The two extreme cases of the product measure 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.
 A: I am a bit surprised to see this question remain unanswered for so long. Since its twin question also has no answer, let me add my answer here. 
The OP has already shown correctly that $\lambda(M)\leq \pi(M)$. So I shall show that $p(M)\leq \lambda(M)$ for every $M\in\mathcal A\rtimes \mathcal B$ where I am using the symbol $\rtimes$ to denote the product $\sigma-$algebra.
Take any $M\in\mathcal A\rtimes\mathcal B$ and any $A\in\mathcal A,B\in\mathcal B$ with $\mu(A)<\infty,\nu(B)<\infty$. Therefore, using the fact that $\pi(A\times B)<\infty$ and that $\pi$ is a measure on $\mathcal A\rtimes \mathcal B$ (that $\pi$ is a measure on $\mathcal A\rtimes \mathcal B$ follows from the definition of $\pi$ as an outer measure and is standard), we have:
$\pi(M\cap A\times B)=\pi(A\times B)-\pi(M^c\cap A\times B)$
Using the inequality that for any $M\in\mathcal A\rtimes \mathcal B$, $\lambda(M)\leq\pi(M)$ we obtain :
$\pi(A\times B)-\pi(M^c\cap A\times B)\leq \pi(A\times B)-\lambda(M^c\cap A\times B)=\lambda(A\times B)-\lambda(M^c\cap A\times B)=\lambda(M\cap A\times B\leq \lambda(M)$
We used $\pi(A\times B)=\lambda(A\times B)$ which is something you can readily check.
In the above, we could do all the subtraction because $\pi(A\times B)<\infty$ is given.
So we got for any $A\in\mathcal A,B\in\mathcal B$ with $\mu(A)<\infty,\nu(B)<\infty$ that $\pi(M\cap A\times B)\leq\lambda(M)$.
Take supremum over all possible $A,B$ on LHS to conclude the result.
