Is the function that gives you the measure of the neighborhood Borel? Let $X$ be a compact metric space (with $\epsilon -$balls $B_{\epsilon }$)
and $\mu $ a Borel probability measure.
Let $a,\epsilon >0.$ 
Is the set $\left\{ x\in X:\mu (B_{\epsilon }(x))\geq a\right\} $ Borel?
or in other words is the function $f(x)=\mu (B_{\epsilon }(x))$ Borel?
 A: We shall need notions of product of measures and Fubini's Theorem. 
Once we have a Borel measure on a topological space $X$, then we can define product topology and Borel sets for $X\times X$ and product measure $\mu\otimes\mu$ also on $X\times X$. 
According to Fubini's Theorem, if $F\in L^1(X\times X)$, then $f(x)=\int_X F(x,y)\,d\mu(y)$ is measurable (and in our case Borel measurable). 
So, if we set 
$$
D_\varepsilon=\{(x,y)\in X\times X: \varrho(x,y)<\varepsilon\}, 
$$
and $F(x,y)=\chi_{D_\varepsilon}$, then $f(x)=\mu\big(B_\varepsilon(x)\big)$ is expressed also as
$$
f(x)\int_X \chi_{D_\varepsilon}(x,y)\,d\mu(y),
$$
and hence it is Borel measurable.
A: Yiorgos is of course perfectly right.
Here is another proof that avoids the use of Fubini's theorem.
Set $A:=\{ (x,y)\in X\times X;\; d(x,y)<\varepsilon\}$.  Then $A$ is an open subset of $X\times X$, so its indicator function $\mathbf 1_A$ is the pointwise limit of an increasing sequence of nonnegative continuous functions $(\phi_n)$ . Explicitely, one may take a sequence of closed sets $(F_n)$ such that $\bigcup_nF_n=A$, choose for each $n$ a continuous function $\psi_n$ such that $0\leq \psi_n\leq 1$, $\psi_n\equiv 1$ on $F_n$ and $\psi_n\equiv 0$ on $(X\times X)\setminus A$, and set $\phi_n:=\max(\psi_1,\dots, \psi_n)$.
Now, since $$f(x)=\mu(B_\varepsilon (x))=\int_{X\times X} \mathbf 1_A(x,y)\, d\mu(y),$$ it follows from the dominated convergence theorem (or the monotone convergence theorem) that 
$$ f(x)=\lim_{n\to\infty} \int_{X\times X} \phi_n(x,y)\, d\mu(y)=\lim_{n\to\infty} f_n(x)\, .$$
The functions $f_n$ are continuous on $X$ (again by the dominated convergence theorem); so $f$ is a Borel function, in fact a lower semi-continuous function because the sequence $(f_n)$ is increasing.
