The $\sigma$-algebra defined on the space of measures $\mathcal{M}(S)$ In Kallenberg's textbook,he defines a space $\mathcal{M}(S)$ that contains all measures in $S$.



In the picture above,he defines a $\sigma$-algebra on $S$ by map$$f_B:\mathcal{M}(S)\to\mathbb{R}$$
$$\mu\to\mu(B)$$
where $B\in\mathcal{S}$,then $f_{B}$ induce a $\sigma$-algebra on $\mathcal{M}(S)$
my question is :
(1)Is this $\sigma$-algebra independent of the choice of $B$?
(2)what does "$(\mu\otimes\nu )(A)$ is measurable " mean in the Lemma 1.35? Does it mean "$f_A$ is measurable"?But I think it contributes nothing to the conclusion.
 A: First of all, the $\sigma$ algebra on $\mathcal{M}(S)$ equals by definition
$$\sigma(\pi_B; B \in \mathcal{S}),$$
i.e. it is the smallest $\sigma$-algebra such that the mappings
$$\pi_B: \mathcal{M}(S) \to \mathbb{R}, \mu \mapsto \mu(B)$$
are measurable for all $B \in \mathcal{S}$. This should clear up the first question.
Now if we restrict the $\sigma$-algebra to the class $\mathcal{P}(S)$ of probability measures, then this yields that
$$\tilde{\pi}_B: \mathcal{P}(S) \to \mathbb{R}, \mu \mapsto \mu(B) \tag{1}$$
is measurable and the $\sigma$-algebra on $\mathcal{P}(S)$ is generated by $\tilde{\pi}_B$, $B \in \mathcal{S}$. Recall the following lemma from measure theory:

Lemma: Let $(E,\mathcal{E})$  be a measure space and $f_i: F \to \mathbb{R}$ a family of mappings. Then a mapping $f:E \to F$ is measurable with respect to the $\sigma$-algebra generated by $f_i$, $i \in I$, if and only if the mapping $f_i \circ f: (E,\mathcal{E}) \to (\mathbb{R},\mathcal{B}(\mathbb{R})$ is measurable for each $i \in I$.

Using this Lemma in our setting shows that $$\mathcal{P}(S) \times \mathcal{P}(T) \ni (\mu,\nu) \mapsto \mu \otimes \nu \in \mathcal{P}(S \times T)$$ is measurable if, and only if, $$\mathcal{P}(S) \times \mathcal{P}(T) \ni (\mu,\nu) \mapsto \mu \otimes \nu(A) \tag{2}$$ is measurable for all $A \in \mathcal{S} \otimes \mathcal{T}$ (the product $\sigma$-algebra of $\mathcal{S}$ and $\mathcal{T}$). If $A= B \times C$ for $B \in \mathcal{S}$, $C \in \mathcal{T}$, then we know that
$$\mu \otimes \nu(A) = \mu(B) \cdot \nu(C).$$
By definition of the $\sigma$-algebra (cf. $(1)$), both mappings $$\mathcal{P}(S) \ni \mu \mapsto \mu(B) \qquad \text{and} \qquad \mathcal{P}(T) \ni \nu \mapsto \nu(C)$$ are measurable. This implies that
$$\mathcal{P}(S) \times \mathcal{P}(T) \ni (\mu,\nu) \mapsto \mu(B) \cdot \nu(C)$$
is measurable (as a product of measurable functions), i.e. $(2)$ holds. Either using some monotone class argument or recalling that it suffices to show measurability on a generator, we conclude that $(2)$ is measurable for all $A \in \mathcal{S} \otimes \mathcal{T}$. This finishes the proof.
A: It is induced by **all* maps $f_B$ of that form (it says "induced by the mappings").
So there is no dependency on $B$.
This means that we have the $\sigma$-field on $\mathcal{M}(S)$ that is the smallest one such that all $f_B$ are measurable (so it contains all $(f_B)^{-1}[A]$ where $A \subset \mathbb{R}$ is measurable, for all $B \in \mathcal{S}$).
Note that $\mathcal{P}(S) = \{ \mu \in \mathcal{M}(S): \mu(S) = 1 \} = \{ \mu \in \mathcal{M}: f_S(\mu) = 1 \} = (f_S)^{-1}[\{1\}]$, so $\mathcal{P}(S)$ is indeed a measurable subset of $\mathcal{M}(S)$. 
As to (2), $\mu \otimes \nu$ is the product measure.
