Measuring closed balls 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 http://en.wikipedia.org/wiki/Hausdorff_distance)
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 $\mu$?
 A: 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. 
