Is the centered maximal function measurable?

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, $x$.

-
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. – mpiktas Jan 4 '11 at 11:22
Continuity of what? Of the function inside the supremum? – user3533 Jan 4 '11 at 11:40
@user3533, as it now stands the $M$ measurability will follow if Radon-Nikodym theorem holds for $\mu$ with respect to $m$. – mpiktas Jan 4 '11 at 13:02