# 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!

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.

• Your proof is very elegant, and yet I have a tiny question about the usage of $\mu(E)<\infty$. It seems that the proof has used it to take advantage of the continuity of probability measure, but that continuity is also part of any measure. Therefore, I am wondering if $\mu(E)<\infty$ is something not really helpful when it seems indeed important? – Salamendrine Mar 1 at 16:18
• @HaoboLi He divides by $\mu(E)$ which would trivialise the measure (to be always $0$) otherwise. – Henno Brandsma Mar 1 at 16:28
• I think in fact that $\mu(E)<\infty$ might be too strong of a condition. To use upper continuity you just need that $\mu(E_n)<\infty$ for some $n$, which I think means this will work also when $\mu(E)=\infty$ and every $x\in E$ has a neighbourhood of finite measure. Upper continuity needs some sort of finiteness of $\mu$. – Keen-ameteur Mar 1 at 16:41