Determining Class of a general Borel measure Let $(X, \mathcal{T})$ be a topological space, and $\Sigma = \Sigma(\mathcal{T})$ the $\sigma$-algebra of Borel sets (that is, the $\sigma$-algebra generated by $\mathcal{T}$).
In Real Analysis and Probability by R. M. Dudley, I have the following definition: Call $\mathcal{C} \subset \Sigma$ a determining class for $(X,\Sigma)$ if: $\mathcal{C}$ generates $\Sigma$ and whenever measures $\mu$ and $\pi$ take equal, finite values on $\mathcal{C}$, then $\mu = \pi$.
The author goes on to show that sufficient conditions for $\mathcal{C}$ to be a determining class are that $X$ is a countable union of elements of $\mathcal{C}$ AND that $\mathcal{C}$ is a semiring: that is, $\emptyset \in \mathcal{C}$ and for any pair of sets $A, B \in \mathcal{C}$, $A\cap B \in \mathcal{C}$ and $A \backslash B$ is a finite union of disjoint elements of $\mathcal{C}$. In particular, these conditions are used to show that a measure defined on the finite-length half-open intervals of $\mathbb{R}$ has a unique extension to the Borel sets of $\mathbb{R}$.
An alternative sufficient condition is that $\mathcal{C}$ is a ring: that is, $\emptyset \in \mathcal{C}$ and for any pair of sets $A, B \in \mathcal{C}$, $A\cup B \in \mathcal{C}$ and $A \backslash B \in \mathcal{C}$.
However, neither of these conditions are particularly topological; no collection of open and closed sets form a ring or a semiring for a general topological space (or even over $\mathbb{R}$). My question is as follows: is there a canonical way to construct a determining class for $(X,\Sigma)$? If not in general, is it possible in special cases?
 A: There is the following formulation:
For a system of sets $\mathscr{E}$ which generates the $\sigma$-Algebra $\mathscr{A}$ on $\Omega$ and is closed under $\cap$ and contains a sequence $(E_n)_{n\in \mathbb{N}}$ such that $\Omega = \bigcup_n^\infty E_n$. Then two measures $\mu, \nu$ which satisfy
$$\mu(E) = \nu(E) \qquad E\in \mathscr{E}$$
and
$$\mu(E_n) = \nu(E_n) < \infty \qquad n \in \mathbb{N}$$
are identical.
Proof: Let $\mathscr{E}'$ be the system of sets $E \in \mathscr{E}: \mu(E) =
\nu(E)<\infty.$ Now, for $E\in \mathscr{E}'$ consider
\begin{align*}
 \mathscr{D}_E:=\left\{ D\in \mathscr{A}: \mu(E\cap D) = \nu(E\cap D) \right\}
\end{align*}
It is straigt forward to show that $\mathscr{D}_E$ is a Dynkin System. Since
$\mathscr{E}$ is closed under $\cap$: $\mathscr{E}\subset \mathscr{D}_E$ and
therefore
\begin{align*}
 \mathscr{A}= \sigma(\mathscr{E}) = \delta(\mathscr{E}) \subset
 \mathscr{D}_E \qquad E\in \mathscr{E}',
\end{align*}
i.e. $\mathscr{D}_E = \mathscr{A}$ for $E\in \mathscr{E}'$. Therefore
\begin{align*}
 \mu(E\cap A) = \nu(E\cap A) \qquad E\in \mathscr{E}', A\in \mathscr{A}.
\end{align*}
In particular this equality holds for $E_n$ with $n\in \mathbb{N}$. W.l.o.g.
the $(E_n)_{n\in \mathbb{N}}$ are pairwise disjoint. Finally, for any
$A\in \mathscr{A}$
\begin{align*}
 \mu(A) = \mu(\bigcup_{n=1}^\infty E_n \cap A) =
 \sum_{n=1}^\infty\mu(A\cap E_n) = \sum_{n=1}^\infty\nu(A\cap E_n) = \dots
 \nu(A),
\end{align*}
which establishes the result.
My source is a german textbook, so I'm not sure if that's helpful. If you still have any questions regarding the proof feel free to ask. I'm very willing to discuss that further, since its also great practice for me.
So if the topological space contains open sets $(E_n)_n$ as above, the open sets uniquely define a measure.
I hope this helps if you didn't alsready know it.
PS: Please consider upvoting my aswer if you find it helpful, I still need reputation 50 to comment on questions.
BS
