Question

Show that the function f :[0, 1] → R defined by
f(x)= { 1, when x is rational ; 2 when x is irrational
is not Riemann integrable.

Answer 1

Answer on Question #69111 – Math – Real Analysis

Question

Show that the function $f\colon [0,1]\to \mathbb{R}$ defined by $f(x) = \left\{ \begin{array}{ll}1, & x\in \mathbb{Q}\\ 2, & x\notin \mathbb{Q} \end{array} \right.$ is not Riemann integrable.

Proof

The definition of Riemann integral is here:

https://en.wikipedia.org/wiki/Riemann_integral#Definition

Let us take an arbitrary partition of interval $[0,1]$: $0 = x_0 < x_1 < \dots < x_n = 1$. Now we shall take the tagged partition $P(x,t)$ of $[0,1]$ in two different ways:

1) $t_i \in [x_i, x_{i+1}] \cap \mathbb{Q}$;

2) $t_i \in [x_i, x_{i+1}]$, and $t_i \notin \mathbb{Q}$.

Then the lower (with respect to the first way) and upper (with respect to the second way) Darboux sums are

$$L(f, P) = \sum_{i=0}^{n-1} \inf_{t \in [x_i, x_{i+1}]} f(t)(x_{i+1} - x_i) = \sum_{i=0}^{n-1} (x_{i+1} - x_i) = 1 - 0 = 1;$$ $$U(f, P) = \sum_{i=0}^{n-1} \sup_{t \in [x_i, x_{i+1}]} f(t)(x_{i+1} - x_i) = \sum_{i=0}^{n-1} 2(x_{i+1} - x_i) = 2 \cdot (1 - 0) = 2.$$

We see that \max_{i\in [0,n - 1]}(x_{i + 1} - x_i)\to 0}\left(U(f,P) - L(f,P)\right) = 2 - 1 = 1\neq 0, so the function $f$ is not Riemann integrable

(see https://www.encyclopediaofmath.org/index.php/Darboux_sum).

Remark. A slightly different way to prove this fact we can find here:

https://en.wikipedia.org/wiki/Riemann_integral#Examples

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