Page 109 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS
P. 109

Measure Theory and Functional Analysis




                    Notes          By the countable subadditivity of  on ,
                                   we have  (E)      (E ) = 0.
                                                n    n
                                   Thus   (E) = 0.
                                   This shows that E is a null set in (X, ,  ).

                                   9.1.2 Complete Measure  Space

                                   Definition: Given a measure   on a  -algebra  of subsets of a set X. We say that the  -algebra 
                                   is complete with respect to the measure  if an arbitrary subset E  of a null set E with respect to
                                                                                       0
                                     is a member of   (and consequently has   (E ) = 0 by  the Monotonicity of  ). When  is
                                                                          0
                                   complete with respect to  , we say that (X, ,  ) is a complete measure space.

                                          Example: Let X = {a, b, c}. Then = { , {a}, {b, c}, X} is a  -algebra of subsets of X. If we
                                   define a set function  on by setting  ( ) = 0,  ({a}) = 1,  ({b, c}) = 0, and  (X) = 1, then  is
                                   a measure on . The set {b, c} is a null set in the measure space (X, ,  ), but its subset {b} is not
                                   a member of . Therefore, (X, ,  ) is not a complete measure space.

                                   9.1.3 Measurable Mapping


                                   Let f be a mapping of a subset D of a set X into a set Y. We write D (f) and R (f) for the domain of
                                   definition and the range of f respectively. Thus
                                                           D (f) = D   X,

                                                           R (f) = {y   Y : y = f (x) for some x   D (f)}   Y.
                                   For the image of D (f) by f, we have f (D (f)) = R (f). For an arbitrary subset E of y we define the
                                   preimage of E under the mapping f by

                                                      –1
                                                     F  (E) : = {x   X : f (x)   E} = {x   D (f) : f (x)   E}.



                                     Notes
                                     1.   E is an arbitrary subset of Y and need not be a subset of R (f). Indeed E may be disjoint
                                          from  (f), in which case f  (E) =  . In general, we have f (f  (E))   E.
                                                               –1
                                                                                         –1
                                                                                            –1
                                     2.   For an arbitrary collection C of subsets of Y, we let f  (C) : = {f  (E) : E   C}.
                                                                                    –1
                                   Theorem 2: Given sets X and Y. Let f be a mapping with D (f)   X and  (f)  Y. Let E and E  be
                                   arbitrary subsets of Y. Then
                                   1.  f  (Y) = D (f),
                                        –1
                                               –1
                                   2.  f  (E ) = f  (Y\E) = f  (Y)\f  (E) = D (f) \ f  (E),
                                                        –1
                                                              –1
                                        –1
                                           C
                                                                          –1
                                                       –1
                                   3.  f  (U   E ) = U   f  (E ),
                                        –1
                                                    
                                        –1
                                                       –1
                                   4.  f  (    E ) =   f  (E ).
                                                    
                                   Theorem 3: Given sets X and Y. Let f be a mapping with D (f)   X and R (f)  Y. If  is a  -algebra
                                   of subsets of Y then f  () is a  -algebra of subsets of the set D (f). In particular, if D (f) = X then
                                                   –1
                                   f  () is a  -algebra of subsets of the set X.
                                   –1
          102                               LOVELY PROFESSIONAL UNIVERSITY
   104   105   106   107   108   109   110   111   112   113   114