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Math Help - the composition of a measurable function with a continuous function is measurable

  1. #1
    Junior Member Dark Sun's Avatar
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    the composition of a measurable function with a continuous function is measurable

    Show that if f is a measurable real-valued function and g is a continuous function defined on (-\infty ,\infty ), then g\circ f is measurable.

    Please help! :'(
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  2. #2
    Junior Member Dark Sun's Avatar
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    This is what I have developed in a moment of clarity; Please tell me if I am right or wrong:

    Certainly every continuous function is measurable, since every interval is measurable (P. 67). Therefore, g is measurable.

    Since f is a real-valued function, restrict g from (-\infty ,\infty ) to the co-domain of f, which is a subset of (-\infty ,\infty ). Since f is measurable, the domain of g restricted to the co-domain of f has a measurable domain, that is, the set of all f(x).

    Also, the set of all elements not in the domain of g|_{f(x)} are (-\infty ,\infty )\sim f(x), which must neccessarily be measurable.

    Therefore, since g is measurable and [(-\infty ,\infty )\sim f(x)]\cup f(x)=(-\infty ,\infty ), the domain of g, and both f(x) and (-\infty ,\infty )\sim f(x) are measurable, then by Problem 21.a., the function g restricted to the co-domain of f must be measurable. Then g(f(x))=g\circ f(x) is measurable since for all x, an element in the co-domain of f is defined, and g restricted to these elements is measurable.
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  3. #3
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    Look at the inverse of g\circ f. So f^{-1}\circ g^{-1}(U) is measurable (U measurable) since, g is continuous and thus measurable. (pull backs of open sets by g^{-1} are open. and open sets generate the \sigma-algebra you are working with).
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