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"Dirichlet function" is meant to be the characteristic function of rational numbers on $[a,b]\subset\mathbb{R}$.

On one hand, a function on $[a,b]$ is Riemann integrable if and only if it is bounded and continuous almost everywhere, which the Dirichlet function satisfies.

On the other hand, the upper integral of Dirichlet function is $b-a$, while the lower integral is $0$. They don't match, so that the function is not Riemann integrable.

I feel confused about which explanation I should choose...

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The Dirichlet function isn't continuous anywhere... –  Ayman Hourieh Jul 6 '13 at 20:44

2 Answers 2

up vote 6 down vote accepted

The Dirichlet function $f$ isn't continuous anywhere. For every irrational number $x$, there is a sequence of rational numbers $\{r_n\}$ that converges to it. We have: $$ \lim_{n\to\infty} f(r_n) = 1 \ne 0 = f(x) $$

Thus, $f$ isn't continuous at irrational numbers. Rational numbers can be handled similarly.

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The Dirichlet function is nowhere continuous, since the irrational numbers and the rational numbers are both dense in every interval $[a,b]$. On every interval the supremum of $f$ is $1$ and the infimum is $0$ therefore it is not Riemann integrable.

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The irrationals are not 'dense' in the rationals and vice versa because those two sets are disjoint. What you meant is that the rationals are dense in $[a,b]$ and so are the irrationals. –  Patrick Da Silva Jul 6 '13 at 20:49
Yes, that's what I meant, I will correct it. –  Fye Jul 6 '13 at 20:52

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