Define $\displaystyle f(x)=x$ if $\displaystyle x$ is rational and $\displaystyle f(x)=0$ if $\displaystyle x$ is irrational.
Is $\displaystyle f(x)$ Riemann Integrable on $\displaystyle [0,1]$?
the lower sum should be zero since in any subdivision, there is an irrational number. and if each subdivision is so small, then upper sum could be really small. so i think the integral is 0
but if you do the samething to Dirichlets funtion, shouldnt the integral be 0 also? but Dirichlets funtion is not Riemann integrable.
i dont see the difference in these functions.
it is not continuous on (0,1], so it is not Riemann integrable.
There is a theorem which finally solve the problem to judge which limitary function is integrable: the set of uncontinuous points forms a "null set",that is, we can use countable numbers of open interval to cover a real numbers' set completely and the sum of the "length" of these interval ($\displaystyle length(a,b)=b-a$)can be arbitrarily close to 0.
A obvious corollary is that if the uncontinuous points set forms an interval (a,b), then the function must not be Riemann integrable.
i see. so the lower and upper sum are not the same. so it is not Riemann integrable.
What if the function is deined as $\displaystyle f(x)=\frac{1}{m}$ if $\displaystyle x=\frac{n}{m}$ where $\displaystyle n,m \in N$ and $\displaystyle f(x)=0$ is $\displaystyle x$ is irrational?
would it be integrable? since we dont know that $\displaystyle f(x)> \frac{1}{2}$on$\displaystyle [\frac{1}{2},1]$, we cant extimate the upper sum.
for every irrational number $\displaystyle x$, we can use a sequence of rational numbers to approach it. In these sequence, the denominator would approach to $\displaystyle \infty$. So the result will have limit 0. So it will be continuous at $\displaystyle x$. So the uncontinous points set would be countable.
Let we list these points by $\displaystyle a_1,a_2,....$, then the sum of the length of the sequence$\displaystyle (a_1-\frac{1}{2^k},a_1+\frac{1}{2^k}),(a_2-\frac{1}{2^{k+1}},a_2+\frac{1}{2^{k+1}})...$ would be $\displaystyle \frac{1}{2^{k-2}}$, which will have limit 0 when $\displaystyle k\rightarrow \infty$. So it would be a "null set". So the function is Riemann integrable