Riemann integration

Reid

1. The problem statement, all variables and given/known data
Prove that the function specified below is Riemann integrable and that its integral is equal to zero.


2. Relevant equations
f(x)=1 for x=1/n (n is a natural number) and 0 elsewhere on the interval [0,1].


3. The attempt at a solution
I have divided the partition into two subintervals, the first with tags different from x=1/n and the second with tags at x=1/n. But, given an epsilon>0, I am not sure how to choose my delta (the norm of the partition) such that the points where the function is not zero doesn't make a contribution.

Or, is my approach all wrong?

Thanks!

quasar987

Maybe not all wrong, but I would say overly complicated. :)

Consider any partition of [0,1]. Note that every subinterval from your partition contains a point not of the form 1/n.

Reid

Yeah, it is sometimes like that if you study independently. :P

So, considering any partition of [0,1]. I should then tag the points different from 1/n, then making all the contributions zero. Right?

quasar987

Wait, a sec, have you seen the result that if a function is discontinuous at a countable number of points then it is integrable?

Reid

No, i have not. But I will definetly look for it now.

Thanks!

quasar987

Well, if your book hasn't covered this yet, try to do without it.

You want to show that the lower and upper integrals are 0. Prove that the lower riemann sums s(f;P) are 0 for any partition P of [0,1]. This will of course imply that the lower integral is 0.

For the upper integral, you want to show that the inf over every P partition of [0,1] of the upper riemann sums S(f,P) is 0. Show that for every epsilon>0, you can always find a partition P' such that S(f;P')<epsilon.

Reid

What you are describing now feels much better, the squeeze theorem. :)

But I don't understand at all, how to deal with the upper integral...? When finding the inf over every partition. I would like to do it in the same way as i treat the lower integral.

:S

quasar987

And I don't understand your question. :P

Reid

Ok. :)

I don't understand this!

Reid

Now, I understand. (I hope so anyway)

When you wrote 'inf' I thought you meant infimum... so I thought that I was really lost since I have never heard of infimum in the context as Riemann integrals. But you must have meant int as in integral, right?

And, yes! I am an analysis-rookie. ;)

quasar987

I sure meant infimum.

How do you define the upper integral? For me, the upper integral is defined as

[tex]\inf_{P\in \mathcal{P}[0,1]}S(f;P)[/tex]

where [tex]\mathcal{P}[0,1][/tex] is the set of all partitions of [0,1] and

[tex]S(f;P)=\sum_{x_i\in P}\max_{x\in[x_i-1,x_i]}f(x)[/tex]

Reid

Sorry!

I don't define the 'upper integral' at all. For me the 'Riemann integral' is defined as a limit of the Riemann sums as the norm tend to zero. That is why I am talking about the partitions and their tags. I can't find any section with upper Riemann sums either. It is only the Riemann sum.

quasar987

I see!

Well in that case it's even simpler! Consider epsilon>0, then for any delta>0, we have that any Riemann sum associated with a partition whose norm is lesser than delta is 0 because in every subinterval of [0,1], there is a point not of the form 1/n!

lurflurf

I would procede thusly
clearly the lower integral is 0
take any partition and choose taggeg point where f=0
consider the upper integral
suppose the norm is h where 1>h>0
The idea is we want to make a large sum
f1h1+f2h2+f3h3+...+fNhN
f=0,1 so we choose f=1 whenever possible
to take maximum advantage consiger our taggged partition
0=x0=<x*1=<x1=<x*2=<...=<x*N-1=<xN-1=<x*N=<xN=1
we would like n for our tagged partition to include points where f=1 when possible
so we begin
1>1-h>...>h+1/2>1/2>-h+1/2>...>h+1/3>1/3>-h+1/3>...
however at some point intervals chosen in this way begin to intersect and an adjustment is needed
we need to know when
1/n-1/(n+1)<2h
elementary algebra tells us this happens when
n>N=floor(-1+sqrt(1+1/(2h)))/2
thus
the upper integral
UI<1/(N+1)+2hN
we want something in h alone
I leave that to you

Reid

Finally, I understand! Thank you so much! :)