Analysis - Riemann integrable function with an infinite number of discontinuity

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pseud0

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The exercise consists in finding a function[MATH] f:[0,1] \to \mathbb{R}[/MATH], Riemann integrable with an infinity of discontinuous points.

I already have two solutions, but I am not able to show that those functions are Riemann integrable.

Function no. 1:
n, natural non zero

[MATH]f(x)=\frac{1}{n^2} [/MATH] if [MATH]x=\frac{1}{n} [/MATH] and 0 if not


Function no. 2:
Let p and q be two integers greater than 1 such as GCD(p,q)=1
[MATH]f(x)= \frac{1}{q}[/MATH] if [MATH]x=\frac{p}{q}[/MATH] and 0 if not

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We studied in class the indicator function of the rationals, as an example of non-Riemann integrable function.
The argument was that it fails Darboux criteria. There is no partition such that [MATH]D(f,P)<\epsilon[/MATH].
This is because, in any subset of the partition the oscillation is 1 (by rational's density). So even if subdivisions are super thin,
[MATH]D(f,P)=R^+ -R^-=1-0=1[/MATH] (the upper Riemann sum minus the lower Riemann sum)
Since [MATH]D(f,p)=1>\epsilon=\frac{1}{2} [/MATH]We conclude by Darboux that the function is not Riemann integrable.
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Back to my problem. I am not able to prove that either of the two candidates are Riemann integrable. I tried to show that for any given epsilon Darboux stands, but I am stuck. Any ideas? Maybe I could suppose it is not and show a contradiction somehow? Can I use the definition of Riemann integral? Are the two cases can be shown by the same method? Is there other example easier to understand?

Thank you for your time,
(And sorry for the poor english, I am learning two langages at the same time: English and Analysis)
 
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As [0,1] is a compact interval as long as the set of discontinuities of f(x) has measure zero and f(x) is bounded it will be Riemann integrable by Lebesgue's integrability condition.

So let \(\displaystyle f(x) = \begin{cases}1 &x \in \mathbb{Q}\\0 &x \not \in \mathbb{Q}\end{cases}\)

As the rationals in [0,1] have measure zero and f(x) is clearly bounded, f(x) will be Riemann integrable on this interval.

Another example is \(\displaystyle f(x) = \begin{cases}1 &x = 1-\dfrac 1 n,~n\in \mathbb{N}\\0 &\text{else}\end{cases}\)

again the set of discontinuities has measure zero.
 
I don't get it. Your first function is not Riemann integrable. It is Lebesgue integrable. I want to show that it is possible for a function to be Riemann integrable while having infinity discontinuity. The Riemann integral and Lebesgue Integral are different operations. Lebesgue generalized Riemann for non-Riemann integrable functions.
 
pseud0 said:
[MATH]f(x)=\frac{1}{n^2} [/MATH] if [MATH]x=\frac{1}{n} [/MATH] and 0 if not
Click to expand...
Start slowly suppose \(f(x)\) is not continuous only at \(c\in(a,b)\).
For any partition of \([a, b]\) we can refine it so that \([c-\varepsilon, c+\varepsilon]\) in now in the partition.
Because that cell has measure \(2\varepsilon\) we can arrange an approximating over the new partition be as small as necessary.
Some authors say to think of putting a tall narrow tent over \(c\). Thus that part of the sum is negligible.
If that makes any sense to you then realize that \(\frac{1}{n}\to 0\) so in any partition of \([0,1]\) the cell containing \(0\) also contains all but a finite collection of the \(\tfrac{1}{n}'s\)
Over that finite collection & \(0\) we can put those tents so the area sum of those discontinuities is negligible.
That is the idea behind proving the function Riemann integrable.
I understand that this may seem vague at first. But you did say that your class had done some proofs.
This is just an adaption of those earlier proofs. Realizing that near zero we make the area small as well as at each discontinuity away from zero also in a small cell.

