Is $ mu(E times F) leq nu(E times F) forall (E,F) implies mu(A) leq nu(A) forall A in mathcal{E} otimes mathcal{F} $ true?
Let $(X,mathcal{E})$ and $(Y,mathcal{F})$ denote two measurable spaces and let $mu,nu$ denote two finite measures on $(X times Y, mathcal{E} otimes mathcal{F})$, where $mathcal{E} otimes mathcal{F}:= sigma(mathcal{E} times mathcal{F})$. Consider the claim
$$
mu(E times F) leq nu(E times F) text{ for all } (E,F) in mathcal{E} times mathcal{F} implies mu(A) leq nu(A) text{ for all } A in mathcal{E} otimes mathcal{F}.
$$
Question: Is this claim true?
Attempt:
Define the set $$
M = {A in mathcal{E} otimes mathcal{F} colon mu(A) leq nu(A)}.
$$
Then we want to show that $M = mathcal{E} otimes mathcal{F}$. Note that clearly $M subseteq mathcal{E} otimes mathcal{F}$. For the other inclusion, we have by assumption that $mathcal{E} times mathcal{F} subseteq M$ so it follows that $mathcal{E} otimes mathcal{F}= sigma(mathcal{E} times mathcal{F}) subseteq sigma (M)$ and hence it would suffice to show $M$ is a $sigma$-algebra.
However I am not sure this is the case. I can’t seem to show that $M$ is closed under complements and countable unions. In particular say for the union part, we can do the following:
$$
mu(bigcup_{n in mathbb{N}}A_n) leq sum_{n in mathbb{N}} mu(A_n) leq sum_{n in mathbb{N}} nu(A_n),
$$
but then we cannot go back to $nu(bigcup_{n in mathbb{N}}A_n)$. So this is where I am stuck.
I have also tried to look at Dynkin’s $pi$-theorem but at least I could not see how this was useful here.
Also I haven’t been able to come up with any counterexample yet.
As a final remark, I will just need to apply this result for $mathcal{E},mathcal{F} = mathcal{B(mathbb{R}^d}),mathcal{B(mathbb{R}^m})$ but I would like to prove it higher generality if this is indeed possible. If this is not possible, are there any conditions we can impose for it to hold? Any feedback/help is much appreciated!
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