Inf of the measure of the neighbourhood of a compact set Let $X$ bea compact metric space, and $\mu$ a Borel probability measure. 
For $A\subset X$ de define $b_{\epsilon}(A):=\left\{  x\mid d(x,A)\leq
\epsilon\right\}  .$
Suppose $A$ is compact and $\mu(A)=0,$ does this mean $\inf\mu(b_{\epsilon
}(A))=0?$
If this is not true is it true if you add extra hypothesis?
 A: Since $\mu$ is a probability measure we have $$\mu(A) = \mu\biggl(\bigcap_{n=1}^\infty b_{1/n}(A)\biggr) = \lim_{n\to\infty} \mu(b_{1/n}(A))$$
and it doesn't matter whether $A$ is a null set or not. We only use that $A$ is closed and compactness of $X$ is not used either.

Further details:


*

*Suppose $A$ is closed. Then $A = \bigcap_{n=1}^\infty b_{1/n}(A)$:
Clearly $A \subseteq b_{1/n}(A)$. On the other hand, if $x \in \bigcap_{n=1}^\infty b_{1/n}(A)$ then for each $n$ there is $a_n \in A$ such that $d(x,a_n) \leq 2/n$ by definition of $b_{1/n}(A)$. Then $a_n \to x$, hence $x \in A$ since $A$ is closed. 

*If $B_1 \subseteq B_2 \subseteq \cdots$ are measurable then $\mu\Big(\bigcup_{n=1}^\infty B_n\Big) = \lim_{n\to\infty} \mu(B_n)$ (easy consequence of $\sigma$-additivity of $\mu$, special case of monotone convergence). 

*Set $B_n = X \setminus b_{1/n}(A)$ then
$$\bigcup_{n=1}^\infty B_n = X \setminus \bigcap_{n=1}^\infty b_{1/n}(A) = X \setminus A$$ 
Since $\mu(X) \lt \infty$  we have $\mu(B_n) = \mu(X) - \mu(b_{1/n}(A))$, so $$\mu(X) - \mu(A) = \mu(X \setminus A) = \lim_{n\to\infty} \mu(B_n) = \lim_{n\to\infty}\left( \mu(X) - \mu(b_{1/n}(A))\right) = \mu(X) - \lim_{n\to\infty}\mu(b_{1/n}(A))$$
and we can rearrange this to 
$$
\mu(A) = \lim_{n\to\infty}b_{1/n}(A)
$$
as claimed above.
