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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?

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    $\begingroup$ No, it's discontinuous everywhere on [0,1] (for all points $x$ there is a sequence of rationals and a sequence of irrationals converging to $x$). $\endgroup$ – Matthew Towers Sep 5 '16 at 8:33
  • $\begingroup$ Possible duplicate of Is Dirichlet function Riemann integrable? $\endgroup$ – Alex M. Jan 15 '17 at 13:39
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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.

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  • $\begingroup$ You should mention that you're using the sequential characterization for continuity. $\endgroup$ – Andres Mejia Sep 5 '16 at 8:53
  • $\begingroup$ But otherwise +1 $\endgroup$ – Andres Mejia Sep 5 '16 at 8:53
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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.)

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  • $\begingroup$ Thank you both for the clear and convincing arguments. $\endgroup$ – Jonas Sep 5 '16 at 10:14
  • $\begingroup$ Thank you both for the clear and convincing arguments. If the indicator fonction of Q is continous if you restrict its domain to the irrationals, then it will be also continous if you restrict the domain of definition to the rationals. Does this continuity make any sense, since there would be countably/uncountably many points at which the indicator function would not be defined ? $\endgroup$ – Jonas Sep 5 '16 at 10:20

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