2
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Salamendrine Mar 1 at 16:18
  • $\begingroup$ @HaoboLi He divides by $\mu(E)$ which would trivialise the measure (to be always $0$) otherwise. $\endgroup$ – Henno Brandsma Mar 1 at 16:28
  • 1
    $\begingroup$ 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 $\mu$. $\endgroup$ – Keen-ameteur Mar 1 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.