question on Integrals

**Kat-M** · August 17th 2009, 04:45 PM

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

**Plato** · August 17th 2009, 05:03 PM

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’?

**Kat-M** · August 17th 2009, 05:17 PM

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.

**ynj** · August 17th 2009, 05:27 PM

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.

**ynj** · August 17th 2009, 05:32 PM

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 $\frac{1}{4}$ !

**Kat-M** · August 17th 2009, 05:43 PM

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.

**ynj** · August 17th 2009, 05:53 PM

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

**ynj** · August 17th 2009, 06:11 PM

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

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