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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