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Math Help - Discontinuity and integration

  1. #1
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    Discontinuity and integration

    Let f(x) = x^{-1/2} if 0 < x < 1, f(x) = 0 otherwise. Let \{r_n\} be an enumeration of rationals, and define g(x) = \sum (1/2^n)f(x-r_n). Show that g is discontinuous at every point and unbounded on every interval.

    Show that if f is a non-negative integrable function, for every \epsilon > 0 there exists E \in M such that m(E) < \infty and \int_{E} f > \int f - \epsilon
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  2. #2
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    Quote Originally Posted by davidmccormick View Post
    Let f(x) = x^{-1/2} if 0 < x < 1, f(x) = 0 otherwise. Let \{r_n\} be an enumeration of rationals, and define g(x) = \sum (1/2^n)f(x-r_n). Show that g is discontinuous at every point and unbounded on every interval.

    Show that if f is a non-negative integrable function, for every \epsilon > 0 there exists E \in M such that m(E) < \infty and \int_{E} f > \int f - \epsilon
    I have no clue about the first one

    For question #2:

    I'm not sure what M is? Is that the domain of the function? Assuming it is

    Let  S_L = {x \in M: f(x) > L }

    Notice that  L*\chi_{S_L} \leq f(x) and thus  \int L*\chi_{S_L} \leq \int f \implies Lm^*(S_L) \leq \int f \implies m^*(S_L) \leq \frac{\int f}{L}

    Since f is integrable, the numerator is bounded and it follows that as L tends to infinity,  m^*(S_L) tends to 0.

    I'll leave it up to you to show that  \int_{S_L} f \rightarrow 0 as  L \rightarrow \infty and so we can find an L large enough such that  \int_{S_L} f < \epsilon and so the set we are looking for is E = M \  S_L

    Note, this used the fact that we can break the integral up into disjoint sets.
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  3. #3
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    Well, since every interval contains a rational, and g blows up at at every rational, g is unbounded in every interval. No idea for discontinuity. That is an absolutely horrible function. Is its graph dense in \mathbb{R}^2?
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