Thread: give an example

Give an example of a function $\displaystyle f: [0,1] \to R $ such that $\displaystyle f \in R[0,1] $ [i.e f is Riemann integrable over [0,1]] $\displaystyle , f(x)>0, \ \forall x \in [0,1] , \ but \ \frac{1}{f} $ is not in $\displaystyle R [0,1] $

Originally Posted by flower3

Give an example of a function $\displaystyle f: [0,1] \to R $ such that $\displaystyle f \in R[0,1] $ [i.e f is Riemann integrable over [0,1]] $\displaystyle , f(x)>0, \ \forall x \in [0,1] , \ but \ \frac{1}{f} $ is not in $\displaystyle R [0,1] $

$\displaystyle f(x)=\left\{\begin{array}{ll}1&\;\;if\,\,x=0\\x&\; \;otherwise\end{array}\right.$

Tonio

Originally Posted by flower3

Give an example of a function $\displaystyle f: [0,1] \to R $ such that $\displaystyle f \in R[0,1] $ [i.e f is Riemann integrable over [0,1]] $\displaystyle , f(x)>0, \ \forall x \in [0,1] , \ but \ \frac{1}{f} $ is not in $\displaystyle R [0,1] $

tonio's answer is of course great, but more generally all you need to do is find a function which has one discontinuity on $\displaystyle [0,1]$ (guess where) such that it's multiplicative inverse is unbounded!