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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 mathforever on 2004-10-10 16:06:53 |
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| Dear Ray
Many thanks for joing our discussion. First of all, as it seems to me we have a small confusion. You wrote:
"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."
as an answer to my argument:
"But by the way, I dimly remember that the function which is discontinous only in the countable number of points is Riemann integrable, and f() is definetely such function!"
So, by f() I meant the function from the entry "Dirichlet's function", which you further call g():
"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"
Thus, my argument is applicable to this function. What you wrote further, is an attempt to prove Riemann integrability directly:
"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."
If I get everything right, you consider the function on the interval [0,1]. Then I have the following questions to the above thing:
1) why you set "p_n (x) = n + 1/n"? did you meant p_n=1?
2) the sentence "since there are at n(n-1)/2 fractions with denominator n" actually mean "... with denominator SMALLER OR EQUAL THEN n"?
But once more. The Reimann integrability of our f() (and your g()) is due to the fact that the function has only countable number of discontinuities. |
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