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Math Help - Discontinuity and Integral

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    Discontinuity and Integral

    Hi,
    I am really confused about discontinuous points and how it affects the integrability. We proved in the lecture that if f is bounded and contunuous except for finite numbers of discontinuous points at [a,b] then f is integrable at [a,b]. The lecturer proved this theorem with induction on the discontinuous points which troubles me a lot. I either don't get it because I am looking at it at the wrong way. When he does induction he says n=0, in other words when we dont have discontinuous points. then he assumes for n and proves for n+1. However when you use induction it's for every n that is natural number so this proof implies that if the function had infinite number of discontinuous points it would still be true since it's for every n? What am I not understanding here?

    Or does he mean, for any given n, he proves it and he says that had the function had n+1 number of discontinuous points it would still be true?

    And what is the reason that if the function has infinite amount of number discontinuous points it's not integrable? As I understand , it would be a problem with the partition. In other words if the function has infinite amount of number discontinuous points then no matter what partition we choose there at least exists two consecutive points in the partition that has discontinuous points. Is my logic correct?

    Thanks
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    Re: Discontinuity and Integral

    I actually found out that riemann integral is integrable at [0,1] so the second thing I said is incorrect.
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    Re: Discontinuity and Integral

    Quote Originally Posted by davidciprut View Post
    Hi,
    I am really confused about discontinuous points and how it affects the integrability. We proved in the lecture that if f is bounded and contunuous except for finite numbers of discontinuous points at [a,b] then f is integrable at [a,b]. The lecturer proved this theorem with induction on the discontinuous points which troubles me a lot. I either don't get it because I am looking at it at the wrong way. When he does induction he says n=0, in other words when we dont have discontinuous points. then he assumes for n and proves for n+1. However when you use induction it's for every n that is natural number so this proof implies that if the function had infinite number of discontinuous points it would still be true since it's for every n? What am I not understanding here?
    Apparently it is induction that you are misunderstanding. If you prove a statement is true for n= 0 and prove "if the statement is true for n= k, then it is true for n= k+1", you have proven, by induction, that it is true for every finite n. You have NOT proven the statement is "true for infinity" (whatever that might mean).

    Or does he mean, for any given n, he proves it and he says that had the function had n+1 number of discontinuous points it would still be true?

    And what is the reason that if the function has infinite amount of number discontinuous points it's not integrable? As I understand , it would be a problem with the partition. In other words if the function has infinite amount of number discontinuous points then no matter what partition we choose there at least exists two consecutive points in the partition that has discontinuous points. Is my logic correct?

    Thanks
    I cannot understand your "logic" enough to say whether it is correct or not. It is certainly NOT correct to say that, even with an infinite number of discontinuities, any of the partition point must be points of discontinuity.
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    Re: Discontinuity and Integral

    Quote Originally Posted by davidciprut View Post
    I actually found out that riemann integral is integrable at [0,1] so the second thing I said is incorrect.
    ??? The Riemann integral of what function?
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    Re: Discontinuity and Integral

    Ok , now I understand. So the induction proves that for any n, discontinuous points we have ,it doesn't affect the integrability of the function. Because the n is a finite numbers.

    You misunderstood what I was saying. I didn't mean that any partition we choose , the points will be discontinuous.
    I am trying to understand how discontinuity affect integrability. Before I explain again I will give an example to express what I want to say.
    For example when they gave us to calculate integrals with Darboux sum, f(x)={1 when x is in [0,1), 3 when x=1 , 2 when x is in (1,2]. So we proved that f is integrable. However when chosing the partition we couldn't chose x=1 as a part of the partition otherwise we wouldn't be able to calculate it. It was problematic. So we chose the partition P={0,1-1/n,1+1/n,2} then we would calculate it and take the limit when n goes to infinity.

    So my point is if there is a function with infinite discontinuous points, any partition that we would chose there exists 2 consecutive and and continuous points which would contain discontinuous pointS. Which means I would have to chose a better partition. In other words I would have to add points in between of every discontinuous points in that interval in my new partition. however even in that partition, between the points we added there exists an interval that has discontinuous points.In other words , because it has infinite discontinuous points I wouldn't be able to create a partition that has a point in between every discontinuous points. Therefore we wouldn't be able to calculate the integral. That's where my ''theory'' came from. However I was wrong because I have a counter example. And that is the riemann function. But what distinguishes this function from other non integrable function with infinite discontinuous points is that this function's discontinious points are countable, which means it's a null set. And therefore it's integrable.

    I am just trying to understand how discontinuous points affect integrability. But I was wrong. Infinite number of discontinuous points don't mean that the function is not integrable.

    So what can we say about discontinuous points and its affect and its geometric interpretations. I will try to think about it.

    Thank you for the feedback. Now I understood what the lecturer did with the induction.

    If there is an information that you can give me about discontinuity and integrability I am happy to hear it. And thank you for the feedback. Much appreciated!
    Last edited by davidciprut; April 24th 2014 at 06:39 AM.
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    Re: Discontinuity and Integral

    I think what the lecturer means is that if it's true for n discontinuities it's true for n+1 discontinuities, then it's true till you run out of discontinuities, ie, it's true for all the finite number of discontinuities of the bounded function.

    In general, an inductive proof can finish before n reaches infinity. The proof applies to an assertion involving n for all values of n for which the assertion applies.

    davidcipruit asks interesting questions. I like the way he thinks.

