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Math Help - How do you show that a monotone increasing real function on a closed interval [a, b]

  1. #1
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    How do you show that a monotone increasing real function on a closed interval [a, b]

    Hello Everyone,
    I have two questions on measure theory:
    1)
    How do you show that a monotone increasing real function on a closed interval [a, b] is Lebesgue measurable function.
    2)
    Suppose A is a subset of the closed interval [0,1] which is dense. (That is, the closure of A is all of [0,1].) suppose U is an open set with A⊆U. Must it be true that m(U) ≥ 1?

    Thank you!
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  2. #2
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    Quote Originally Posted by Kelles View Post
    Hello Everyone,
    I have two questions on measure theory:
    1)
    How do you show that a monotone increasing real function on a closed interval [a, b] is Lebesgue measurable function.
    It happens to be integrable. First note f is bounded since f(a) \leq f \leq f(b). (Also assume that f(a)\not = f(b) because otherwise [tex]f[/math[ would be a constant function, and there is nothing to prove).

    Let U(f,P) be the upper Darboux sum with respect to partition P of [a,b] and L(f,P) be the lower Darboux sum with respect to partition P. To show this function is integrable we need to show for any \epsilon >0 we can find a partition P such that U(f,P)-L(f,P) < \epsilon. Note, since f is monotone it means U(f,P) - L(f,P) = \sum_{k=1}^n [f(x_k)-f(x_{k-1})](x_k - x_{k-1}). Let \delta be \min_{1\leq k \leq n} (x_k - x_{k-1}). Then we see that U(f,P)-L(f,P) \leq \sum_{k=1}^n [f(x_k)-f(x_{k-1})]\delta = [f(a)-f(b)]\delta. Thus, if \delta < \epsilon/[f(b)-f(a)] then U(f,P)-L(f,P)<\epsilon. This means given any \epsilon > 0 we can create equal subdivision of the interval until the interval length \leq \delta and then we are have U(f,P)-L(f,P)<\epsilon. Thus f is integrable.
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  3. #3
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    Is integrability of a function means that the function Lebesgue measurable ?

    If a function is integrable, does it mean that the function Lebesgue measurable ?

    Thank you very much.
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  4. #4
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    Quote Originally Posted by Kelles View Post
    If a function is integrable, does it mean that the function Lebesgue measurable ?
    I would assume so. I never studied measure theory, but from I understand Lebesque integrability is a generalization of Riemann integrability. Thus, if a function is Riemann integrable then it is Lebesque integrable.
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