Show that Borel measure on compact metric is small when diameter of the set is small I am quite new to Borel measure and I am stuck when handling the following question:
Let $(E,d)$ be compact metric space and assume $(E,\mathcal{B}(E),\mu)$ is a measure space. Moreover, if the measure satisfies
1) $\mu(E)< \infty$
2) $\forall x\in E,\ \mu(\{x\}) = 0$
How can we show that $\forall \epsilon>0,\ \exists \delta>0\ s.t.\ \forall X\in \mathcal{B}(E),\ diam X<\delta \implies \mu(X)< \epsilon$? 
I was trying to reach a contradiction from the fact that singletons are of measure zero, but I don't know how to apply the fact that $E$ is compact.
Thanks!
 A: For probability measures you have the property that if $E_{n+1}\subseteq E_n$, then
$$ \mathbb{P}\left( \cap E_n \right)=\lim \mathbb{P}(E_n). $$
$\frac{\mu(\cdot)}{\mu(E)}$ is a probability measure, and therefore satisfies that property. Assume towards contradiction that there exists $\epsilon>0$ such that for all $\delta>0$ there exists $F_\delta\in \mathcal{B}(E)$ satisfying
$$ \mu(F_\delta)\geq \epsilon \quad \text{and} \quad \text{diam}(F_\delta)<\delta. \tag{1} $$
In particular there exists a sequence of sets $F_n$ satisfying $\mu(F_n)\geq \epsilon$ and $\text{diam}(F_n)<\frac{1}{2^n}$. By the triangle inequality, for all $n$ there exists $x_n\in F_n$ such that $\overline{B}(x_n,\frac{2}{2^n})\supseteq F_n$, where $\overline{B}(x,\delta)$ is the closed ball at radius $\delta$ around $x$. Since $E$  is compact $x_n$ has a subsequence converging to some $x_0\in E$.
Verify that $\{ \overline{B}(x_{n_k},\frac{2}{2^{n_k}}) \}$ is decreasing, and by Cantor's intersection theorem
$$ \cap_k \overline{B}(x_{n_k},2^{1-n_k}) =\{ x_0 \}.$$
Notice that $F_{n_k}\subseteq \overline{B}(x_{n_k},2^{1-n_k}) $ while $\mu(F_{n_k})\geq \epsilon$. You can use all this to get a contradiction.
