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If $ \mu $ is a finite Borel measure on $R^n $ and if $B_1(x)$ denotes an open ball of radius 1 centered at x, is it true that for compact subset $K$ of $R^n$ there is a point $x_0$ in $K$ such that $\mu(B_1(x_0)) = inf \{ \mu( B_1(x))| x \in K \} $? what if we replace inf by sup?

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Is $B_1$ an open or a closed ball? –  Nate Eldredge Nov 7 '12 at 4:09
    
@NateEldredge open ball.. Sorry for not accurating that –  Detectives Nov 7 '12 at 4:13
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up vote 0 down vote accepted

I'm assuming $\mu$ is a positive measure. Let $\chi_x$ be the indicator function of $B_1(x)$. If $x_n \to x$, $\liminf_{n \to \infty} \chi_{x_n}(y) \ge \chi_x(y)$ so by Fatou's lemma, $\liminf_{n \to \infty} \mu(B_1(x_n)) \ge \int \liminf_{n \to \infty} \chi_{x_n}(y)\ d\mu(y) \ge \mu(B_1(x))$. Thus $x \to \mu(B_1(x))$ is lower semicontinuous. A lower semicontinuous function on a compact set attains its minimum.

If you replace inf by sup, consider $n=1$ where $\mu$ is the sum of Lebesgue measure on $[0,2]$ and a unit point mass at $0$.

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