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Let $(X,d)$ be a metric space and $\mu$ a probability measure. Let $f(x)=\mu(B(x,r))$. Is this map upper semi continuous?

I have some other assumptions since this is part of a larger proof but I'm wondering if this question is trivial or do I need to use some of my other assumptions from the proof.

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First of all it depends on $\mu$. For example a probability measure can be discrete and in this case the function will be upper semi-continuous at some points. – Nikita Evseev Sep 23 '12 at 1:40
up vote 0 down vote accepted

Assuming $B(x,r)$ denotes the open ball about $x$ of radius $r$, this map is lower semicontinuous but need not be upper semicontinuous.

Note that if $d(x,y) < r$ then $d(x_n,y) < r$ for sufficiently large $n$, so $y \in B(x_n, r)$. This means $\liminf_{n \to \infty} 1_{B(x_n, r)}(y) \ge 1_{B(x, r)}(y)$. Now we can use Fatou's lemma to see $$\liminf_{n \to \infty} \int 1_{B(x_n, r)}\,d\mu \ge \int \liminf_{n \to \infty} 1_{B(x_n, r)}\,d\mu \ge \int 1_{B(x,r)}\,d\mu.$$ That is, $\liminf_{n \to \infty} f(x_n) \ge f(x)$.

To see it need not be upper semicontinuous, take $X = \mathbb{R}$, $\mu = \delta_0$, $x_n = 1 - 1/n$, $x=1$, $r=1$. Then $\mu(B(x_n, r)) = 1$ for all $n$ but $\mu(B(x,r)) = 0$.

In fact, $f$ is upper semicontinuous (hence continuous) at $x$ if and only if $\mu(\partial B(x,r)) = 0$.

If $B(x,r)$ was meant to be the closed ball, then $f$ is indeed upper semicontinuous. The proof is essentially the same as above.

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