$\mu \otimes \nu(A \times B)=\mu(A)\nu(B)$

Let $(\Omega,\mathcal{A},\mu)$ and $(X,\mathcal{C},\nu)$ be to measure spaces with s-finite measures (so $\mu(\Omega)>0$ and $\mu(X)>0$).

Show that $\mu \otimes \nu(A \times B)=\mu(A)\nu(B)$ for $A\times B \in \mathcal{A} \times \mathcal{C}$ and that if they are $\Sigma-$finite measures then $\mu \otimes \nu$ is uniquely determined by its values on $A\times B \in \mathcal{A} \times \mathcal{C}$

I know that they are finite it is simply that the measure is $<\infty$. Further, it is s-finite if $\mu=\sum_{n=1}^{\infty}\mu_n$ where each $\mu_n$ is a finite measure. For it to be $\Sigma$-finite if the underlying set is a countable union of measurable sets with finite measure which is of course a weaker condition than finite.

I do not know how to proceed to solve the question with the definitions. For the uniqueness I was thinking about Fubini's theorem.

$(\mu \otimes \nu)(A \times B) = \mu(A)\nu(B)$ is part of the definition of the product measure (my definition uses Caratheodory's theorem to extend this rule to a measure on the product sigma-algebra).

As for uniqueness, suppose $\mathcal{A}$ is an algebra that generates a sigma-algebra $\mathcal{F}$. Suppose $\mu, \nu$ are measures on the measurable space $(X, \mathcal{F})$ that agree on $\mathcal{A}$. Suppose there exist $A_j \in \mathcal{A}$ with $\mu(A_j) = \nu(A_j) < \infty$ such that $X = \bigcup_{j = 1}^{\infty}A_j$. Then $\mu = \nu$. When $\mu(X) = \nu(X) < \infty$, this is a consequence of the monotone class theorem. Otherwise, apply the previous case to each $A_j$ (after making them disjoint). I think the $\pi-\lambda$ theorem can also work here.