# Thread: examining a riemann integral

1. ## examining a riemann integral

Let g(x)=0 if $\displaystyle x \in [0,1]$ is rational and $\displaystyle g(x)=1/x$ if $\displaystyle x \in [0,1]$ is irrational. Explain why $\displaystyle g$ is not riemann integrable.

2. Originally Posted by Chandru1
Let g(x)=0 if $\displaystyle x \in [0,1]$ is rational and $\displaystyle g(x)=1/x$ if $\displaystyle x \in [0,1]$ is irrational. Explain why $\displaystyle g$ is not riemann integrable.
when taking dissections every interval contains rational and irrational points, so the lower sums are always equal to zero for every dissection but the upper sums are always greater than zero.

Bobak

edit: sorry that isn't a complete justification because the upper integral could could be arbitrary small ( i.e. $\displaystyle < \epsilon \ \ \forall \epsilon > 0$ ) however it is easy to show that 1 is a lower bound on the upper integral.

3. Also for a function to be Riemann integrable doesn't it have to be bounded? So $\displaystyle g(x_n)\to\infty$ as $\displaystyle x_n\to{0}$ and $\displaystyle x_n$ is irrational.