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Homework 3 Q1.2

Homework 3 Q1.2

av Nils Tiemer -
Antal svar: 14

If we define a function
f(x) =

    x if x is rational

    1/2 if x is irrational


Isn't this function continuous at 1/2 and by the proposed statement in the question, then be an element of the set of integrable functions with respect to alpha- and alpha+. But shouldn't we be able with a similar argument to 1.1 show that this function is not integrable to either alpha- or alpha+, thereby disproving the statement?

Som svar till Nils Tiemer

Re: Homework 3 Q1.2

av Sofia Tirabassi -
No this is not continuos. This is 1/2 the Dirichlet function which is discontinuos at every point, and it is indeed discontinuos at every point. To see this, let $a$ be a real number.
Suppose first that $a$ is rational. Then, fix $\epsilon=1/3<1/2$, we have that for any $\delta>0$ there is an irrational number t in (a-\delta,a+\delta). And we have that |f(a)-f(t)|=1/2>\epsilon even if |a-t|<\delta. Thus $f$ is not continuos at $a$
If $a$ is not rational, then the argument is the same. Take the same \epsilon, then, for every positive \delta there are irrational numbers t in (a-\delta,a+\delta). And we have that |f(a)-f(t)|=1/2>1/3 even if |a-t|<\delta. Thus $f$ is not continuos at $a$
I hope this helps.
Som svar till Sofia Tirabassi

Sv: Re: Homework 3 Q1.2

av Max Karlsson -
Hello Sofia! Me and Nils are still trying to verify this with each other.

We are looking into your argument:

"Then, fix $\epsilon=1/3<1/2$, we have that for any $\delta>0$ there is an irrational number t in (a-\delta,a+\delta). And we have that |f(a)-f(t)|=1/2>\epsilon even if |a-t|<\delta. Thus $f$ is not continuos at $a$"

However, the function we have defined is equal to 1/2 if the number is irrational. This would make f(a) - f(t) = 0, if a = 1/2. Since a is a rational number, our function we have defined would evaluate it to "1/2" - and not 0. Hence the difference f(a) - f(t) = 0. If however t was a rational number; then it should be possible to find a delta s.t f(a) - f(t) < epsilon (?)

Nils Tiemer and I believe you have missinterpretaded our function. We have made a similar function to example 4.27(b) in Baby Rudin. If our f(x) is not continous at x=1/2 , why would the function at 4.27(b) be continous at x=0 (page 94) be?

Is our function called the Dirichet function?

I apologize in advance if my english is insufficient. Please if I need to clarify something, dont hesitate to point that out. /Max
Som svar till Max Karlsson

Re: Sv: Re: Homework 3 Q1.2

av Sofia Tirabassi -
Hello! In my argument I proved that the function is not continuous at Every point. I first proved that it is not continuous at every irrational point (not the case of 1/2) in the second case I proved that it is not continuous at rational points (1/2 case).
For your convenience I will spell put the argument when a=1/2.
Let epsilon=1/3 for every delta ther is an IRRATIONAL number t in (1/2-delta, 1/2+delta). Then we have that
|f(t)-f(a)|=1/2>epsilon even if |t-a| Thus f is not continuous at 1/2

Is this better?
Som svar till Nils Tiemer

Sv: Homework 3 Q1.2

av Matteo Brandt -
Hi,
I might be wrong but I think the statement in 1.2 must include that f can not have infinitely many discontinuities. For reference, look at theorem 6.10 in Rudin. Your function should indeed be continuous at x=1/2 since for any epsilon we can choose delta = epsilon. For irrational x, dY(f(x), f(1/2)) = 0 and for rational x, dY(f(x), f(1/2)) < |1/2 +- epsilon - 1/2| = epsilon. So it looks like you have found a valid counterexample to the statement. However your function has infinitely many discontinuities (it is discontinuous at all other points) so it does not satisfy the prerequisites of 6.10. Could this be the case?
Som svar till Matteo Brandt

Re: Sv: Homework 3 Q1.2

av Sofia Tirabassi -

The exercise is a slight modification (where we have 0 instead of 1/2)of exercise 3 in Rudin part a and b (see attached file), in which only bounded is asked. The function f is integrable both with beta1 and beta2 iff f(0+)=f(0)=f(0-) this means that the limit of f for x that goes to 0 exists and ut is equal f(0), which means that the function is continuous.

But an answer that with the additional assumption of finitely many discontinuity points will be  accepted.

Finally, I do not understand your comment about finitely many discontinuity points. The function f(x)=0 for x irrational and f(x)=1/2 for x rational is nit continuous at any point...

