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Unit 10: Measurable Functions




                                                                                                Notes


             Note  Egoroff’s theorem can be stated as: almost every where convergence implies almost
             uniform convergence.
          10.1.10 Riesz  Theorem


          Let {f } be a sequence of measurable functions which converges in measure to f. Then there is a
              n
          subsequence  f   which converges in measure to f a.e.
                      n k
          Proof: Let { } and { } be two sequences of positive real numbers such that       0 as n    and
                    n      n                                           n
                   .
              n
           n 1
          Let us now choose an increasing sequence {n } of positive integers as follows.
                                              k
          Let n be a positive integer such that

                                     m x E f (x) f(x)
                                             n
          Since f     f in measure for a given   > 0 and  > 0,  a positive integer n  such that
               n                       1        1                    1
                                m x E, f (x) f(x)        ,  n  n
                                         n 1        1   1    1

          Similarly, let n  be a positive number such that  m x E, f (x) f(x)  ,  n  n  and so
                      2                                 n 2        2    2   2   1
          on.
          In general let n  be a positive number such that
                      k
                           m x : x E, f (x) f(x)       and that n  n  .
                                      n k        k    k        k   k 1

          We shall now prove that the subsequence  f   converges to f a.e.
                                              n k
               k                                    
          Let A  =   x : x E, f (x) f(x)  i  ,k N and A  A .
                                                         k
                            n i
                  i k                                k 1
          Clearly, {A } is a decreasing sequence of measurable sets and m (A ) <  .
                   k                                            1
          Therefore, we have

          m (A) =  limm(A )
                        k
                 k
          But m (A )        0 as k  .
                 k      i
                     i k
          Hence m (A) = 0.

          Now it remains to show that {f } converges to f on E – A. Let x    E – A.
                                   n                        o





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