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Math Help - Measurable Functions

  1. #1
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    Measurable Functions

    Let f:[0,1] --> (0,  \infty ) be a (Lebesgue) integrable function
    and define g to be:  g(x) = \sqrt{f(x)} . Show that g is integrable.

    Answer:

    If g is a measurable function, I can easily show that it's integrable. It's my reasoning that g is measurable that may be suspect, I claim that:

    Let O be an open set. We know that:  f^{-1}(O) = U is a measurable set.

    Thus,  g^{-1}(O) = \sqrt{U}, where  \sqrt{U} = { \sqrt{x}: x \in U}

    So it remains to show that U measurable implies  \sqrt{U} is measurable, which i was able to show but is a bit long, so I'll skip it.

    Does this seem right, though? Or did i make it way too complicated?
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  2. #2
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    Quote Originally Posted by southprkfan1 View Post
    Does this seem right, though? Or did i make it way too complicated?
    I think you did. f is measurable and \sqrt{\,} is continuous on [0,\infty) and so it is measurable, hence  \sqrt{\,} \circ f is measurable.
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