Riemann integrability

Let $f:[0, 1] \rightarrow \mathbb{R}$ be defined as:

$$f(x)= \left\{\begin{matrix}
 0&,  &x \in [0,1]\cap \left ( \mathbb{R} \setminus \mathbb{Q} \right ) \\
x_n &,  &x=q_n  \in [0,1] \cap \mathbb{Q} \\
\end{matrix}\right.$$

where $x_n$ is a sequence such that $\lim x_n =0$ and $0\leq x_n \leq 1$ and $q_n$ is an enumeration of the rationals of the interval $[0, 1]$.

Prove that $f$ is Riemann integrable and that $\displaystyle \int_0^1 f(x)\, {\rm d}x =0$.

Solution



We define $\displaystyle f_n = \left\{\begin{matrix}
x_n &,  &  x \in \{ q_1, \dots, q_n\}\\
0 & , & x \notin \{ q_1, \dots, q_n \}
\end{matrix}\right.$.

Apparantely $f_n$ are Riemann integrable and $\displaystyle \int_0^1 f(x)\, {\rm d}x=0$. Also $f_n \rightarrow f$ uniformly since $f_n, \; f$ coincide except of some points $q_m$ ($m>n$) . Hence:

$$0\leq \left | f(x)-f_n(x) \right |\leq \sup \left \{ x_m \mid m>n \right \} \xrightarrow{n \rightarrow +\infty}0 $$

(since $\lim x_n =0$)

Therefore $f$ is Riemann integrable and from theory we have the interchange of integral and limit, hence:

$$\int_{0}^{1}f(x)\, {\rm d}x = \int_{0}^{1}\lim f_n (x)\, {\rm d}x = \lim_{n \rightarrow +\infty}\int_{0}^{1}f_n(x)\, {\rm d}x = \lim 0 = 0$$

and the exercise is complete.

The exercise can also be found in mathematica.gr