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Math Help - Prove a function is integrable

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    Prove a function is integrable

    Q: Suppose that the function h:[a,b]\mapto is integrable and that the function g:[a,b]\mapto\mathbb{R} satisfies the equality h(x)=g(x) except at the single point z\epsilon[a,b]. Prove that g is integrable and \int^b_{a}g=\int^b_{a}h.
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    Quote Originally Posted by tttcomrader View Post
    Q: Suppose that the function h:[a,b]\mapto is integrable and that the function g:[a,b]\mapto\mathbb{R} satisfies the equality h(x)=g(x) except at the single point z\epsilon[a,b]. Prove that g is integrable and \int^b_{a}g=\int^b_{a}h.
    Since h is integrable on [a,b] it is discontinous at countably many points, by removing one point from the interval the function h becomes g which is still discontinous at countably many points. Thus, h is integrable.
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    Discontinuous? How do you know h(x) is discontinuous?
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    You don't know that it is definately discontinuous, but because it is integrable then it can only be discontinuous at most at a finite number of places.
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    I don't know such a theorem yet, the professor had not gone over that with us. Is there another way to prove this problem?

    Actually, is there are proof for that theorem that you can share with me?
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    If you consider the integral from a to b on h(x) then that would be equal to the integrals of a to c, added to c to b. This is quite straight forward, and let this one differing point be c.

    Then

    Integral from a to b of h(x)

    = Integral from a to c-dx of h(x) added to integral from c+dx to b of h(x)
    [as dx -> 0]

    = integral from a to b of g(x)

    Where g(x) will then only differ at the point c.

    That's pretty much a hand waving arguement I think though.
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    Actually it can be discontinuous a set of measure zero.
    But I suspect the person who set this question expected something much less powerful. Here is a quick proof of what I think is meant by this problem.

    Suppose that z in an interior point of (a,b). There is a e>0 such that (z-e,z+e) is a subset of [a,b]. Now by additively g is integrable on each of these subintervals: [a,z-e], [z-e,z+e] and [z+e,b]. Only on [z-e,z+e] have you changed the definition g so we can control the variation that interval, after all its length is only 2e. If we make e small enough the sum remains the same.
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    Thanks, I used this theorem to complete this problem. I will show it to the professor tomorrow and see if that is what he wants.

    Thanks again!
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    Quote Originally Posted by TheBrain View Post
    it can only be discontinuous at most at a finite number of places.
    It can still be infinitely many points also.
    For example, if "f" is defined on [a,b] and discontinous on all rational points and continous on all irrational points (assuming it is possible) then it is still integrable.
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    So a function can be integrable even in that case? If there a proof for it? I really want to see it.

    Thanks.

    Man, to be honest, lately I can't solve a single problem without helps from this forum or the professor. Am I reaching my limit?
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    Quote Originally Posted by tttcomrader View Post
    So a function can be integrable even in that case? If there a proof for it? I really want to see it.

    Thanks.

    Man, to be honest, lately I can't solve a single problem without helps from this forum or the professor. Am I reaching my limit?
    Keep at it. Sometimes you just get to a point where you need to be shown how to do many things, but you get past it at some stage. Once you get to a point where you've had enough "exposure" you'll be able to do all the things someone once had to show you and then YOU can "wow" the more inexperienced students!

    -Dan
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    Quote Originally Posted by tttcomrader View Post
    Am I reaching my limit?
    No, for example, you where able to understand about what I said about rational discontinuities. And furthermore were amazed by the statement. If you were at the limit you would not have understood what I said nor would have found the statement amazing. When you start to reach a point in math when you are amazed by some statements (for example, there exists a function which is continous everywhere and differenciable nowhere!!) then you are going good.
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    Thanks, guys, that means a lot, the other day the professor was saying that I'm gotta need to "click" at some point before I can serioiusly consider going for a PHD in Real Analysis.

    Let's hope I "click" soon.

    KK
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