This is a poor format in which to try to explain this idea. But again, simply put: each term of the approximation is multiplied the length of the cell in the partition containing it. By making the areas around each of the discontinuities small we can approximate total area.
 
pseud0 said:
Function no. 2:
Let p and q be two integers greater than 1 such as GCD(p,q)=1
[MATH]f(x)= \frac{1}{q}[/MATH] if [MATH]x=\frac{p}{q}[/MATH] and 0 if not
Click to expand...
Wouldn't f be 0 except when x=p/q, in which case f would be 1/p or undefined? You would not even have a single discontinuity point if p>q, as p/q would not even be in [0,1].
 
I just opened a second account because I am not currently home and cannot login if I am not on my PC.

The way I see it:
It is implied that p<q, because the interval of interest is [0,1]. If x is rational then f(x)=1/q and is always well defined. Since, if x is a rational number then there is an unique p and unique q such as x=p/q and GCD(p,q)=1

However, I have kind of abandon that function and worked on the other.

But my proof fail. This is the begining of my argument:

We want to show that function f is Riemann integrable.

[MATH]f:[0,1] \to \mathbb{R}[/MATH][MATH]f(x)=1/n^2[/MATH] if [MATH]x=1/n[/MATH] and f(x)=0 if not
n, natural strictly greater than 1

To do so we will use the Darboux crtiteria.

Darboux :
f(x) is Riemann integrable on [0,1] if and only if
for any given epsilon greater than zero, it exists a partition P such that D(f,p) is smaller than epsilon.

For any given epsilon it exists a even number k such that 1/k is smaller than epsilon.
Let P={0, 1/k, 2/k, 3/k, ..., 1)
In the subdivisions [MATH]I_i[/MATH]=[i/k, (i+1)/k] for i>(k/2) , there is no discontinuity of f. The oscillation for those subdivisions is 0.

For the others, the oscillation of f is smaller or equal to [MATH]i^2/k^2[/MATH] .

So [MATH]D(f,p)=\sum_{k=1}^{k/2} (Sup_{I_i} f - Inf_{I_i} f )[/MATH]|P|

[MATH] \leq \sum_{k=1}^{k/2} (i^2/k^2) ((i+1)/k - i/k) \\ =(1/k^3) \sum_{k=1}^{k/2} i^2\\ =(1/k^3) (k/2)((k/2)+1)(2(k/2)+1)/6[/MATH]
If the last phrase is smaller than 1/k for all k then the proof would be done. I checked for some high k rapidly. It seems false. I'll look more into it later when I'll be home.

And again,
Thank you all for your help
 
I must start by saying that I have never studied analysis before. So please double check this carefully, and forgive me if this is wrong because I'm learning too. Following the logic of pka's post#4, for function 1, I think the idea is to choose partitions:-

P={0, x, <--> 1/(m-1)-ε, 1/(m-1)+ε, <--> 1/(m-2)-ε, 1/(m-2)+ε, <--> ... <--> 1/2-ε, 1/2+ε, <--> 1/(1)-ε, 1}

... I have emphasized the large partition widths with <-->. All other partition widths are very small. Like the tents that pka talked about. In all the large width partitions the function's value is 0. In the very small width partitions the function has discontinuities - and a +ve upper value - but it ends up being multiplied by the small width.

The first partition 0 to x covers all the cases where the adjacent "1/k" gaps become closer than 2ε

So (1/(k-1)-ε) - (1/k+ε) < 0
Therefore k < (1 + sqrt(1 + 2/ε))/2
Let m = floor( (1 + sqrt(1 + 2/ε))/2 )
and x = 1/m

Dealing with the first and last partitions as special cases, I get:-

[math] U_{f,P} = \left(\frac{1}{m^2}\right) x+\sum_{k=2}^{m-1}{\left(\left(\frac{1}{k^2}\right)\left(2\epsilon\right)\right)}+\frac{1}{1}\epsilon [/math]
[math]=\frac{1}{m^3}+2\epsilon \sum_{k=2}^{m-1}{\frac{1}{k^2}} +\epsilon [/math]
I'm pretty sure this tends to zero as epsilon tends to zero. The sum from 1 to infinity of 1/k^2 is pi^2/6, but this finite result is multiplied by 2ε
 
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