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Math Help - Characteristic Function

  1. #1
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    Characteristic Function

    Define  f as follows:  f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \ \text{is rational} \\ 1 \ \ \ \ \text{if} \ x \ \text{is irrational} \end{cases}

    Prove  f is not Riemann integrable using the "addition" laws.


    Proof. So we can break up  f as follows:  f(x) = f_{R}(x) + f_{I}(x) . These are the rational and irrational parts of  f respectively. Suppose for contradiction that  f is Riemann integrable on some given interval  [a,b] . Then  f_{R}(x) and  f_{I}(x) are Riemann integrable. But they are not. Contradiction.  \diamond

    Is this correct?
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  2. #2
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    Quote Originally Posted by manjohn12 View Post
    Define  f as follows:  f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \ \text{is rational} \\ 1 \ \ \ \ \text{if} \ x \ \text{is irrational} \end{cases}

    Prove  f is not Riemann integrable using the "addition" laws.
    I am not really sure what you mean by "addition laws". It happens to be really easy to prove this by using the definition of the Riemann integral. For any subinterval [x_{j-1},x_j] (in the partition) there exists a rational and an irrational point. If for any partition we choose only the irrational points then the Riemann sum has value \Sigma_{j=1}^n (1)(x_{j-1}-x_j) = b-a. If for any partition we choose only the rational points then the Riemann sum has value \Sigma_{j=1}^n (0)(x_{j-1} - x_j) = 0. The problem is that b-a and 0 cannot be made arbitrary close to one another with a fine enough parition (since it is independent of ||P||).
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    I am not really sure what you mean by "addition laws". It happens to be really easy to prove this by using the definition of the Riemann integral. For any subinterval [x_{j-1},x_j] (in the partition) there exists a rational and an irrational point. If for any partition we choose only the irrational points then the Riemann sum has value \Sigma_{j=1}^n (1)(x_{j-1}-x_j) = b-a. If for any partition we choose only the rational points then the Riemann sum has value \Sigma_{j=1}^n (0)(x_{j-1} - x_j) = 0. The problem is that b-a and 0 cannot be made arbitrary close to one another with a fine enough parition (since it is independent of ||P||).
    See here. Yes that is the standard way of proving it.
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  4. #4
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    So would the contradiction be correct?
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  5. #5
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    Instead of the traditional way of doing it, would breaking it up into a "rational part" and an "irrational part" be the best way to proceed?
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