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作者  lightwith (none) 站內  97RA_I
標題  Remark 1.1
時間  2008/11/15 Sat 18:56:24

972201003


We note that if a real-valued function f is Riemann integrable on [a,b],
then f is Lebesgue integrable on [a,b]. But the reverse is not true always.
The following is such  a example.        ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄ ̄


Let f(x)= { 1 , if x is an irrational number in [0,1]
          { 0 , otherwise

    and

    g(x)= { 0 , if x is an irrational number in [0,1]
          { 1 , otherwise

    Let P: 0 = x_1 < x_1 < x_2 < ...< x_n =1 be any partition of [0,1]

    Then    sup   f(x) = 1   and     inf   f(x) = 0 ,for all j=1..n
         x  ≦x≦x               x  ≦x≦x
          j-1     j               j-1     j

    and hence the upper Riemann sum of w.r.t P, U(P,f)=1
              the lower Riemann sum of w.r.t P, L(P.f)=0

    ∴  f is not Riemann integrable.

    Similarly, g is not Riemann integrable.


    Let E  = {x: x is a rational number of [0,1] } and
         1
        E  = {x: x is an irrational number of [0,1] }
         2

    then f and g are all simple functions satisfying

        f(x) = 0.χ   (x) + 1.χ   (x)
                    E_1           E_2

        g(x) = 1.χ   (x) + 0.χ   (x)
                    E_1           E_2

   By the definition of the Lebesgue integral for simple function,

   we obtain that f and g are Lebesgue integrable and

        ∫    f = 0.m(E ) + 1.m(E ) = 1
         [0,1]          1          2

        ∫    g = 1.m(E ) + 0.m(E ) = 0
         [0,1]          1          2


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□ Modify: 2008/12/08 Mon 10:47:41  webpcf4.mcl.math.ncu.edu.tw 修改

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