# Showing a specific piecewise function is not Riemann integrable

I have the following problem.

Consider the function $$f:[0,1]\rightarrow \mathbb{R}$$ defined by

$$f(x)= \begin{cases} x^3, & x \notin \mathbb{Q};\\ 0 , & x \in \mathbb{Q}. \\ \end{cases}$$

Show that $$f$$ is not Riemann integrable on [0,1].

The way I went about it is as follows:

Let $$P$$ be an arbitrary partition of the interval $$[0,1]$$ given by

$$x_{0}=0 < x_1 < ...

Now since the Lower Riemann Sum with respect to a partition $$P$$ is defined as

$$L(f,P) := \sum_{i=1}^{n} \inf\{f(x); x \in [x_{i-1},x_i] \}(x_{i}-x_{i-1})$$ and in any one interval $$[x_{i-1},x_i]$$ there exists a rational number which would make $$\inf\{f(x); x \in [x_{i-1},x_i] \} \equiv 0$$.

Thus since $$P$$ was an arbitrary partition, we will have that the Lower Riemann Integral defined as

$$\mathcal{L}( f) := \sup\{L(f,P); P \text{ is a partition of } [0,1]\}$$

will be $$\mathcal{L}( f) \equiv 0.$$

We have similar definitions for the Upper Riemann Sums and Upper Riemann Integral respectively:

$$U(f,P) := \sum_{i=1}^{n} \sup\{f(x); x \in [x_{i-1},x_i] \}(x_{i}-x_{i-1})$$

$$\mathcal{U}( f) := \inf\{U(f,P); P \text{ is a partition of } [0,1]\} .$$

I would argue as follows. Since in all the intervals $$[x_{i-1},x_i]$$ there will always be an irrational number, say $$l>0$$ such that $$\sup\{f(x); x \in [x_{i-1},x_i] \}>0$$ and the distance $$x_i-x_{i-1}>0$$ then we must have that $$U(f,P)>0$$.

Thus since $$P$$ was artbitrary, $$\mathcal{U}( f)>0$$ so $$\mathcal{U}( f) \neq \mathcal{L}(f).$$ Thus, $$f$$ is not integrable on $$[0,1]$$.

Is my argument correct?

• Your argument is correct! – Shashi Dec 10 '17 at 17:55

$$\mathcal{U}\left(f\right)=\int_{0}^{1}x^{3}{\rm d}x=\frac{1}{4}$$
Just a comment about terminology - to my knowledge $\mathcal{U}\left(f\right)$ and $\mathcal{L}\left(f\right)$ are known as Darboux's integrals. See here.