$\sigma$-field generated by Borel sets

Question

I what sense in the snippet is the $\sigma$-field $\cal M$ on $M$ generated by all the functions $$m\to m(C)$$ for $C$ Borel;how does $\cal M$ look like ?

Answer 1

$M$ is the set of all counting measures on $(\mathbb{R},\mathcal{B}) $, where $\mathcal{B}$ is the Borel $\sigma$-field.

For each $C \in \mathcal{B}$, define a function $f_C : M \rightarrow \mathbb{R}$ such that $f_C(m)=m(C)$.

Let $\mathcal{M}$ be the $\sigma$-field generate by the functions in $\{ f_C : C \in \mathcal{B}\}$.

It means that $\mathcal{M}$ be the $\sigma$-field generate by the sets in $\{ f_C^{-1}(D) : C \in \mathcal{B} , D\in \mathcal{B} \}$, in other words, $\mathcal{M}$ be the $\sigma$-field generate by the sets in $\{ \{m \in M : m(C) \in D \} : C, D \in \mathcal{B} \}$.

Using the properties of $\sigma$-field and Borel $\sigma$-field, we can simplify to: $\mathcal{M}$ be the $\sigma$-field generate by the sets in $\{ \{m \in M : m(C) > a \} : C \in \mathcal{B} \textrm{ and } a \in \mathbb{R}\}$.

Remark 1: Let us prove, step by step, that $$\{ f_C^{-1}(D) : C, D \in \mathcal{B} \}= \{ \{ m \in M : m(C) \in D \} : C, D \in \mathcal{B} \}$$

Note that, given any $ C, D \in \mathcal{B}$,
$$m \in f_C^{-1}(D) \Leftrightarrow (m \in M \textrm{ and } f_C(m) \in D) \Leftrightarrow (m \in M \textrm{ and } m(C) \in D)$$

So, for all $ C, D \in \mathcal{B}$, $$f_C^{-1}(D) = \{ m \in M : m(C) \in D \} \tag{1}$$ It follows that $$\{ f_C^{-1}(D) : C, D\in \mathcal{B} \}= \{ \{ m \in M : m(C) \in D \} : C, D \in \mathcal{B} \}$$

Remark 2: Let us prove, step by step, that \begin{align*} \sigma(\{ \{ m \in M : m(C) \in D \} : & \, C, D \in \mathcal{B} \})= \\ &=\sigma(\{ \{m \in M : m(C) > a \} : C \in \mathcal{B} \textrm{ and } a \in \mathbb{R}\}) \end{align*}

First, let $\Gamma= \{(a,+\infty) \subseteq \mathbb{R} : a \in \mathbb{R}\}$. Note that $\mathcal{B}$ is generated by $\Gamma$, that is $\mathcal{B}=\sigma(\Gamma)$.

So, we must prove: \begin{align*} \sigma(\{ \{ m \in M : m(C) \in D \} : & \,C\in \mathcal{B} \textrm{ and } D\in \sigma(\Gamma) \})= \\ &=\sigma(\{ \{m \in M : m(C) \in D \} : C \in \mathcal{B} \textrm{ and } D \in \Gamma \}) \end{align*}

It means, using $(1)$: $$ \sigma(\{ f_C^{-1}(D) : C\in \mathcal{B}, D\in \sigma(\Gamma) \})= \sigma(\{ f_C^{-1}(D) : C \in \mathcal{B}, D \in \Gamma \}) $$

Clearly, $$\{ f_C^{-1}(D) : C \in \mathcal{B}, D \in \Gamma \} \subseteq \{ f_C^{-1}(D) : C\in \mathcal{B}, D\in \sigma(\Gamma) \}$$

So, we have,

$$\sigma(\{ f_C^{-1}(D) : C \in \mathcal{B}, D \in \Gamma \}) \subseteq \sigma (\{ f_C^{-1}(D) : C\in \mathcal{B}, D\in \sigma(\Gamma) \})$$

Now, let us write $\Psi =\sigma (\{ f_C^{-1}(D) : C\in \mathcal{B}, D\in \Gamma \})$.

For each $C\in \mathcal{B}$, consider $\Sigma_C = \{E \subseteq \mathbb{R}: f_C^{-1}(E) \in \Psi\}$. Using the properties on the inverse image and the fact that $\Psi$ is a $\sigma$-algebra, it is easy to prove that $\Sigma_C$ is $\sigma$-algebra. It is also clear that $\Gamma \subset \Sigma_C$. So $\sigma(\Gamma) \subseteq \Sigma_C$.

It means that, for each $D\in \sigma(\Gamma)$ , $f_C^{-1}(D) \in \Psi$. It means, for all $C\in \mathcal{B}$, $$\{f_C^{-1}(D) : D\in \sigma(\Gamma)\} \subseteq \Psi$$ So $$\{ f_C^{-1}(D) : C\in \mathcal{B}, D\in \sigma(\Gamma) \} = \bigcup_{C\in \mathcal{B}}\{f_C^{-1}(D) : D\in \sigma(\Gamma)\} \subseteq \Psi$$

Since $\Psi$ is a $\sigma$-algebra, we have

$$\sigma (\{ f_C^{-1}(D) : C\in \mathcal{B}, D\in \sigma(\Gamma) \}) \subseteq \Psi$$