Prove the uniqueness of product measure? Assume $\sigma$-finiteness, if a measure $m$ on $\mathscr{A}\times\mathscr{B} $satisfies $m(A\times B)=\mu(A)\nu(B)$, and I want to prove that there is only one measure statisfy this (say, the product measure $ \mu \times \nu$).
I knwo that this can be done using monotone class theorem (prove something is a monotone class, thus it is $\mathscr{A}\times\mathscr{B} $ itself), but I have trouble with constructing the monotone class. Anyone can give a hint on how to construct the monotone class? Thanks!
 A: The uniqueness of the product measure is a direct consequence on the following well-known result on the uniqueness of measures:

Let $\varrho_1$, $\varrho_2$ be measures on a measurable space $(X,\mathcal{A})$ and suppose that $\mathcal{A}$ is generated by a $\cap$-stable family $\mathcal{G}$ (i.e. $\mathcal{A} = \sigma(\mathcal{G})$ and $F,G \in \mathcal{G} \implies F \cap G \in \mathcal{G}$). If $$\varrho_1(G) = \varrho_2(G) \qquad \text{for all $G \in \mathcal{G}$}$$ and if there exists a sequence $(G_j)_j \subseteq \mathcal{G}$, $G_j \uparrow X$ such that $\varrho_1(G_j)+\varrho_2(G_j)<\infty$ for all $j$, then $$\varrho_1(A) = \varrho_2(A) \qquad \text{for all $A \in \mathcal{A}$}.$$

Now suppose that $m_1$ and $m_2$ are two measures which satisfy $$m_1(A \times B) = m_2(A \times B) = \mu(A) \nu(B) \qquad \text{for all $A \in \mathcal{A}$, $B \in \mathcal{B}$.}$$


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*Define $\mathcal{G} :=\mathcal{A} \times \mathcal{B}= \{A \times B; A \in \mathcal{A}, B \in \mathcal{B}\}$. Show that $\mathcal{G}$ is $\cap$-stable and recall (or show) that $\mathcal{G}$ generates the product $\sigma$-algebra $\mathcal{A} \otimes \mathcal{B}$.

*Using the $\sigma$-finiteness of $\mu$ and $\nu$ show that there exists a sequence $(G_j)_j \subseteq \mathcal{G}$, $G_j \uparrow X \times Y$, such that $m_1(G_j)+m_2(G_j)<\infty$.

*Apply the result on uniqueness of measures.

