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Math Help - Riemann Integrability of 1/f

  1. #1
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    Riemann Integrability of 1/f

    I am studying for a qualifying exam and have come across this problem from a past exam. I would like to know how to complete it if anyone can help.

    We have f\colon [0,1] \to \mathbb{R} given as a Riemann integrable function whose values are all greater than 1, and must show that 1/f is also Riemann integrable.

    I have shown the case when the original f is bounded, as Riemann integrability implies continuity almost everywhere, which carries to 1/f (which is also bounded) and is hence Riemann integrable.

    But I cannot reckon the situation involving an unbounded f without appealing to the ugly partition definition of Riemann integrability. Is there a nicer way? I am inclined to think so because it's a qual problem designed to be completely solved in twenty minutes.
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  2. #2
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    If f is bounded below by 1, then 1/f is bounded above by 1. If in addition, f's discontinuities has measure zero, then the same must be true for 1/f, hence 1/f is Riemann integrable. So what if f is unbounded? If f goes to infinity somewhere, that just means 1/f goes to zero somewhere.

    This is just off the top of my head.
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  3. #3
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    Thanks for replying. Unfortunately I do not see how that conclusion guarantees that 1/f behaves suitably for integrability. For what it's worth, the domain was explicitly given as [0,1]. Of course, given continuity, we are furnished with a bound, but in the noncontinuous case perhaps there is another application of compactness that will complete the proof (note this was necessary when showing the result using the definition of Riemann integrability-- I was just hoping for something nicer).
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  4. #4
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    Never mind. Apparently Riemann integrability is only defined for bounded functions to begin with, so f can be assumed bounded from the hypothesis. I have been working so much with the potential pathological behavior of Lebesgue integrable functions that I completely forgot this convenient little restriction. I already handled the bounded case, so the problem is solved. It's amazing how many times pulling out my old undergrad text (RIP W. Rudin) has rescued me. It was never that easy to read before. Thanks again Ackbeet.
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