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Measure Theory and Functional Analysis




                    Notes          At the p  stage, we get a measurable set
                                         th
                                                      n
                                   A    E  with m (A ) <    and a positive integer N  such that
                                    p   p–1       p    p                    p
                                                      2
                                                1
                                   |f  (x) – f (x)| <     n  N  and x  E  where
                                    n                   p        p
                                                p
                                                             E = E  – 1 – A .
                                                              p   p     p

                                   Let                       A =    A p ,
                                                                  p 1


                                   then                   m (A)     m(A )
                                                                        p
                                                                  p 1
                                                                  n
                                   But                   m (A ) <
                                                                 2p
                                                             p

                                                                     n
                                                          m (A) <    p
                                                                    2
                                                                  p 1
                                                                        1    1
                                         1                         a    2    2
                                   But      is a G.P. series so  S =           = 1
                                         2 p                     1 r     1   1
                                      p 1                              1
                                                                         2   2
                                                          m (A) <
                                                                    
                                   Also,                  E – A = E –   A  p
                                                                     p
                                                                  
                                                               =     (E A )
                                                                         p
                                                                   p
                                                                 
                                                               =    (E  p 1  A )
                                                                           p
                                                                  p
                                                                 
                                                               =    E p
                                                                  p
                                   Let x   E – A. Then x   E    p and so
                                                       p
                                                                          1
                                                             |f  (x) – f (x)| <   ,   n   N .
                                                               n                   p
                                                                          p
                                                        1
                                   Let us choose p such that    <  , we get
                                                        p

                                                       |f  (x) – f (x)| <   x   E – A and n   N  = N
                                                         n                            p



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