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Riemann Lebesgue Theorem

  1. May 8, 2010 #1
    1. The problem statement, all variables and given/known data
    We have a corollary that if f(x) is in the set of Riemann Integrable functions and g(x) is continuous, then g(f(x)) is also a riemann integrable function

    Show that if g(x) is piecewise continuous then this is not true

    2. Relevant equations
    Hint: take f to be a ruler function and g to be a characteristic function

    3. The attempt at a solution

    So piecewise continuous means (intuitively) that the function must be defined separately, but still has no gaps on the x-axis. So if f is riemann integrable, and g is piecewise continuous, then g(f(x))'s discontinuity points are the same as g(x)'s. Now I have to prove that there are uncountable many of them. Don't know where to go with this
  2. jcsd
  3. May 8, 2010 #2


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    Homework Helper

    as a counter example, how about g(x) = 0 if x = 0, and 1 otherwise and f(x) the ruler (thomae's function)
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