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Math Help - Correct proof?

  1. #1
    Junior Member
    Joined
    Nov 2010
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    Correct proof?

    Theorem:
    If f is bounded in [a,b] and the number of discontinuity points of f is finite then f is integrable in [a,b].

    Proof:
    Using induction:
    When there are 0 discontinuity points then f i continuous and therefore f is integrable in [a,b].
    Suppose the claim is true when we have k or less discontinuity points, and we'll prove it for a case where we have k+1 discont. points.
    Let x be a discont. point of f in [a,b]. Let e be such that x is the only discont. point in the interval [x-e,x+e].
    Then there are k or less discont. points in [a,x-e], so according to the inductive assumption f is integrable in [a,x-e]. We can make the same argument for the intervals [x-e,x+e] and [x+e,b], and so since f is integrable in [a,x-e] and [x-e,x+e] and [x+e,b], f is integrable in [a,b].

    Is my proof correct? Thanks
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  2. #2
    Newbie
    Joined
    Jul 2011
    Posts
    10

    Re: Correct proof?

    Quote Originally Posted by durrrrrrrr View Post
    Theorem:
    If f is bounded in [a,b] and the number of discontinuity points of f is finite then f is integrable in [a,b].

    Proof:
    Using induction:
    When there are 0 discontinuity points then f i continuous and therefore f is integrable in [a,b].
    Suppose the claim is true when we have k or less discontinuity points, and we'll prove it for a case where we have k+1 discont. points.
    Let x be a discont. point of f in [a,b]. Let e be such that x is the only discont. point in the interval [x-e,x+e].
    Then there are k or less discont. points in [a,x-e], so according to the inductive assumption f is integrable in [a,x-e]. We can make the same argument for the intervals [x-e,x+e] and [x+e,b], and so since f is integrable in [a,x-e] and [x-e,x+e] and [x+e,b], f is integrable in [a,b].

    Is my proof correct? Thanks

    I think it is correct but I cannot be completely sure
    Nemesis
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