Let $(X_{1},\mathcal{A}_{1},\mu_{1})$ and $(X_{2},\mathcal{A}_{2},\mu_{2})$ be $\sigma-$finite measures, then there exists one $\sigma-$finite measure $\pi:=\mu_{1}\otimes \mu_{2}$ on $\mathcal{A}_{1} \otimes \mathcal{A}_{2}$ (i.e. the product measure) such that
$1.$ $\mu_{1}\otimes \mu_{2}(A_{1}\times A_{2})=\mu_{1}(A_{1})\mu_{2}(A_{2})$ for any $A_{1} \in \mathcal{A}_{1}, A_{2} \in \mathcal{A}_{2}$
$2.$ Then $\forall A \in \mathcal{A}_{1} \otimes \mathcal{A}_{2}:\int_{X_{1}}\mu_{2}(A_{x_{1}})d\mu_{1}(x_{1})=\int_{X_{2}}\mu_{1}(A_{x_{2}})d\mu_{2}(x_{2})$
I struggle to find how $1.$ and $2.$ are related to each other. I mean the only difference I see is that in $1.$ we are looking at $A_{1} \in \mathcal{A}_{1}, A_{2} \in \mathcal{A}_{2}$ rather than $A \in \mathcal{A}_{1} \otimes \mathcal{A}_{2}$. What important concept am I missing here? Is $1.$ simply a special case of $2.$ since $\mathcal{A}_{1}\times \mathcal{A}_{2}\subseteq \mathcal{A}_{1} \otimes \mathcal{A}_{2}$, as $\mathcal{A}_{1} \otimes \mathcal{A}_{2}$ is the smallest $\sigma-$Algebra such that the projection map is measurable?
The actual product measure is given in (2).
If a measure on $\mathcal{A_1} \otimes \mathcal{A_2}$ satisfies (1), then there is only one such measure and it is given by the formula in (2).
So both things yield the same thing: a description of the unique product measure.