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Math Help - Is this function integrable?

  1. #1
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    Is this function integrable?

    If f is a bounded function from [a,b] into the real numbers that is
    integrable on [a,b], then for any c such that a < c < b, is f
    integrable on [a,c] and is f integrable on [c,b]?

    I want to say no, since if I pick f(x)= \frac {1}{x} , then f is integrable from [-2,2] but not if you move the boundary to 0.

    Is that right? Thanks.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Question

    I have also observed same question in my book..
    Last edited by ADARSH; November 5th 2008 at 08:03 AM.
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  3. #3
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    So I guess this theorem is true then?

    Here is my proof so far:

    Let  \epsilon > 0 be given. Since f is integrable, then there exists a partition P= \{ x_0 = a, x_1, x_2 , . . . , x_{N-1}, x_N = b \} on [a,b] such that U(f,P)-L(f,P)< \epsilon

    Define M_j=sup \{ f(x) : x \in [x_j,x_{j+1} \} and m_j=inf \{ f(x) : x \in [x_j,x_{j+1} \} , so we have U(f,P)-L(f,P)= \sum ^{N-1}_{j=0}(M_j)(x_{j+1}-x_{j}) - \sum ^{N-1}_{j=0}(m_j)(x_{j+1}-x_{j}) = \sum ^{N-1}_{j=0}(M_j-m_j)(x_{j+1}-x_{j})

    Now let c=x_k \ \ \ \ \ k \in (0,N) , then U(f,P)-L(f,P)= \sum ^{N-1}_{j=0}(M_j-m_j)(x_{j+1}-x_{j}) =
     \sum ^{k}_{j=0}(M_j-m_j)(x_{j+1}-x_{j}) + \sum ^{N-1}_{j=k+1}(M_j-m_j)(x_{j+1}-x_{j}) < \epsilon

    Define P_1 = \{ x_0, x_1 , . . . , x_k = c \} and  P_2 = \{ c = x_k, x_{k+1}, . . . , x_N \}

    But we know that
     \sum ^{k}_{j=0}(M_j-m_j)(x_{j+1}-x_{j}) + \sum ^{N-1}_{j=k+1}(M_j-m_j)(x_{j+1}-x_{j}) =
     \sum ^{k}_{j=0}(M_j)(x_{j+1}-x_{j}) - \sum ^{k}_{j=0}(m_j)(x_{j+1}-x_{j}) +  \sum ^{N-1}_{j=k+1}(M_j)(x_{j+1}-x_{j}) - \sum ^{N-1}_{j=k+1}(m_j)(x_{j+1}-x_{j}) =
    U(P_1,f)-L(P_1,f)+U(P_2,f)-U(P_2,f) < \epsilon

    I need to show that  U(P_1,f)-L(P_1,f) < \epsilon and  U(P_2,f)-U(P_2,f) < \epsilon

    Any hints? Thanks.
    Last edited by tttcomrader; October 23rd 2008 at 10:13 AM.
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  4. #4
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    Note that the theorem says that the function is bounded.
    I donít know what definition of integrable you are using.
    However, if you are using some variation of subdivisions we can always include c as an endpoint in a refinement of the subdivision.
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  5. #5
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    Quote Originally Posted by tttcomrader View Post
    If f is a bounded function from [a,b] into the real numbers that is
    integrable on [a,b], then for any c such that a < c < b, is f
    integrable on [a,c] and is f integrable on [c,b]?
    If f is integrable on [a,b] then f is integrable on [c,d]\subset [a,b].
    This is a known result.
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