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Math Help - integrable function

  1. #1
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    integrable function

    Suppose  f is Riemann integrabe on  [a,b] . Let  \xi be any number in  [a,b] , and let  r be any real number different from  f(\xi) . Define a function  g as follows:  g(x) = \begin{cases} f(x) \ \ \ \ \text{if} \ x \neq \xi \\ r \ \ \ \ \ \ \ \ \  \text{if} \ x = \xi \end{cases} . Prove that  g is Riemann integrable on  [a,b] and that  \int_{a}^{b} g = \int_{a}^{b} f .

    Proof. Let  \epsilon>0 . Choose  \delta < \frac{\epsilon}{2|r|} . Let  P = \{x_{0}, x_{2}, \ldots, x_{n} \} be a partition of  [a,b] with  ||P|| < \delta . Choose  x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*} arbitrarily such that  x_{i-1} \leq x_{i}^{*} \leq x_i . Then  \mathcal{R}(g,P) = \sum_{i=1}^{n} g(x_{i}^{*})(x_{i}-x_{i-1}) .



    So  \mathcal{R}(g,P) = \begin{cases} f(x_{i}^{*})(x_{i}-x_{i-1}) \ \ \ \ \  \ \ \ \ \  \ \ \ \ \ \ \ \ \text{if} \ x_{i}^{*} \neq \xi \ \text{for any} \ i \\ r(x_{i}-x_{i-1}) \ \ \ \ \  \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \ \ \text{if} \ x_{i}^{*} = \xi \ \text{for exactly one} \ i \\ r(x_{i+1}-x_{i})+r(x_{i}-x_{i-1}) \ \ \ \ \ \text{if} \  \xi = x_{i+1}^{*} = x_{i}^{*} \ \ \ \ \ \text{for some} \ i \end{cases}


    So  |\mathcal{R}(g,P)| \leq |\mathcal{R}(f,P)|+2|r|||P|| . Thus  |\mathcal{R}(g,P)|-|\mathcal{R}(f,P)| \leq 2r||P|| . Then  ||\mathcal{R}(g,P)|-|\mathcal{R}(f,P)|| \leq |\mathcal{R}(g,P)-\mathcal{R}(f,P)| \leq 2|r|||P|| < \epsilon . Hence  g is Riemann integrable on  [a,b] and  \int_{a}^{b} g = \int_{a}^{b} f .  \blacksquare


    Is this correct?
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  2. #2
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    It seem correct in gereral. (I have not carefully read it).
    But I notice you may want \delta  < \frac{\varepsilon }{{2\left| r \right| + 1}} to avoid divison by zero.
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  3. #3
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    Let f,g be integrable on [a,b] then f\pm g is integrable on [a,b] and \smallint_a^b (f\pm g) = \smallint_a^b f \pm \smallint_a^b g.

    If you know this fact then you can use it to prove your problem.

    Lemma: Let f be a zero function on [a,b] except at finitely many points, then f is integrable with \smallint_a^b f = 0.

    Proof: Let the non-zero points be z_1,...,z_k so that f(z_j)\not = 0 and let M = \max_{ 1\leq j\leq k} |f(z_j)|. For \delta >0 and a partition P=\{x_0,x_1,...,x_n\} form the Riemann sum R=\sum_{j=1}^n f(t_j) \Delta_j where ||P|| < \delta. Notice that | R| \leq \sum_{j=1}^n |f(t_j)| \Delta_j \leq M\cdot k \delta. This can be made arbitrary small if \delta is small enough. Thus, \smallint_a^b f = 0.

    To prove you problem, consider f-g and use the lemma above.
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