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Edited Question: Let $\mu$ be some measure on $\mathbb{R}^d$ and $m$ the Lebesgue measure. Define the maximal function $$ M(x) = \sup_{r>0} \; \frac{\mu(B(x,r))}{m(B(x,r))} .$$ Why is $M$ a measurable function?

Old Question: In class we defined the "maximal function" similar to how it's defined here, but instead of taking the supremum on all balls containing $x$, we took all balls centered at $x$.

It was then claimed this function is measurable, but I can't see why. I do understand that with the definition given by my wikipedia link, the preimage of $(a,\infty]$ is open and thus the maximal function is measurable. But why is the centered maximal function measurable? (I missed this lesson, and my friends say the professor said it's obvious, so the answer is probably easy).

EDIT: I meant to gives this link to wikipedia, where the maximal function is defined with balls containing, not necessarily centered at, $x$.

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    The definition in wikipedia page uses the balls centered around x. Note that the measurability comes from continuity and then measurability of supremum, so you do not actually need to concern yourself about preimages.2011-01-04
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    Continuity of what? Of the function inside the supremum?2011-01-04
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    So what is your question? The wikipedia page about centered function gives the answer why it is measurable. Is the explanation in wikipedia page not helpful?2011-01-04
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    @user3533, yes continuity of function inside the supremum. The maximal function takes the supremum by one argument of two argument continuous function. Such functions are measurable.2011-01-04
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    My question was: Why is the function inside the supremum continuous? But now I see it. My confusion arised from the fact that in class the definition was not with an integral, but with another measure, i.e: divide the measure of the ball by some measure m by the Lebesgue measure of the ball. Is this function still continuous? Seems to me like it can "jump" if there's a point with nonzero measure.2011-01-04
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    @mpiktas: Edited the question to reflect my current misunderstanding. Thanks for the help so far.2011-01-04
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    @user3533, as it now stands the $M$ measurability will follow if Radon-Nikodym theorem holds for $\mu$ with respect to $m$.2011-01-04
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    Thanks. Will ask the professor if that's what was meant. So a counterexample really does exist if $\mu$ has a point of nonzero measure, right?2011-01-04

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