Piecewise Continuity and Integrability...

Question

Let $f: [0, 1] \to \mathbb{R}$ be defined by $f(x) = x$, if $x \in \mathbb{Q}$ and $f(x) = -x$, if $x \in \mathbb{R}\setminus\mathbb{Q}$. Show that $f$ is not Riemann integrable.

Proof Idea: There is a theorem that says: if $f$ is continuous on a closed and bounded interval, then $f$ is Riemann integrable over that interval. So, my idea is to show that $f$ is not continuous so that it is not Riemann integrable.

To this end, we must find $\epsilon_0 > 0$ and $c \in [0, 1]$ such that for all $\delta > 0$ with $x \in [0, 1]$ satisfying $|x - c| < \delta$, then $|f(x) - f(c)| \geq \epsilon_0$. I want to use the density of the irrationals somewhere, but I am not sure where it fits. I know that

$$|f(x) - f(c)| = |x - c|, \: \text{or} \: |f(x) - f(c)| = |x + c|, \: \text{or} \: |f(x) - f(c)| = |-x + c|,$$

depending on if $x, c$ are rational or irrational.

Also, I know that there exists a rational $r \in (c - \delta, c + \delta)$ such that $|x - r| > \delta. Then,

$$|f(x) - f(x)| \leq |f(x) - f(r)| + |f(r) - f(c)| = |x - r| + |r - c| < |x - r| + \delta > \delta + \delta = 2\delta,$$

so take $\epsilon_0 = 2\delta$, yeah?

Edit: I just added the above "Also, I know...".

Any help would be much appreciated.

Answer 1

Note that the theorem goes only in one direction. In other words proving that $f$ isn't continous on closed bounded region doesn't imply that $f$ isn't Riemann integrable. Indeed, note that $f$ having a single jump discontinuity isn't continuous, yet it is Riemann integrable. Hence your proof isn't valid.

You can prove the claim by computing the lower Riemann sum and the upper Riemann sum. For the upper one we have that:

$$R_n = \frac 1n \sum_{i=0}^{n-1} \sup_{x \in \left[\frac in, \frac{i+1}{n}\right]} f(x) = \frac 1n \sum_{i=0}^{n-1} \frac {i+1} n = \frac 1{n^2} \frac{n(n+1)}{2} = \frac{n+1}{2n}$$

For the lower sum we have that:

$$r_n = \frac 1n \sum_{i=0}^{n-1} \inf_{x \in \left[\frac in, \frac{i+1}{n}\right]} f(x) = \frac 1n \sum_{i=0}^{n-1} \frac {-(i+1)} n = \frac 1{n^2} \frac{-n(n+1)}{2} = \frac{-(n+1)}{2n}$$

Now note that $\lim R_n = \frac 12$, while $\lim r_n = - \frac 12$. As both limits don't agree we conclude that $f$ isn't Riemann integrable on $[0,1]$

Answer 2

There is indeed a theorem that says that if $f$ is continuous on a closed and bounded interval, then $f$ is Riemann integrable. That theorem says nothing about what happens if $f$ is not continuous. So, your proof doesn't work.

You could approach the problem proving that every superior sum is greater than $\frac12$ and that every inferior sum is smaller than $-\frac12$.