    Edit: Example. Suppose you have an assertion for n=1 to 1million, but you don't want to prove it for each n. If true for n implies true for n+1, then it's true for all million assertions.
    Last edited by Hartlw; April 24th 2014 at 10:24 AM.
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    Re: Discontinuity and Integral

    Quote Originally Posted by davidciprut View Post
    If there is an information that you can give me about discontinuity and integrability I am happy to hear it. And thank you for the feedback. Much appreciated!
    A function can be highly discontinuous and still be integrable.

    Define $f$ on $[0,1]$ as $f(x) = \left\{ {\begin{array}{rl} {0,}&{x = 0} \\ {{n^{ - 1}},}&{{{(n + 1)}^{ - 1}} < x \le {n^{ - 1}}} \end{array}} \right.$

    That function has infinity many points of discontinuity but is still integrable. Can you prove it?
    What is its value?
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    Re: Discontinuity and Integral

    The function makes no sense. What is n? Is f(x) single valued on the interval? Is it defined for all x on the interval? Can you show that?
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    Re: Discontinuity and Integral

    Quote Originally Posted by Hartlw View Post
    The function makes no sense. What is n? Is f(x) single valued on the interval? Is it defined for all x on the interval? Can you show that?
    If you answer advanced question, then be prepared for fairly standard notation.
    Anyone who have worked with integrals can read that example.
    It is from Belding & Mitchell's Foundations of Analysis.
    That function in the standard literature is known as an infinite descending staircase.
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    Re: Discontinuity and Integral

    What's the point of offering up an undefined function and then giving an inaccessible reference? How does that address the OP?

    "infinite descending staircase" doesn't appear in google.

    I have a source that refutes your assertion but it's too advanced for you.
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    Re: Discontinuity and Integral

    Quote Originally Posted by Plato View Post
    A function can be highly discontinuous and still be integrable.

    Define $f$ on $[0,1]$ as $f(x) = \left\{ {\begin{array}{rl} {0,}&{x = 0} \\ {{n^{ - 1}},}&{{{(n + 1)}^{ - 1}} < x \le {n^{ - 1}}} \end{array}} \right.$

    That function has infinity many points of discontinuity but is still integrable. Can you prove it?
    What is its value?
    I get 2/3.
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    Re: Discontinuity and Integral

    It looks like the function is defined for lim n-> infinity.

    It is not defined for irrational numbers on the interval so it does not satisfy requirement of definition of integral.

    How did you get 2/3?

    The posts are legitimate if:

    1) You can show f is defined for irrational numbers.
    2) You can show how you got 2/3.

    Otherwise its all blah blah.
    Last edited by Hartlw; April 24th 2014 at 12:34 PM.
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    Re: Discontinuity and Integral

    Plato and Hartlw, firstly I wanted to say thanks for the feedback, I appreciate that you even take your time to reply. But... With all due respect, this is not a competition or showing who is better in math. We are all here to learn, well, at least I am. So please stop insulting each other because it becomes really uncomfortable.
    This is a general problem in the forum, that people are responding each other like they are insulting one another or they are competing for reasons that I can't understand. Let's just respect each other and learn from each other. I am pretty sure the reason that the forum was established was to discuss math subjects rather than competing... I don't mean to offend anyone.

    Let's go back to the subject.
    Plato, The function that you gave is really interesting. I will first try to understand the function then calculate it and write an answer for that. And hopefully you can give me a feedback to my answer. Then I will think why it's integrable. And come to a conclusion.

    Hartlw, my trouble with the example is , let's say P(n) is a property. And we want to prove that from 1 to million all n that is natural number off course , P(n) is true. But the way the induction goes is you prove the base case and than you say let's assume that it is true for any n than you prove that it's true for n+1, (1<n<1000000 , less and equal) .Let's assume we prove that it is true for n+1. But then you will get 1,000,001 is also true, because the last number that you can give is million. and therefore 1,000,002 is true etc..
    So when we prove with induction it's for all natural numbers as I see it. I like the analogy with dominoes in induction. Because that's what it is , you prove for k, then you assume for n and prove for n+1 and therefore for all n>k or n=k it is true. With you example you meant, if I wanted to prove a property that it is true for 1, to million I would still use induction?

    But now I get what the instructer meant. He meant that for any number n of discontinuous points the function would still be ingtegrable in that interval. Which means it's finite. I looked at it at the wrong way at first. Induction says that the property is true for all n>k or n=k, (k=base case) but it doesn't mean that it has infinite discontinuous points. I don't know why I got so confused with it.
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    Re: Discontinuity and Integral

    Quote Originally Posted by romsek View Post
    I get 2/3.
    That is close.
    $\sum\limits_{n = 1}^\infty {\dfrac{1}{{{n^3} + {n^2}}}} $

    $\{0\}\cup\{n^{-1} :n\in\mathbb{Z}^+\}$ is an infinite partition of $[0,1]$, from right to left.
    The function is a step function. Step functions are fundamental in integral studies.
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    Re: Discontinuity and Integral

    Quote Originally Posted by Hartlw View Post
    It looks like the function is defined for lim n-> infinity.

    It is not defined for irrational numbers on the interval so it does not satisfy requirement of definition of integral.

    How did you get 2/3?

    The posts are legitimate if:

    1) You can show f is defined for irrational numbers.
    2) You can show how you got 2/3.

    Otherwise its all blah blah.
    it's just a series of steps .. what does irrational numbers have to do with anything?

    I did screw it up. I read it for some reason as powers of 2. You can see the integral is just an infinite series by partitioning [0,1] for a Riemann integral at the points of discontinuity.

    The correct series is

    $\displaystyle{\sum_{n=1}^\infty}\dfrac {1}{n^2(n+1)}$

    which sums to $\dfrac {\pi^2}{6} - 1$
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