But I hope this helps

Mvh

Sofia

Bilaga Screenshot_20240726_202228_Samsung Notes.jpg
Som svar till Sofia Tirabassi

Sv: Re: Sv: Homework 3 Q1.2

av Matteo Brandt -
Hi Sofia,

I see how we need f(1/2-) = f(1/2) for f to be in R(alpha+) and f(1/2+) = f(1/2) for f to be in R(alpha-) and so for the intersection of the two we need f(1/2-) = f(1/2) = f(1/2+), so f must be continuous at x = 1/2. However looking at the function that Nils and Max proposed as a counterexample, I'm wondering if we don't also need f to have a finite amount of discontinuities?

Regarding this comment, I think you misread their example; it is not the Dirichlet function, but a similar one with f(x) = x for rational x, not f(x) = 0. Isn't this function continuous at x = 1/2, yet not Riemann integrable, or am I missing something?
Som svar till Matteo Brandt

Re: Sv: Re: Sv: Homework 3 Q1.2

av Sofia Tirabassi -

Yes I misread the function! Luckilybfor me, that is still not continuos at 1/2 (see attached note). However that function is continuous in 0. Which gives you that f(x-1/2) is continuos at 1/2.

But I do not think it gives a counterexample.

I hope this helps

If you have further questions keep asking :)

Sofia

Som svar till Matteo Brandt

Re: Sv: Re: Sv: Homework 3 Q1.2

av Sofia Tirabassi -

Hello again,

In the attached file I shof that f(x-1/2) is integrable with respect to both alphas. So you do not really need that f has finitely many singularities.


In the Rudin text he uses finitely many singularities to see that f is not just continuous in s, but in an interval around s. In fact if the singularities are finite, the set of singular points is closed and around a non singular point there is a whole neighborhood of non singular points.


However, you do not need this assumption, as I showed in one of my replies to another post, if I have an interval around a smooth point and I reduce its length. The sup and inf of the function tends to f(s).


Som svar till Sofia Tirabassi

Sv: Re: Sv: Re: Sv: Homework 3 Q1.2

av Matteo Brandt -
Hi Sofia,

Thank you for your answer. The function proposed was f(x) = x for x rational and f(x) = 1/2 for x irrational but your example is basically the same. Turns out that the function just looks less "nice" than it actually is since it's indeed integrable :)

Does this mean however that the assumption of finitely many discontinuities is not necessary in theorem 6.10 in Rudin? Is it only assumed to make the proof easier? I'm still a bit confused about that.
Som svar till Matteo Brandt

Re: Sv: Re: Sv: Re: Sv: Homework 3 Q1.2

av Sofia Tirabassi -

I do not have Rudin on hand, I think on one side the proof is easier, on the other side I think that in the 6.10 it is nit assumed Bounded, while it is assumed in the exercise

Som svar till Matteo Brandt

Re: Sv: Re: Sv: Re: Sv: Homework 3 Q1.2

av Sofia Tirabassi -

I do not have Rudin on hand, I think on one side the proof is easier, on the other side I think that in the 6.10 it is nit assumed Bounded, while it is assumed in the exercise

Som svar till Sofia Tirabassi

Sv: Re: Sv: Re: Sv: Re: Sv: Homework 3 Q1.2

av Matteo Brandt -
Hi again,

Bounded is also included in 6.10. Everything looks to be the same which is why I'm confused.

The theorem states: Suppose f is bounded on [a, b], f has only finitely many points of discontinuity on [a, b], and alpha is continuous at every point at which f is discontinuous. Then f is in R(alpha).

I have not used the theorem in my solution of 1.2 because of the finite discontinuities required, but it seems like without it, 1.2 can be solved with the theorem. Nevertheless, I don't want us to spend too much of our time figuring out this detail. It just seems strange that Rudin included a "weaker" version of the theorem.
Som svar till Matteo Brandt

Re: Sv: Re: Sv: Re: Sv: Re: Sv: Homework 3 Q1.2

av Sofia Tirabassi -

We were drowning in a glass of water:

Theorem 6.10 works for every increasing alpha, the exercise assumes that alpha is a step function, that is a very nice increasing function. Upper semicontinuous or lower semicintinuous.


So it is natural that some hypotheses can be discarded in this case. Theorem 6.10 gives you an implication. With its exercise Rudin want first to make sure that you understood the proof of 6.10, since its solution is similar. Then he want to tell you that some assumptions can be relaxed if we limit our alphas.

Mvh

Sofia

Som svar till Sofia Tirabassi

Sv: Re: Sv: Re: Sv: Re: Sv: Re: Sv: Homework 3 Q1.2

av Matteo Brandt -
Hi Sofia,

It makes sense now. We can not guarantee that it works for f with infinite discontinuities for all alpha, but for the alpha in 1.2 it does. So we can not use the theorem because it's too general for our case. Thank you for the clarification!