# Math Help - question on Integrals

1. ## question on Integrals

Define $f(x)=x$ if $x$ is rational and $f(x)=0$ if $x$ is irrational.
Is $f(x)$ Riemann Integrable on $[0,1]$?

2. Originally Posted by Kat-M
Define $f(x)=x$ if $x$ is rational and $f(x)=0$ if $x$ is irrational.
Is $f(x)$ Riemann Integrable on $[0,1]$?
Given any subdivision, $\mathcal{D}$, of the interval $[0,1]$ what is the upper sum and the lower sum with respect to $\mathcal{D}$, ?

Can that difference be made ‘small’?

3. ## subdivisions

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.

4. 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 ( $length(a,b)=b-a$)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.

5. Originally Posted by Kat-M
and if each subdivision is so small, then upper sum could be really small.
really? in $[\frac{1}{2},1]$, any upper sum would be greater than $\frac{1}{4}$!

6. ## upper sum

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 $f(x)=\frac{1}{m}$ if $x=\frac{n}{m}$ where $n,m \in N$ and $f(x)=0$ is $x$ is irrational?

would it be integrable? since we dont know that $f(x)> \frac{1}{2}$on $[\frac{1}{2},1]$, we cant extimate the upper sum.

7. Originally Posted by Kat-M
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 $f(x)=\frac{1}{m}$ if $x=\frac{n}{m}$ where $n,m \in N$ and $f(x)=0$ is $x$ is irrational?

would it be integrable? since we dont know that $f(x)> \frac{1}{2}$on $[\frac{1}{2},1]$, we cant extimate the upper sum.
for every irrational number $x$, we can use a sequence of rational numbers to approach it. In these sequence, the denominator would approach to $\infty$. So the result will have limit 0. So it will be continuous at $x$. So the uncontinous points set would be countable.
Let we list these points by $a_1,a_2,....$, then the sum of the length of the sequence $(a_1-\frac{1}{2^k},a_1+\frac{1}{2^k}),(a_2-\frac{1}{2^{k+1}},a_2+\frac{1}{2^{k+1}})...$ would be $\frac{1}{2^{k-2}}$, which will have limit 0 when $k\rightarrow \infty$. So it would be a "null set". So the function is Riemann integrable

8. I am sorry I do not follow your idea but use another theorem..but this theorem is really convenience when judging the integrability