Relation between a set's diameter and its measure Let $X$ be a compact metric space, and $\mu$ be a finite positive Borel measure on $X$. Suppose that $\mu(\{x\})=0$ for every $x \in X$. Prove that for every $\varepsilon>0$ there is a $\delta>0$ such that, if $E$ is any Borel subset of $X$ having diameter less $\operatorname{than} \delta$, then $\mu(E)<\varepsilon$. I think that if $\mu(\{x\})>0$, then no matter how small the diameter is, $\mu(E)>0$, where $E$ is open and contains $x$. Is this the right direction to prove it? If so, is it enough for me to say this is true for $E$ being open without taking $E$ to be other kinds of Borel sets like closed sets or countable union/intersection of open/closed sets?
 A: You basically need to prove that the measure of $\delta$-balls vanish uniformly at $\delta\to 0$. This then proves the statement, as any set of diameter smaller than $\delta$ is a subset of a ball of radius $\delta$ (that ball is by so much larger that any point of the set can be chosen as middle point. You could go smaller, but does it matter?). Your approach is kind of the other direction, as you’d prove that if singletons have positive mass the wanted property is wrong.
For a fixed center this is clear, for as measures are continuous from above for sets of finte measure you have $\lim_{\delta\to0}\mu(B_\delta(x_0)) = \mu(\{x_0\})$.
But how do we do this uniformly (so that $\delta$ does not depend on $x_0$)? Consider a particular $\delta$. If $x_n$ is a sequence of middlepoints that converges towards some $x$ (note $X$ is compact) so that $\mu(B_\delta(x_n))$ converges towards the supremum, then $\mu(B_{\delta+\rho}(x))\geq \sup_y\mu(B_\delta(y))$ for any $\rho>0$, so in fact $B_\delta(x)$ is of maximal measure.
Now take $\delta_n = 1/n$ and obtain such points $x_{\delta_n}=x_n$ that $\mu(B_{\delta_n}(x_n))=\max_y\mu(B_{\delta_n}(y))$. The set of these $x_n$ is at most countable, so we have again
$$ \lim_{\delta\to0} \mu\left(\bigcup_n B_\delta(x_n)\right) = \mu(\{x_n:n\in\mathbb N\}) = 0$$
Thus we get: For any $\epsilon$ there is some $\delta_n$ so that
$$ \mu\left(\bigcup_n B_{\delta_n}(x_n)\right) < \epsilon$$
But then for all $y\in X$ we have
$$\mu(B_{\delta_n}(y))\leq \mu(B_{\delta_n}(x_n)) \leq \mu\left(\bigcup_n B_{\delta_n}(x_n)\right) < \epsilon$$
EDIT: In fact we do not need compactness: If $x_n$ is a sequence so that $\mu(B_\delta(x_n))$ converges to the supremum, then we can find a point $x$ within that sequence so that for some (arbitrary) $\rho>0$ we have for all $y$ that $\mu(B_\delta(y))\leq \mu(B_\delta(x))+\rho$.
So then find such points $x_{\delta_n}=x_n$ that this condition holds with $\rho_n=1/n$. Then we can proceed the same way. We then take a $\delta_n$ so that
$$ \mu\left(\bigcup_n B_{\delta_n}(x_n)\right) < \epsilon/2$$
and that $1/n<\epsilon/2$. Then for any $y$ we get
$$ \mu(B_{\delta_n}(y)) \leq \mu(B_{\delta_n}(x_n))+1/n<\epsilon/2+\epsilon/2 = \epsilon$$
Also in fact we do not need that $\mu$ is finite, but it suffices to have local finiteness and some sort of robustness to translation.
