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Let $(X,\parallel \cdot \parallel)$ be Banach and $$\mathcal{BC}(X)=\{A\subset X\colon A \text{ is closed, bounded and non-empty}\}.$$ The natural metric on this space is the Hausdorff distance $d_H$ (see

Let $C_x(r)$ be the closed ball around $x\in X$ with radius $r$. How can I show that the map $f\colon X\to\mathcal{BC}(X)$ where $f(x)=C_x(r)$ is continuous w.r.t. $d_H$?

And is the map $g\colon X\to\mathbf{R}$ with $g(x)=\mu(C_x(R))$ Borel measurable for a Borel measure $\mu$?

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Let $x_1,x_2\in X$ and $r_1,r_2>0$ with $\|x_1-x_2\|\leq \varepsilon/2$ and $|r_1-r_2|\leq \varepsilon/2$. We will bound the distance between the two closed balls $C_1:=C_{x_1}(r_1)$ and $C_2:=C_{x_2}(r_2)$.

Let $w\in C_1$. If $w\in C_2$, then $\inf_{z\in B_2} \|w-z\|=0$. Let's assume that $w\notin C_2$, that is, $\|w-x_2\| > r_2$. Then we have $$r_2\leq \|w-x_2\|\leq \|w-x_1\|+\|x_1-x_2\|\leq r_1+\varepsilon/2 \leq r_2+\varepsilon.$$

Letting $w^*=x_2+{r_2\over \|w-x_2\|} (w-x_2)$ we have $w^*\in B_2$ and $\|w-w^*\|\leq \|w-x_2\|-r_2\leq \varepsilon.$ so $\inf_{z\in B_2} \|w-z\|\leq \varepsilon.$

Combining the two cases, and taking the supremum over $w\in C_1$, we deduce that $\sup_{w\in C_1}\inf_{z\in C_2} \|w-z\|\leq \varepsilon$. By symmetry we get $d_H(C_1,C_2)\leq \varepsilon$.

From this you can show that $(x,r)\mapsto B_x(r)$ is jointly continuous from $X\times (0,\infty)$ to $\mathcal{BC}(X)$. This is a bit stronger than the result you wanted.

Since $C_x(R)=x+C_0(R)$, your second question is answered here: How to prove that $f (x) = \mu (x + B)$ is measurable? assuming that $\mu$ is a non-negative, finite measure.

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I just wanted to let you know that the question is now on MO (posted by the same person or from the same computer) – t.b. Mar 30 '12 at 20:03
Thanks. Maybe the OP will get a better answer there. Mine is the first thing that popped into my head. – Byron Schmuland Mar 31 '12 at 13:08
As noted on MO, this turned out to be from a homework assignment – user16299 May 11 '12 at 0:46

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