Question

Show that the function f: [0,1]→ R defined by

F(x) = { 2 , when x is rational

{ 3 , when x is irrational

is not riemann integrable on[0,1]

Answer 1

ANSWER.

Let $P=\left \{ 0=t_{0}<t_{1} <...<t_{n}=1\right \}$ be a partition of the segment [0,1]. Denote $M_{k}=sup \left \{ f(t):t\in\left [t _{k-1},t_{k} \right ]\right \} ,$ $m_{k}=inf \left \{ f(t):t\in\left [t _{k-1},t_{k} \right ]\right \} , k=1,...,n$ .

Since in any segment there are rational numbers and irrational numbers , then $M_{k}=3, m_{k}=2$ for all $k=1,...,n$ . So the upper Darboux sum $U(f,P)$ respect to $P$ is the sum

$U(f,P)=\sum _{1}^{n}M_{k} \left ( t_{k}-t_{k-1} \right )=\sum _{1}^{n}3\left ( t_{k}-t_{k-1} \right )=3(1-0)=3$

The lower Darboux sum $L(f,P)$ is the sum

$L(f,P)=\sum _{1}^{n}m_{k} \left ( t_{k}-t_{k-1} \right )=\sum _{1}^{n}2\left ( t_{k}-t_{k-1} \right )=2(1-0)=2$ .

Therefore, the upper Darboux integral

$U(f)= inf\left \{ U(f,P):P\, is \, \, partition \, \, of\, \, [0,1] \right \}=3,$

the lower Darboux integral

$L(f)= sup\left \{ L(f,P):P\, is \, \, partition \, \, of\, \, [0,1] \right \}=2.$

The upper and lower Darboux integrals for $f$ do not agree , so $f$ is not integrable.