What is a martingale measure? And in particular what is a $L^2(P)$ valued sigma-finite measure?

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My text book defines a Martingale measure in the following way:

Let $(\Omega, \mathscr F, \{\mathscr F_t\}_{t \ge 0}, P)$ be a filtered probability space. A Processes $\{M_t(A)\}_{t \ge 0, \; A\in \mathscr B(\mathbb R^n) }$ is a martingale measure with respect to the filtration $\{\mathscr F_t\}_{t\ge 0}$ if:

1: $M_0(A)=0$ for any $A \in \mathscr B(\mathbb R^n)$

2: If $t\ge 0 $ then $M_t$ is a sigma finite $L^2(P)$ valued signed measure

3: For all $A \in \mathscr B(\mathbb R^n)$, $\{M_t(A)\}_{t \ge 0}$ is a $0$-mean martingale with respect to the filtration $\mathscr F$.

I have a question about the meaning of (2). How should I think of this? For fixed $t \ge 0$, is $M_t(A)$ the measure of $A$ for any $A \in \mathscr B( \mathbb R^n)$? Or is $E\big[\big(M_t(A)\big)^2\big]$ the measure of $A$?

Also what woud it mean to be a sigma finite $L^2(P)$ valued measure? Does that mean that we can find a collection of countable sets, $\{B_n\}$, such that these sets cover $\mathbb R^n$ and that $E\big[\big(M_t(B_n)\big)^2\big] < \infty $?

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(2) just means that the countable additivity holds in $L^2(\Omega)$ instead of almost surely, so for disjoint $A_n$'s in $\mathcal{B}(\mathbb{R}^n)$ you have \begin{align} \lim_{n\to \infty} E\left[ \left( M_t\left(\cup_{k=1}^{\infty} A_k\right) - \sum_{k=1}^n M_t(A_k)\right)^2 \right] = 0 \tag{*}\label{*} \end{align} for each $t \geq 0$.

I think you need an additional assumption of $$ A\cap B = \emptyset \implies M_t(A)\text{ and }M_t(B)\text{ are independent, for each } t\ge 0 $$ to get countable additivity (from \eqref{*}) under disjoint union for $E(M_t(\cdot)^2)$.

The assumption $$ M_t(\emptyset) = 0 \text{ a.s. for each } t\ge 0$$ will ensure that $E(M_t(\emptyset)^2) =0$. From these two assumptions together one can conclude that $E(M_t(\cdot)^2)$ is a measure on $\mathbb{R}^n$.

[Note that the second assumption is also responsible for finite a.s. additivity of each $M_t$, again from \eqref{*}.]

Or, you can just demand that $E(M_t(A)^2)$ be a measure of $A$; see chapter 4 of Applebaum's book Lévy Processes and Stochastic Calculus for this approach.