Given the indicator function of the rationals on $(0,1)$.

Could anyone help me understand why the measure of the set of discontinuity points is not zero?

It is pretty obvious to me that this function is not Riemann integrable. So according to Lebesgue theorem, the set of discontinuity points is not zero.

But it seems to me that the number of "jumps" in the function should be of size $|Q|$, and thus countable and with measure zero.

What am I missing here?

  • $\begingroup$ What do you believe the set of discontinuity points is for this function? $\endgroup$
    – lulu
    May 3, 2019 at 15:18
  • $\begingroup$ All the rational point are discontinuity points (aren't they?) $\endgroup$ May 3, 2019 at 15:45
  • 2
    $\begingroup$ As are all the irrational points. $\endgroup$
    – lulu
    May 3, 2019 at 15:47

1 Answer 1


This function doesn't have any points of continuity in $(0,1)$. Let's call the function $f$. If you take any $x\in (0,1),$ you can build a sequence of rational numbers $(x_n)$ in $(0,1)$ which converges to $x$ and a sequence of irrational numbers $(y_n)$ in $(0,1)$ which converges to $x$. Then $f(x_n)$ is the constant sequence $1,$ while $f(y_n)$ is the constant sequence $0$. So one of these sequences does not converge to $f(x)$; hence $f$ is not continuous at the point $x$.

In other words, the set of discontinuities contains the whole interval $(0,1),$ which doesn't have measure zero.


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