Answer to Question #152982 in Calculus for Ankita

Question

A function f:[a,b]–R is defined as
f(x) = 4 if x is rational
f(x) = -3 if x is irrational
Check whether f is integrable on [a,b]. Justify your answer

Answer 1

f is not Integrable if $a \neq b$

Proof :-

Let a partition be, $P$

Where $P : a=x_0<x_1<...<x_n=b$

Let, $I_j= [x_{j-1},x_j]$ , $M_j= \sup_{I_j} f$ , $m_j=\inf_{I_j}f$

Now $\forall j\in \{1,2,...,n\}$ $\exists$ a rational and an irrational number $\in I_j$

$\Rightarrow M_j=4$ and $m_j=-3$ $\forall j\in \{1,2,...,n\}$

$\therefore U(f, P) = \sum_{j=1}^{\infty}|I_j|M_j=\sum_{j=1}^{\infty}|I_j|\times4$

$=4\times\sum_{j=1}^{\infty}|I_j|=4(b-a)$

and $L(f, P) = \sum_{j=1}^{\infty}|I_j|m_j=\sum_{j=1}^{\infty}|I_j|\times(-3)$

$=(-3)\times\sum_{j=1}^{\infty}|I_j|=3(a-b)$

So, Lower Integral, $L_1= \sup_{P}L(f,P)=\sup_{P}3(a-b)=3(a-b)$

and, Upper Integral, $L_2= \sup_{P}U(f,P)=\inf_{P}4(b-a)=4(b-a)$

Now if $a\neq b,$

$L_1=3(a-b)\neq 4(b-a)=L_2$

Hence, $f$ is not integrable on $[a,b]$ , if $a\neq b$

If $a=b,$ clearly f is integrable and $\int_a^bf=0$ .