0

Setup: $(X,\tau)$ is a second-countable topological space with a countable base $\mathcal{B}$. Write $\Sigma$ to be the Borel $\sigma$-algebra, which is also generated by $\mathcal{B}$ by the second-countable assumption.

Now suppose that $\mu: \mathcal{B}\to \mathbb{R}$ is a function which satisfies the properties that a measure should satisfy, i.e., (1) $\mu(B)\geq 0$ for every $B\in \mathcal{B}$ with $\mu(\emptyset)=0$ and (2) for every pairwise disjoint countable sequence $B_i\in \mathcal{B}$, there holds $$ \mu\left(\bigcup_{i\in \mathbb{N}} B_i\right)=\sum_{i\in\mathbb{N}}\mu(B_i). $$ Question: Then is it true that $\mu$ defines a unique Borel measure?

I know that if $\mu$ and $\nu$ were both Borel measures which agree on the base $\mathcal{B}$, then $\mu=\nu$ (see Borel measure is uniquely determined by its values on the base for under the second-countable assumption). But this assumes that $\mu$ and $\nu$ are already defined on the entire Borel set.

Sorry if this is a somewhat trivial question... It has been a while since I've thought about measure theory stuff.

CA13
  • 2,559
  • There is Caratheodory's extension theorem which answers your question about uniqueness, although I think for the existence we might need to have some condition which makes $\mu$ to agree on intersections of elements of $\mathcal{B}$ – Jakobian Jul 26 '24 at 17:15
  • @Jakobian Ok yes this is coming back to me a bit now... So I guess what would really be necessary is to be able to say that $\mu$ is is countably additive on the Ring of sets (or algebral of sets) generated by the base $\mathcal{B}$. The function $\mu$ I am working with is given as a limit of other measures $\mu_i$ where the limit is only known to exist on the members of $\mathcal{B}$... but if I only have to check that it plays nicely with finite intersections/unions, then this should follow immediately by interchanging the limits – CA13 Jul 26 '24 at 18:23
  • You asked: "is it true that $\mu$ defines a unique Borel measure?" Without any other additional assumption, the answer is NO. Here is a counter-example. Let $X = \Bbb N$. For any $i \in \Bbb N$, let $U_i = { n \in \Bbb N : n \geqslant i }$. Let $\mathcal{B} = {U_i}_{i \in \Bbb N}$. It is easy to see that $\mathcal{B}$ is the base of second-countable topology. Define $\mu(\emptyset)=0$ and $\mu(U_i)=1$, for all $i \in \Bbb N$. (continue) – Ramiro Jul 26 '24 at 22:56
  • Clearly, $\mu$ satisfies: (1) $\mu(B)\geq 0$ for every $B\in \mathcal{B}$ with $\mu(\emptyset)=0$ and (2) for every pairwise disjoint countable sequence $B_i\in \mathcal{B}$, there holds $$ \mu\left(\bigcup_{i\in \mathbb{N}} B_i\right)=\sum_{i\in\mathbb{N}}\mu(B_i). $$ The condition (2) is vacuously true. However, $\mu$ cannot be extended to a ($\sigma$-additive) measure on the $\sigma$-algebra generated by $\mathcal{B}$. – Ramiro Jul 26 '24 at 22:57

0 Answers0