Lebesgues integrability criteria for Riemann integrability I am stuck in a simple question which should be easy.
The indicator function of $\mathbb Q$   on the compact interval $[0,1]$ is not Riemann integrable. The proof is very easily done by taking the upper and lower sums which are not equal.  Yet I cannot understand why the Lebesgue inegrability condition does not seem to work: a bounded function $f: [a,b] \to \mathbb R$ is Riemann integrable if and only if its points of discontinuity form a set with Lebesgue measure zero.
So the question is: does the set od discontinuities for the indicator function of $ \mathbb Q$ on $[0,1]$ have Lebesgue measure zero? The problem arose when I was reading a book  which stated that the indicator function is continuous on the set of the irrationals, which itself is un uncountable set! So what is wrong?
 A: The indicator function $1_{\Bbb Q}$ is discontinuous everywhere.
If $x \in \Bbb Q$, then consider the sequence $(x_n)_{n \ge 1}$ given by $x_n = x + \frac {\sqrt 2} n$. This is clearly a sequence of irrational numbers with $x_n \to x$. Then $1_{\Bbb Q} (x) = 1$, but $\lim 1_{\Bbb Q} (x_n) = \lim 0 = 0$, so $1_{\Bbb Q}$ is discontinuous in $x$, so the points of $\Bbb Q$ are points of discontinuity for $1_{\Bbb Q}$.
If $x \notin \Bbb Q$, then consider the sequence $(x_n)_{n \ge 1}$ of its approximations from below (for instance, if $x = \sqrt 2$, then $(x_n)_{n \ge 1} = \{1, 1.4, 1.41, 1.414, 1.4142, \dots \}$). Then $1_{\Bbb Q} (x) = 0$, but $\lim 1_{\Bbb Q} (x_n) = \lim 1 = 1$, so $1_{\Bbb Q}$ is discontinuous in $x$, so the points of $\Bbb R \setminus \Bbb Q$ are points of discontinuity.
It follows that every point of $\Bbb R$ is a point of discontinuity for $1_{\Bbb Q}$, so the measure of the points of discontinuity of $1_{\Bbb Q}$ in an interval $(a,b)$ is $b-a > 0$ so $1_{\Bbb Q}$ is not Riemann-integrable.
A: The indicator function for $\mathbb Q $ is everywhere discontinuous on $[0,1] $ and so it is not riemann integrable. On the other hand, the function is only nonzero on $\mathbb Q $, a set of measure zero, so its lebesgue integrable has zero value. I.e:
$$\int_{[0,1]} \chi_{\mathbb Q}\, dm=0. $$
To see that the rationals have measure zero, just note that the set of rationals is countable, and so by countable additivity, $$m (\mathbb Q)=m (\bigcup_{n \in \mathbb N} q_n)= \sum_{n \in \mathbb N} m (q_n)=0. $$
Intuitively, you can find an arbitrarily "small" open cover for $\mathbb Q$, since each point has no dimension, and there are only countably many.
Certainly, the function $\chi_{ \mathbb Q} $ is continuous if you restrict its domain to the irrationals, since it is identically zero on that set, and constant functions are continuous (note that a constant function's preimage is either empty, or the whole set, both which are open in any topology.)
