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Thread: 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 : $\displaystyle
    \int ^a_b f= \int ^a_b g
    $
    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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    My advice would be to split the interval up into the following parts:
    $\displaystyle [a,c-\epsilon],[c-\epsilon,c+\epsilon],[c+\epsilon,b]$.

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

    $\displaystyle \int_a^b(f(x)-g(x))dx\le 2\epsilon|F+G|$
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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 $\displaystyle c $ in the problem. It's just saying that for $\displaystyle c$ 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 $\displaystyle \int f=\int g $ given that $\displaystyle f(x)=g(x)$ except at $\displaystyle x=c$
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    Quote Originally Posted by putnam120 View Post
    My advice would be to split the interval up into the following parts:
    $\displaystyle [a,c-\epsilon],[c-\epsilon,c+\epsilon],[c+\epsilon,b]$.

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

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

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

    $\displaystyle \int_{c-\epsilon}^{c+\epsilon}|f(x)-g(x)|dx\leq M\int_{c-\epsilon}^{c+\epsilon}dx=2M\epsilon$
    And you make $\displaystyle \epsilon$ 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 $\displaystyle h=f-g$, then $\displaystyle h=\left\{\begin{array}{lr}k&:x=c\\0&:x\neq c\end{array}\right\}$

    Calculate the lower Riemann sum of $\displaystyle h$. Let $\displaystyle a=x_1<x_2<...<x_n=b$ be a partition of $\displaystyle [a,b]$ and let $\displaystyle [x_{i-1},x_i]$ be the segment containing $\displaystyle c$. $\displaystyle \min h(x)$ on $\displaystyle [x_{i-1},x_i] = 0$, because it contains a point that is not $\displaystyle c$, 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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    Yes sorry for the ambiguity.
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    We can prove something stronger. If $\displaystyle f$ is integrable and $\displaystyle g$ is a function that differs at countably many points distinct from $\displaystyle f$ such that they converge then $\displaystyle g$ is integrable with same value as $\displaystyle f$. 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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