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Unit 12: General Convergence Theorems




                                                                                                Notes
                                       =    (f  f)   (f  f)  as (E – A)   A =
                                             n        n
                                         E A       A

                                               1    f   f
                                         2mE        n
                                             E A  A

                                             m(E A) 2M 1
                                         2mE
                                                         A

                                             mE 2M mA  as m (E – A)   mE
                                         2mE

                                       <    2M
                                         2     4M


                                       =
                                         2  2
                                       =


          Thus                  f n  f  <
                               E   E
          But   was arbitrary


                                lim f  =  f
                                n    n
                                   E     E
          12.1.5 Fatou’s Lemma

          If {f } is a sequence of non-negative measurable functions and f (x)   f (x) almost everywhere on
             n                                              n
          a set E, then

                                     f    lim f
                                              n
                                         n
                                    E       E
          i.e.                       f   lim inf f
                                         n      n
                                    E          E
          Proof: Since integrals over sets of measure zero are zero.

               Without loss of generality, we may assume that the convergence is everywhere. Let h be
               a bounded  measurable function with h   f and h  (x) = 0 outside a set E    E of finite
               measure.
               Define a function h  by
                              n
               h (x) = Min. {h(x), f (x)}
                n             n
               then h (x)   h(x) and h (x)   f (x)
                    n            n    n
               h  is bounded by the boundedness of h and vanishes outside E  as x   E – E   h(x) = 0
                n
                  h (x) = 0 because
                  n


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