Page 260 - DMTH401_REAL ANALYSIS
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Real Analysis




                    Notes                                       Î Î
                                   = [U(T,f) – L(T,f)] + [U(T,g) – L(T,g)]  <  +  = Îfor T occurring in (5). This shows that f + g Î R(a,b)
                                                                 2  2
                                                      b             b       b
                                   It remains to show that  [f(x) g(x)]dx+  =   [f(x) dx +   g(x)
                                                      
                                                      a             a       a
                                   Now
                                   b          b
                                                               +
                                                         £
                                                 +
                                     +
                                                                         +
                                                                   £
                                     (f g)(x)dx =    (f g)(x) dx U(P,f g) U(P,f) U(P,g)                    (6)
                                              a
                                   for any partition P of [a,b]. Given any Î > 0 we can find a partition P of [a,b] such that
                                                 b
                                          U(P,f) <    f(x)(x) dx+ Î /2
                                                 a
                                                 b
                                          U(P,g) <   g(x) dx+ Î /2                                          (7)
                                                 a
                                   Substituting (7) in (6), we obtain
                                          b           b       b
                                             +
                                            (f g)(x) dx <   f(x) dx +    g(x) dx+ Î                       (8)
                                          a           a       a
                                   Since (8) holds for arbitrary Î > 0, we obtain
                                          b           b       b
                                             +
                                            (f g)(x) dx £   f(x) dx +    g(x) dx                          (9)
                                          a           a       a
                                   Replacing f and g by –f and –g in (9) we obtain
                                          b            b          b
                                                                   -
                                            -
                                              -
                                                         -
                                            ( f g)(x) dx £   { f (x)} dx +   { g(x)} dx
                                   or
                                          b             b      b
                                             +
                                            (f g) (x) dx £ -   f(x) dx -   g(x) dx
                                          a             a      a
                                   This is equivalent to
                                          b           b       b
                                             +
                                            (f g) (x) dx     f(x) dx +    g(x) dx
                                          a            a      a
                                   Combining (9) and (10), we get
                                          b           b      b
                                             +
                                                             
                                            (f g) (x) dx =   f(x) dx g(x) dx
                                          a           a      a
                                   Which proves the theorem.
                                   Theorem 12: If f Î R(a,b) and g Î R (a,b), then f – g Î R(a, b) and

                                   b           b       b
                                     -
                                     (f g) (x) dx =   f(x) dx -   g(x) dx.
                                   a           a       a
                                   Proof: Since g E R [a,b], therefore -g Î R [a,b] and
                                   b          b
                                    - [g(x)]dx = -   g(x) dx






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