Let’s start off with two measure spaces, $ (X,\mathcal{A},\mu) $ and $ (Y,\mathcal{B},\nu) $, and suppose that we want to form their product measure space. It can happen that more than one product measure exists: One has the measures arising from the Carathéodory construction on the $ \sigma $-algebra $ \mathcal{A} \times \mathcal{B} $ corresponding to the outer measures $ \pi(M) \stackrel{\text{df}}{=} \inf \left( \left\{ \sum_{n \in \mathbb{N}} \mu(A_{n}) \nu(B_{n}) ~ \middle| ~ M \subseteq \bigcup_{n \in \mathbb{N}} A_{n} \times B_{n} \right\} \right) $ and $ p(M) \stackrel{\text{df}}{=} \sup(\{ \pi(M \cap A \times B) \mid \mu(A),\nu(B) < \infty \}). $
Now, in this question, one exercise is to show that any measure $ \lambda $ on $ \mathcal{A} \times \mathcal{B} $ such that $ \forall A \in \mathcal{A}, ~ \forall B \in \mathcal{B}: \qquad \lambda(A \times B) = \mu(A) \nu(B) $ satisfies $ p(M) \leq \lambda(M) \leq \pi(M) $ for all $ M \in \mathcal{A} \times \mathcal{B} $.
One direction is easy: $ \displaystyle M \subseteq \bigcup_{n \in \mathbb{N}} (A_{n} \times B_{n}) $ implies that $ \lambda(M) \leq \sum_{n \in \mathbb{N}} \lambda(A_{n} \times B_{n}) = \sum_{n \in \mathbb{N}} \mu(A_{n}) \nu(B_{n}). $ I also know that for sets $ W \in \mathcal{A} \times \mathcal{B} $ such that $ \pi(W) < \infty $, one has $ p(W) = \pi(W) $. What I cannot show is the reverse, i.e., $ \lambda(W) \geq \pi(W) = \lambda(W) $, which would then imply the above inequality. Does someone have a hint for me?
Thanks in advance.