# Showing that the characteristic function of $\Bbb Q$ is not Riemann Integrable

So I know that the characteristic function of the rationals is not Riemann integrable and we can show this by showing that the upper and lower sums are different.

But I have a theorem in my notes which states (we have not proved) that a function, $f$, is Riemann integrable iff the set of all points where $f$ is not continuous is a set of zero measure.

So I was wondering if the following reasoning is ok:

We know that $[0,1]$ has measure $1$ and the set $\mathbb{Q}$ is a set of zero measure. So it must be the case that $[0,1] - \mathbb{Q}_{[0,1]}$ is of measure $1$, so $f$ is not Riemann integrable.

(I know that this question is not very well informed, but I have this theorem and so I can use this in my exam as a quick way to show a function is not Riemann integrable, I would be more than happy for more information though.)

Thanks for any help

• The fact that $\mathbb Q$ has measure $0$ is not important here. What is important is that $f$ is discontinuous at every point of $[0,1]$, whether rational or not, and $[0,1]$ has nonzero measure. – Robert Israel May 6 '12 at 18:50
• Ah thanks very much, yes that should have been obvious but as I say I am kind of working on shaky ground (this kind of thing wasn't really in the scope of our course) – hmmmm May 6 '12 at 18:57

## 1 Answer

As pointed by Robert Israel in a comment:

The fact that $\Bbb Q$ has measure $0$ is not important here. What is important is that $f$ is discontinuous at every point of $[0,1]$, whether rational or not, and $[0,1]$ has nonzero measure.