Page 10 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS
P. 10
Unit 1: Differentiation and Integration: Differentiation of Monotone Functions
Notes
Let I be any interval in M and let be the supremum (least upper bound of the lengths of the
1 1
intervals in M disjoint from I (i.e., which do not have any point common with I ).
1 1
Obviously < as m (0) < .
1
1
1
Now we choose an interval I from M, disjoint from I , such that (I ) . Let be the
2 1 2 1 2
2
supremums of lengths of all those intervals of M which do not have any point common with I
2
or I obviously < .
2 2
In general, suppose we have already chosen r intervals I , I , … I (mutually disjoint). Let be
1 2 r r
the supremums of the length of those intervals of M which do not have any point in common
with I (i.e., which do not meet any of the intervals I , I , … I . Then m (0) < .
r
i
1
2
r
i 1
r r
Now if E is contained in I , then Lemma established. Suppose I i E . Then we can find
i
i 1 i 1
1
interval I s.t. (I ) which is disjoint from I , I , … I .
r+1 r 1 r 1 2 r
2
Thus at some finite iteration either the Lemma will be established or we shall get an infinite
1
sequence <I > of disjoint intervals of M s.t. (I ) and < , n = 1, 2, 3 ….
r r 1 r r
2
Note that < > is a monotonically decreasing sequence of non-negative real numbers.
r
Obviously, we have that I r 0 ) m(0) hence for any arbitrary > 0, we can
(
r
i 1 r 1
find an integer N s.t.
1
(I )
r 5 .
r N 1
N
Let a set F = I r .
r 1
N
The lemma will be established if we show that m* (F) < . For, let x F, then x I r x is an
r 1
N
an interval I in M s.t. x I and (I)
element of E not belonging to the closed set I r is so
r 1
N
small that I does not meet the I , i.e.
r
r 1
I I = , r = 1, 2, … N.
r
Therefore we shall have (I) 2 (I ) as by the method of construction we take
N n 1
1
I .
n 1 N
2
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