J) = ∆. Since
m∗ I ≥ ∆ for each ∆, it follows m∗ I = ∞ = l(I).
References
Royden, H. L. Real analysis. Third edition. Macmillan Publishing Company,
New York, 1988.
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