Folland 3.12. For j=1,2 let $\mu_j, \nu_j$ be $\sigma$-finite measures on $(X_j,M_j)$ such that $\nu_j << \mu_j$. Show $\nu_1\times \nu_2 << \mu_1 \times \mu_2$.
This is answered in many places but I still have some confusion indirected to the problem itself, and I hope you can confirm my understanding or point out my misunderstanding. Appreciated.
We can first look at $(X,M_1,\nu_1)$ and $(X,M_2,\nu_2)$ two measure spaces. The $\sigma$-algebra generated by measurable rectangles ($A\times B$, where $A\in M_1, B\in M_2)$ is $M_1\times M_2$. Define the premeasure to be $\omega_1(A\times B) = \nu_1(A)\nu_2(B)$, where $A\times B$ is a rectangle, that is, (finite or countable) disjoint union of rectangles. Then this premeasure generates an outer measure whose restriction on $M_1\times M_2$ is a measure, and since $\nu_1,\nu_2$ are $\sigma$-finite, this is a unique extension of $\omega_1$, and we denote this product measure $v_1\times v_2$.
Similarly, still on $\sigma$-algebra $M_1\times M_2$, we now define the premeasure to be $\omega_2(A\times B) = \mu_1(A)\mu_2(B)$. Then this premeasure generates an outer measure whose restriction on $M_1\times M_2$ is a measure, and since $\mu_1,\mu_2$ are $\sigma$-finite, this is a unique extension of $\omega_2$, and we denote this product measure $\mu_1\times \mu_2$.
In sum, on the $\sigma$-algebra $M_1\times M_2$, there could be many different measures. The uniqueness mentioned in Theorem 1.14 in Folland is just about the unique extension of premeasure. But there could be many premeasures defined on that same algebra, which then generates different outer measures, so we can have many different measures on $M_1\times M_2$. Is that correct?
Moreover, when we are talking about Borel $\sigma$-algebra on $\mathbb{R}$, use different increasing and right continuous functions $F$, we can define many different premeausres, which leads to different outer measures, and thus different Borel / Lebesgue-Stieltjes measures. When we talk about the uniqueness of Borel / Lebesgue-Stieltjes measure, again we are associated to one particular function $F$.