Proving Riemann Integrability of a Function to Zero

[Reid]

Homework Statement

Prove that the function specified below is Riemann integrable and that its integral is equal to zero.

Homework Equations

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

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 definitely 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. :)

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.

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

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

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

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

 

[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! :)