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Thread: Showing a function is measurable

  1. #1
    Senior Member Dinkydoe's Avatar
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    Showing a function is measurable

    Let $\displaystyle X=(S,d) $ a banach space. I want to show the following: For any borel measure $\displaystyle \mu$, the map $\displaystyle \psi: X\to \mathbb{R}$ given by $\displaystyle x\mapsto \mu(\overline{B}_x(r)) $ is measurable.

    I believe this can be done in a number of ways, one of which is proving that the set $\displaystyle U_c:=\left\{x:\psi(x)\leq c\right\}$ for any $\displaystyle c$ is contained in $\displaystyle \mathcal{B}(X) $(borel sigma-algebra of $\displaystyle X$)

    I think proving this, can not be done without using the fact that $\displaystyle d$ is induced by some norm $\displaystyle \left\|\cdot \right\|$. If we take for example $\displaystyle X= (\mathbb{R}, d_E)$ with Euclidian metric, ($\displaystyle \mu$ lebesgue measure) it is obvious $\displaystyle U_c$ is contained in $\displaystyle \mathcal{B}(X)$ as $\displaystyle U_c = X$ or $\displaystyle U_c = \emptyset$ for any $\displaystyle c$. But for any random metric $\displaystyle d$, this is not obvious...

    A norm-induced metric has (as I recall) 2 extra properties. 1. translation invariance $\displaystyle d(x+a,y+a)=d(x,y).$ and 2. $\displaystyle d(\alpha x,\alpha y) = |\alpha|d(x,y)$

    I hope what I think is correct here...so suppose $\displaystyle \psi(x)= \alpha$ for some $\displaystyle \alpha >0$ then for any $\displaystyle y\in X$ we have $\displaystyle \psi(y)=\alpha$ as well...

    It's just a hunch, i have no clue actually...as I don't know what properties $\displaystyle \mu$ might have. (except for the standard properties)

    Any idea what I'm missing here?

    Thank you

    (good to see u guys back online though...btw: can it be implemented that $ x^2 $ is valid latex-code here, instead of wraps..it's sometimes
    quite bothersome to write tags everywhere :P)
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  2. #2
    Super Member girdav's Avatar
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    Re: Showing a function is measurable

    Can yo use Fubini's theorem?
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