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Math Help - Riemann Integrable Function

  1. #1
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    Riemann Integrable Function

    Suppose f and g are Riemann Integrable Functions on[a,b] and f(x)=g(x) if x not equal to c.
    show that : <br />
\int ^a_b f= \int ^a_b g<br />
    Last edited by math43; May 5th 2009 at 08:38 AM.
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  2. #2
    Super Member fardeen_gen's Avatar
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    I googled away to glory and maybe this will help:
    Darboux theory

    EDIT: Can the experienced answerers shed light on this problem?
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  3. #3
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    My advice would be to split the interval up into the following parts:
    [a,c-\epsilon],[c-\epsilon,c+\epsilon],[c+\epsilon,b].

    The integrals on the last 2 parts are the same. For the region in the middle since f,g are assumed Riemann integrable they are bounded. So say |f|\le F, |g|\le G. Thus you have that

    \int_a^b(f(x)-g(x))dx\le 2\epsilon|F+G|
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  4. #4
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    where the $c$ coming from?
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  5. #5
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    Quote Originally Posted by hkito View Post
    where the $c$ coming from?
    Well c was given in the problem!
    Do you understand the question?
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  6. #6
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    Well I don't see any c in the problem. It's just saying that for c not equal to x. For me it's not clear enough.
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  7. #7
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    It says that as long as x does not equal c you know that f(x)=g(x). However, you are not guaranteed that f(c)=g(c).
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  8. #8
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    Okay,
    so the problem is show that \int f=\int g given that f(x)=g(x) except at x=c
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    Quote Originally Posted by putnam120 View Post
    My advice would be to split the interval up into the following parts:
    [a,c-\epsilon],[c-\epsilon,c+\epsilon],[c+\epsilon,b].

    The integrals on the last 2 parts are the same. For the region in the middle since f,g are assumed Riemann integrable they are bounded. So say |f|\le F, |g|\le G. Thus you have that

    \int_a^b(f(x)-g(x))dx\le 2\epsilon|F+G|

    Since f(x)=g(x) on [a,c-\epsilon]\cup[c+\epsilon]
    you just need to show that \int_{c-\epsilon}^{c+\epsilon}f=\int_{c-\epsilon}^{c+\epsilon}g
    For that you since you know that f and g are integrable, so f-g is bounded , say by some  M.

    \int_{c-\epsilon}^{c+\epsilon}|f(x)-g(x)|dx\leq M\int_{c-\epsilon}^{c+\epsilon}dx=2M\epsilon
    And you make \epsilon go to zero
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  10. #10
    Super Member redsoxfan325's Avatar
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    There's a really handy theorem in my book that says that "if a function has finitely many points of discontinuity, then it is Riemann-integrable". That's very nice for this problem, because if h=f-g, then h=\left\{\begin{array}{lr}k&:x=c\\0&:x\neq c\end{array}\right\}

    Calculate the lower Riemann sum of h. Let a=x_1<x_2<...<x_n=b be a partition of [a,b] and let [x_{i-1},x_i] be the segment containing c. \min h(x) on [x_{i-1},x_i] = 0, because it contains a point that is not c, so therefore the lower Riemann sum and the whole integral is zero.
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    Quote Originally Posted by redsoxfan325 View Post
    There's a really handy theorem in my book that says that "if a function has finitely many points of discontinuity, then it is Riemann-integrable".
    If I remember correctly this can be extended to saying that if a function is discontinuous on a set of measure 0 then it is Riemann-integrable.
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  12. #12
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    Quote Originally Posted by putnam120 View Post
    If I remember correctly this can be extended to saying that if a function is discontinuous on a set of measure 0 then it is Riemann-integrable.
    (If this is true - it seems to be, but I don't have a proof lol) then it has to be with Lebesgue measure, isn't it ?
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  13. #13
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    Yes sorry for the ambiguity.
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  14. #14
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    We can prove something stronger. If f is integrable and g is a function that differs at countably many points distinct from f such that they converge then g is integrable with same value as f. The nice thing is that the proof is similar as in the finite case. Because when you have convergence it means that within a a small enough neighborhood of the limit all the points lie there. So you form a partition with thin enough rectangles around these values.
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