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| ``Re: How to integrate the Dirichlet function''
by rspuzio on 2004-10-10 13:33:19 |
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| Careful. even though your function only differs from zero at a countable number of points, that does not mean that it is discontinuous at a countable number of points! In fact, your function is discontinuous at an uncountable number of points --- it is discontinuous at every point on the real axis. Remember that continuity mans that lim_{x -> y) f(x) = f(y). However, this limit does not exist for any y in the case of your f. To see why, go back to the epsilon-delta definition of limit. For every epsilon, there has to exist a delta such that |f(x) - f(y)| < epsilon when |x-y| < delta. But what if we chose epsilon to be a half? No matter how small we choose delta, we can find an x such that |x-y| < delta but |f(x) - f(y)| = 1. So this function is not continuous for any y.
Therefore, there is no contradiction with the theorem you cite about functions which are integrable in the sense of Riemann and Darboux. Your function is not Riemann integrable and it is nowhere continuous. In fact, your function is a standard counterexample used to show the superiority of Lebesgue's integral over Riemann's integral.
By the way, as for your explanation of why f is Riemann integrable using discretizations of the interval, the problem with it is that, in Riemann's definition of his integral, he requires you to look at all Riemann sums such that the distance between adjacent points at which the function is being discretized is smaller than a given amount. Unfortunately, once you include all such sums, you find out that they do not converge either. That doesn't mean your argument is completely worthless, though. What it means is that you have to be more selective about which sums are allowed as approximations of the integral. Henstock figured out a bettter way of choosing sums which gies rise to the so-called Generalized Riemann integral. So your argument shows that your function can be integrated by using a generalized Riemann integral. Again there is no contradiction because the Generalized Riemann integral can be shown to be equivalent to the Lesbegue integral.
Hope this helps,
Ray |
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