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

$$ x_{0}=0 < x_1 < ... <x_{n-1}<x_{n}=1. $$

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?

  • $\begingroup$ Your argument is correct! $\endgroup$
    – Shashi
    Dec 10, 2017 at 17:55

1 Answer 1


Your argument seems right. You can even argue that

$$\mathcal{U}\left(f\right)=\int_{0}^{1}x^{3}{\rm d}x=\frac{1}{4}$$

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.


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