# Every compact subset of $\mathbb R^1$ is the support of a Borel measure

I know this already has an answer here, though it is very cryptic. So, I'm making this post for solution/proof-verification (it is not a duplicate) - I've come up with a measure on Borel sets of $$\mathbb R$$ corresponding to a compact set. Please help me fill in the gaps, if any.

Prove that every compact subset of $$\mathbb R^1$$ is the support of a Borel measure.

Here's my work:
Suppose $$A \subset \mathbb R$$ is compact. Compact metric spaces are separable, so there exists a set $$B = (b_n)_{n\in\mathbb N} \subset A$$ which is dense in $$A$$ (i.e. $$\overline B = A$$ w.r.t. the subspace topology on $$A$$). Define $$\mu$$ on Borel sets of $$\mathbb R$$ as follows, where $$\delta_p$$ is the dirac measure at point $$p\in\mathbb R$$. Note that $$\delta_p(X) = 1$$ if $$p\in X$$ and $$\delta_p(X) = 0$$ if $$p\not\in X$$, for every $$X\subset\mathbb R$$. Also we know that the support of $$\delta_p$$ is $$\{p\}$$, and the support of a finite sum of measures is the (finite) union of their supports. If $$C$$ is a Borel set in $$\mathbb R$$, $$\mu(C) = \sum_{n=1}^\infty \frac{\delta_{b_n}(C)}{2^n}$$ To check that $$\mu$$ is a measure, we need to show that $$\mu$$ is countably sub-additive. I think it suffices to notice the countable sub-additivity of Dirac measures. Unfortunately, I'm unable to characterize the support of $$\mu$$, because it is only in the finite case that we are allowed to union over the supports. However, I do feel that the above construction suffices.

Could I please get some help in completing my proof? Thank you!

Follow-up questions:

1. What is special about Borel sets in this construction? Can we define $$\mu$$ over a larger $$\sigma$$-algebra? There must be something special about Borel sets, otherwise, the statement would probably not be framed this way.
2. How do we deal with the case when $$K = \varnothing$$? What's the corresponding measure?
3. If $$B$$ is finite, I think we cannot work with the finite sum. Instead, if we have distinct $$b_1,b_2,\ldots,b_N$$, we can define $$b_n := b_N$$ for $$n > N$$? (See this link). Please clarify.
4. The book gives questions 11 and 12 together (image below for reference) - so are we supposed to use the definition of support from Q11 in Q12? Hopefully, it is equivalent to the one here.

To show that $$\mu$$ is a measure, note that $$\mu (\emptyset ) = 0$$ is immediate, and $$\sigma$$-additivity follows from the $$\sigma$$-additivity of your Dirac masses. That is, if $$X = \bigsqcup_{k=1}^\infty B_k$$, where the union is disjoint and the $$B_k$$'s are Borel, then $$\mu (X) = \sum_{n=1}^\infty \frac{\delta_{b_n}(X)}{2^n} = \sum_{n=1}^\infty \frac{1}{2^n} \sum_{k=1}^\infty \delta_{b_n}(B_k) = \sum_{k=1}^\infty \sum_{n=1}^\infty \frac{1}{2^n} \delta_{b_n}(B_k) = \sum_{k=1}^\infty \mu (B_k)$$ Note that we can interchange the order of summation because all the summands are non-negative.

To see that the support of $$\mu$$ is $$A$$, first observe that $$A$$ is already a closed set, so that if we take an open set $$U$$ that does not hit $$A$$ then we must show that $$\mu (A) = 0$$. But that is immediate because such a $$U$$ cannot hit any of the points $$\{b_n\}$$. Conversely, if $$K$$ is a closed proper subset of $$A$$, then we have that $$b_n \not\in K$$ for some $$n$$, and so $$\mu (K^c) \geq \delta_{b_n} (K^c) > 0$$. Since the support of a measure is the smallest closed set whose complement is $$\mu$$-null, we conclude that $$A$$ is the support of $$\mu$$.

• I think your proof of the support is incomplete. You've only shown what cannot lie inside the support, not what lies inside it. Commented May 23, 2021 at 16:34
• I've updated the solution to clarify this point. Commented May 23, 2021 at 16:42
• Thanks @Jose! I have added some follow-up questions (the proof as it is currently seems slightly incomplete); I'd be grateful if you could answer them. Commented May 23, 2021 at 16:52
• 1) The support of a measure is a topological concept; the Borel $\sigma$-algebra plays nicely with the usual topology in $\mathbb{R}$. 2) The zero measure. 3) If $B$ is finite, then so is $A$, so just take the counting measure. Commented May 23, 2021 at 16:59
• Could you explain (1) in more detail? It is not clear. I mean, what will go wrong if we have any other $\sigma$-algebra in mind? Commented May 23, 2021 at 17:07

The fact that $$\mu$$ is a measure can be proved by using theorems from RCA Rudin itself. As $$K$$ is compact, there exists a closed interval $$H$$ such that $$K \subset H$$. $$H$$ is compact. $$K$$ is compact w.r.t subspace topology on $$H$$. This implies separability of $$K$$. As $$K$$ is also a metric space, there is a countable dense set $$B = \{b_n\}_{n \in \mathbb{N}}$$ w.r.t subspace topology on $$H$$. Now define the counting measure $$\lambda$$ on $$H$$ as follows:

$$E \in \mathcal{B}(H), \; \lambda(E) = \#E \cap B$$

Verifying that $$\lambda$$ is a measure is easy I suppose. Define the function $$f : H \rightarrow [0,\infty]$$ by: $$f(x) = \begin{cases} \frac{1}{2^{-n}}, \; x = b_n \\ 0, & \text{otherwise} \end{cases}$$

Observe that $$f = \sum_1^\infty \frac{1}{2^{-n}} \chi_{\{b_n\}}$$. The sets $$\{b_n\}$$ are borel measurable and so are the functions $$\chi_{\{b_n\}}$$. Hence $$f$$ is measurable by:

Th 1.9(c):

and Th 1.14:

Define the measure $$\mu$$ by: $$E \in \mathcal{B}(H), \mu(E) = \int_Efd\lambda = \sum_{b_n \in E} \frac{1}{2^{-n}}$$

$$\mu$$ is a measure by Th 1.29:

$$\mu(H) = \int_H f d\lambda = \sum_{b_n \in H} \frac{1}{2^{-n}} = \sum_{b_n \in K} \frac{1}{2^{-n}} = \mu(K) = \sum_1^\infty \frac{1}{2^{-n}} = 1$$. Suppose $$G$$ is a proper subset of $$K$$ and is compact. $$G$$ is closed as a subset of $$H$$. Then $$\exists x \in K, x \in H-G$$. As $$H-G$$ is open and $$B$$ is dense in $$K$$, $$\exists b_n \in B, b_n \in H-G$$, hence $$b_n \not \in G$$, hence $$\mu(G) \leq 1 - \frac{1}{2^{-n}} < \mu(K)$$.