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

borel-measuresborel-setscompactnessmeasure-theorysolution-verification

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.

enter image description here

Best Answer

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 $\mu$.

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