# Thread: division of sections ,bounds definition question

1. ## division of sections ,bounds definition question

f(x)=1 for $x=\frac{1}{n}$
f(x)=0 for otherwise
A)
prove that for 0<a<1 f is integrabile an $\int_{a}^{1}f(x)dx=0$
B)prove that for every $\epsilon>0$there is a division P of the [0,1] section so
$S(P)<\epsilon$
C)prove that f is integrabile on [0,1] and $\int_{0}^{1}f(x)dx=0$

how i tried:
part A i have solved
its very easy i just couddnt see that 1/n is behind a and till 1 we have n-1 fractures
so its 0 exsept finite number of points so the integral is zero

i am stuck on part B:

i was told to look at $[\frac{\epsilon}{2},1]$section so from part A $\int_{\frac{\epsilon}{2}}^{1}f(x)dx=0$and because f is integrabile on the section then $\int_{\frac{\epsilon}{2}}^{1}f(x)dx=inf\{S(P)|P=di viding[\frac{\epsilon}{2},1]}$
“there exists deviding Q of $[\frac{\epsilon}{2},1]$ so $S(Q)<\frac{\epsilon}{2}$ (other wise $\frac{\epsilon}{2}$ would be lower bound of the upper sums and the integral will not be zero.”

why the integral would be zero???

i cant see how to get it
i know that if its integrabile then the integral equals the lower bound of the upper sums
but why it will be zero?

2. ## Re: division of sections ,bounds definition question

Choose your intervals so that the endpoints are 1/2, 1/3, 1/4, etc.

3. ## Re: division of sections ,bounds definition question

why??
how it will show that the integral will not be zero
?

4. ## Re: division of sections ,bounds definition question

It won't! The integral is 0!!!
(C) specifically says that!

5. ## Re: division of sections ,bounds definition question

Originally Posted by transgalactic
f(x)=1 for $x=\frac{1}{n}$
f(x)=0 for otherwise
B)prove that for every $\epsilon>0$there is a division P of the [0,1] section so
$S(P)<\epsilon$
C)prove that f is integrabile on [0,1] and $\int_{0}^{1}f(x)dx=0$
i am stuck on part B:
Let $\delta=\frac{\epsilon}{2}$ then the a positive integer, $J$, such that if $n\ge J$ then $\frac{1}{n}<\delta$.
Now note that the interval $[0,\delta]$ has an upper sum which is less that $\delta^2$.
Now cover the $J-1$ points $1,\frac{1}{2},\cdots,\frac{1}{J-1}$ that add up to a number less than $\frac{\epsilon}{2}$.
Now we have a partition and the upper sum is less than $\epsilon$

6. ## Re: division of sections ,bounds definition question

A similar thing in slightly different words.
Originally Posted by transgalactic
i was told to look at $[\frac{\epsilon}{2},1]$section so from part A $\int_{\frac{\epsilon}{2}}^{1}f(x)dx=0$and because f is integrabile on the section then $\int_{\frac{\epsilon}{2}}^{1}f(x)dx=inf\{S(P)|P=di viding[\frac{\epsilon}{2},1]}$
“there exists deviding Q of $[\frac{\epsilon}{2},1]$ so $S(Q)<\frac{\epsilon}{2}$ (other wise $\frac{\epsilon}{2}$ would be lower bound of the upper sums and the integral will not be zero.”
Since $\inf\{S(P)\mid P\ \mbox{is a division of}\ [\epsilon/2,1]\}=0$, there exists a division Q such that $S(Q) <\epsilon/2$. This follows from the definition of infimum. For the segment $[0,\epsilon/2]$, choose any division. Since the value of f(x) does not exceed 1, the upper sum of the segment $[0,\epsilon/2]$ does not exceed $\epsilon/2$. For a division for the whole segment [0, 1] take the union of any division on $[0,\epsilon/2]$ and Q. Then the corresponding sum will be $< \epsilon$.