Prove that $\mu \times \lambda$ is a measure on $\mathcal{S} \times \mathcal{T}.$

Question

Let $(X, \mathcal{S}, \mu)$ and $(Y, \mathcal{T}, \lambda)$ be $\sigma$-finite measure spaces. If $Q \in \mathcal{S} \times \mathcal{T}$ we define $$\mu \times \lambda(Q) = \int_{X} \lambda(Q_{x})d\mu(x) = \int_{Y}\mu(Q^y)d\lambda(y).$$ Prove that $\mu \times \lambda$ is a measure on $\mathcal{S} \times \mathcal{T}.$

Could anyone give me an idea of how to use the given 2 equalities in the proof?

Also, I found this question here Measurability of product measures $ \{\mu \in M: (\mu \times \mu)(A) \in B\} \in \mathscr{M}$ is the solution of the given question similar to the solution of this question.

Note: I am studying from Royden and Fitzpatrick "Real Analysis" fourth edition.

Answer 1

Hint: Show and use that for $(E_n)_{n\geq 1}\subseteq \mathcal{S}\times \mathcal{T}$ one has $$ \Big(\bigcup_{n\geq 1} E_n\Big)_x=\bigcup_{n\geq 1}(E_n)_x $$