Proposition 1:- The outermeasure of an interval is it's length part 1

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Proposition 1 The outer measure of an interval is its length.
Proof We begin with the case of a closed, bounded interval [a, b]. Let c 0. Since the open
interval (a - E, b + E) contains [a, b] we have m* ([a, b]) ((a - E, b+ E)) = b - a+ 2E. This
holds for any c 0. Therefore m * ([a, b]) b - a. It remains to show that m* ([a, b]) b - a.
But this is equivalent to showing that if {Ik}k 1 is any countable collection of open, bounded
intervals covering [a, b], then
00
E f(Ik) b - a. (1)
k=1
By the Heine-Borel Theorem,4 any collection of open intervals covering [a, b] has a finite
subcollection that also covers [a, b]. Choose a natural number n for which {Ik}k=1 covers
[a, b]. We will show that
n
I f(Ik)?b-a, (2)
k=1
and therefore (1) holds. Since a belongs to uk=1 Ik, there must be one of the Ik's that contains
a. Select such an interval and denote it by (al, bl ). We have a1 a b1. If b1 b, the
inequality (2) is established since
n
Ef(Ik)bl-alb-a.
k=1
Otherwise, b1 E [a, b), and since b10 (al, b1), there is an interval in the collection {Ik)k=1,
which we label (a2, b2), distinct from (al, bi), for which b1 E (a2, b2); that is, a2 b1 b2.
If b2 b, the inequality (2) is established since
n
I f(Ik) (b1-al)+(b2-a2)=b2-(a2-bl)-alb2-alb-a.
k=1
We continue this selection process until it terminates, as it must since there are only n
intervals in the collection {Ik}k=1. Thus we obtain a subcollection {(ak, bk )}k
1 of {Ik}k=1 for
which
while
at a,
ak+1 bk for l k N -1,
and, since the selection process terminated,
bN b.
Thus
n N
I f(Ik) 2f((ai,bi))
k=1 k=1
_ (bN -aN)+(bN-1 -al)
=bN-(aN-bN-1)-...-(a2-bl)-al
bN - al b - a.