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![[parent]](http://images.planetmath.org:8080/images/uparrow.png) Viewing Message
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| ``Re: How to integrate the Dirichlet function''
by rspuzio on 2004-10-10 14:12:29 |
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| To further illustrate the point, I thought another example might be helpful. Define a function g as follows: If x is irrational, let g(x) = 0. If x is rational, we can write x in lowest terms as m/n with n>0 and gcd(m,n) = +1 or -1. In this case, let g(x) = 1/n
In this case, g is continuous at all irrational points and discontinuous at all rational points. The reason for this is as follows: Suppose that x is rational. No matter how small we choose delta, there will exist an irrational number y such that |x-y| < delta. Then |g(x) - g(y)| = 1/n. Suppose that x is irrational. For any n, choose delta = min_m |x - m/n|. Then |g(x) - g(y)| < 1/n when |x-y| < delta.
In this case, g, unlike f, has a countable number of discontinuities. Moreover, g is Riemann-Darboux integrable. To see this, we can, for every n, construct a piecewise constant function p_n such that g < p_n and int_0^i p_n (x) dx < 2/n. Here is a description of p_n: About every fraction with denominator smaller than n, construct an interval of length n^{-3}. If x lies inside one of these intervals, set p_n (x) = n + 1/n. Otherwise, set p_n(x) = 3/(2n). Since there are at n(n-1)/2 fractions with denominator n (and some of them are equal because of cancellation, so the number of rational numbers is smaller), we have int_0^1 p_n (x) dx < 3/(2n) + n(n-1)/(2n^3) < 3/(2n) + 1/(2n) = 2/n.
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