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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

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2 Answers 2

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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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  • $\begingroup$ I think your proof of the support is incomplete. You've only shown what cannot lie inside the support, not what lies inside it. $\endgroup$ Commented May 23, 2021 at 16:34
  • $\begingroup$ I've updated the solution to clarify this point. $\endgroup$ Commented May 23, 2021 at 16:42
  • $\begingroup$ 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. $\endgroup$ Commented May 23, 2021 at 16:52
  • $\begingroup$ 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. $\endgroup$ Commented May 23, 2021 at 16:59
  • $\begingroup$ 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? $\endgroup$ Commented May 23, 2021 at 17:07
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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):

Th 1.9(c) RCA Rudin

and Th 1.14:

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:

Th 1.29 RCA Rudin

$\mu(H) = \int_H f d\lambda = \sum_{b_n \in H} \frac{1}{2^{-n}} = \sum_{b_n \in K} \frac{1}{2^{-n}} = \mu(K) = \sum_1^\infty \frac{1}{2^{-n}} = 1$. 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 $\mu(G) \leq 1 - \frac{1}{2^{-n}} < \mu(K)$.

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