# give an example

• Feb 16th 2010, 09:05 AM
flower3
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 $\displaystyle R [0,1]$
• Feb 16th 2010, 09:12 AM
tonio
Quote:

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
• Feb 16th 2010, 12:45 PM
Drexel28
Quote:

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!