Rudin's Real Complex Analysis Chapter 8, exercise 7, asks us to verify that if $(X,\mu _1,\mathfrak M_1)$ and $(Y,\mu_2, \mathfrak M_2)$ are $\sigma$-finite measure spaces, then there exists a unique $\sigma$-finite measure $\lambda$ on the product space such that $$\lambda(A \times B)=\mu_1(A)\mu_2(B)$$ for every $A\in \mathfrak M_1, B\in \mathfrak M_2$.
Applying the Hahn-Kolmogorov Theorem with the fact that the collection $\mathcal A$ of finite disjoint unions of measurable rectangles is an algebra and that $\mathfrak M_1\times \mathfrak M_2$ is generated by $\mathcal A$ proves the statement.
BUT, I am wondering whether there is another proof of the statement. Particularly, I am curious about whether there is a proof not relying on the concept of pre-measure and not using Hahn-Kolomogorov Theorem since neither of them is stated in Rudin's book.
Any hint would be really appreciated. Thank you.
You can apply the monotone class theorem. First suppose $(\mu_1\times\mu_2)(X\times Y)=\lambda(X\times Y)<\infty$, and put $$\mathcal C=\{E\in\mathfrak M_1\times \mathfrak M_2:(\mu_1\times\mu_2)(E)=\lambda(E)\}.$$ Clearly this contains the algebra of all finite disjoint unions of products $A\times B$ with $A\in\mathfrak M_1$, $B\in\mathfrak M_2$. Show that $\mathcal C$ is a monotone class, and by the monotone class theorem $\mathcal C=\mathfrak M_1\times\mathfrak M_2$.
If now $\mu_1,\mu_2$ are $\sigma$-finite, obtain $A_n\in\mathfrak M_1$, $B_n\in\mathfrak M_2$ with $\mu_1(A_n)<\infty$, $\mu_2(B_n)<\infty$, $X=\cup_nA_n$, and $Y=\cup_nB_n$. Appply the previous argument to the $\sigma$-algebras restricted to $A_m\times B_n$, then take unions to obtain the result.