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I have a problem and a solution here please let me know

  1. Feb 24, 2007 #1
    I have a problem here and a solution. Not sure if I am on the right track. Please let me know if this proof is right.

    we are given:f :[a,b]-R, f is integrable and bdd below(f is greater
    than t(i use t instead of delta) for all x belongs to [a,b] ), t is
    greater than 0
    claim: 1/f is integrable.
    solution: f is integrable implies f is bdd - f is bdd above and below
    implies 1/f is bdd below and above - 1/f is bdd-{1}
    f is greater than t and t is greater than 0 for all x
    - f is greater than 0 for all x. - 1/f can not be infinity for all x-
    1/f is continuous.{2}
    from {1} & {2} 1/f is a bdd continuous function .
    hence 1/f is integrable.
  2. jcsd
  3. Feb 24, 2007 #2
    How can you conclude that 1/f is continuous when you don't even know that f is?
  4. Feb 24, 2007 #3
    f might not be bounded. I would consider the integral in two parts: One where f<2t and the other where f>= 2t (say). That is one integral where f is big and one where f is not so big.

    As far as how to split up the interval that way would depend on the kind of integral you are doing. Lebesgue - no problem: If you are only dealing with Riemann - may have to turn to (piecewise) continuity to separate the interval accordingly.
  5. Feb 25, 2007 #4


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    f(x) identically equal to 0 satisfies all the conditions you give but 1/f certainly is not bounded!
  6. Feb 25, 2007 #5

    matt grime

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    Halls, one of the conditions was f(x)>t>0 for all x in [a,b].
  7. Feb 26, 2007 #6
    what is bdd?
  8. Feb 26, 2007 #7


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    murshid islam: bdd= bounded.

    matt grime- I really do need to learn to read, don't I!
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