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Lets say we start off with two measure spaces

$(X, \mathcal{A}, \mu), (Y, \mathcal{B}, \nu)$

and suposse want to form the product measure. It can happen that there is more than one 'product' measure, i.e. one has the measure arising from the Caratheodory construction on some $\sigma$-alg. $\mathcal{C}$ corresponding to the outer measure

$\pi(M) := \text{inf}\{\sum_{n \in \mathbb{N}} \mu(A_n) \nu(B_n) : M \subset \cup_{n \in \mathbb{N}} A_n \times B_n\}$


$p(M) := \sup(\pi(M \cap A \times B) : \mu(A) < \infty, \nu(B) < \infty)$

Now in this question: Uniqueness of product measure (non $\sigma$-finite case), one excercise is to show that for any measure $\lambda_1$ on $\mathcal{C}$ that satisfies $\lambda_1(A\times B) = \mu(A) \nu(B)$ also satisfies the rule $p(M) \leq \lambda_1(M) \leq \pi(M)$ for all $M \in \mathcal{C}$.

One direction is easy: $M \subset \cup_{n \in \mathbb{N}} A_n \times B_n$ implies $\lambda_1(M) \leq \sum_n \lambda_1(A_n \times B_n) = \sum \mu(A_n) \nu(B_n)$. Also i know already, that on sets $W \in \mathcal{C}$ such that $\pi(W) < \infty$, one has $p(W) = \pi(W)$. What i am still not managing is to show the reverse, namely that $\lambda_1(W) \geq \pi(W) = \lambda(W)$ which would then imply the above inequality. Does someone have a hint for me?

Thanks in advance.

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This question seems to be related to this one:… Unfortunately I do not have an answer to any of the two – Tom Jun 27 at 15:13

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