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Thread: Measurable Functions

  1. #1
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    Measurable Functions

    Let f:[0,1] --> (0,$\displaystyle \infty $) be a (Lebesgue) integrable function
    and define g to be: $\displaystyle 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: $\displaystyle f^{-1}(O) = U$ is a measurable set.

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

    So it remains to show that U measurable implies $\displaystyle \sqrt{U}$ 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 $\displaystyle \sqrt{\,}$ is continuous on [0,\infty) and so it is measurable, hence $\displaystyle \sqrt{\,} \circ f$ is measurable.
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