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Riemann integrable functions continuous except on a set of measure zero?

  1. Dec 16, 2009 #1
    Is it true that a function is Riemann integrable on a bounded interval only if it's equal to a continuous function almost everywhere? I'd imagine this is the case, given the Riemann-Lebesgue lemma, which says that a function is RI iff its set of discontinuities has measure zero. (So the "continuous function" is then just f restricted to the complement of its set of discontinuities.) But I might be wrong. Help?
     
  2. jcsd
  3. Dec 16, 2009 #2
    I've just discovered this is incorrect. Consider the function

    [tex]
    f(x) = \begin{cases}
    1 & \text{ if } 0\leq x \leq 1/2\\
    0 & \text{ if } 1/2 < x \leq 1
    \end{cases}
    [/tex]

    Then [tex]f[/tex] is continuous almost everywhere, but it cannot be equal to a continuous function almost everywhere by an argument involving inverse images of open sets, etc. Bummer.
     
  4. Dec 16, 2009 #3

    Hurkyl

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    However, for every e>0, there exists a set A of measure less than e and a continuous function g such that, outside of A, f and g are equal.
     
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