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P. 321

Unit 26: Lebesgue Measure




          Theorem 8: Littlewood’s 3rd Principle/Egoroff’s Theorem                               Notes
          Let E  M with m(E) < ¥, f: E  (–¥, ¥) be measurable and { } n   be a sequence of measurable
                                                           f
                                                           n
          functions on E such that
                                         f   f  a.e. on E.
                                         n
          Then for any  > 0 there is a (measurable) subset S of E with m(S) <  such that
                                     f   f  uniformly on E\S.
                                     n



             Notes Again, the condition m(E) < ¥ cannot be dropped. Otherwise f  =    and f = 0
                                                                      n  [n, ¥)
             would be a counter example.
          Proof: We claim that for any  > 0 and  > 0, there exists A  E with m(A) <  and N   such that
                           |f (x) – f(x)| <   whenever n  N and x  E \A.
                             n
          Be careful the above statement is not saying that f   f uniformly on E \A since A depends on
                                                  n
           and .
          To prove our claim, we let
                                   G  = {x  E : |f (x) – f(x)|  }
                                    n          n
          and
                                   G = lim G  :=    E ,  where E  =    G .
                                            n      n           n     k
                                               n               k n
                                                                  
          Note that if x  G then x  E  for all n  , it follows that f (x)  f(x). Since the set of all x such
                                 n                       n
          that f (x)  f(x) is of measure zero, we have m(G) = 0. Note also that m(E  ) < ¥ and E  “decreases”
              n                                                    1        n
          to G, so limm(E ) = m(G) = 0 by Lemma 4. There is N   such that m(E ) < . This N, together
                       n                                            N
          with A : = E , proved our claim.
                    N
                                                                 k
          Now, let  > 0 be given. Apply the above result to = 1/k and  = /2 , we obtain A  with m(A )
                                                                             k       k
               k
          < /2  and N    such that
                     k
                                       1
                           |f (x) – f(x) | <  whenever n  N  and x  E \A .
                            n                            k           k
                                       k
          Let S =  k A ,  then m(S)   å ¥ k 1 m(A ) <  and |f (x) – f(x)| < 1/k whenever n  N  and x  E\S.
                                   =
                     k
                                                  n
                                                                            k
                                        k
          Hence, f   f uniformly onE\S.
                 n
          Self Assessment
          Fill in the blanks:
          1.   Every open subset V of  is a ........................... of disjoint open intervals.
          2.   The family of all measurable sets is denoted by M. We will see later M is a -algebra and
               translation-invariant containing all intervals. The set function m:  M  [0, ¥] defined by
                                 m(E) = m*(E)  for all E  M
               is called .....................................
          3.   Let X be a .............................. and Y be a topological space. A function f: X   Y is called
               measurable if f (V) is a measurable set in X for every open set V inY.
                           –1
          4.   Let E  M, f: E  [–¥, ¥] and g: E  [–¥, ¥]. If f = g almost everywhere on E then the
               ................................ of f and g are the same.



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