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Suppose that $\mu$ is a $\sigma$-finite measure on $\mathcal{X}\equiv\bigotimes_{i=1}^n\mathcal{X}_i$. Let $\Pi$ denote the set of all $\sigma$-finite product measures on $\mathcal{X}$. Define $$ \mathcal{A}\equiv\{\nu\in \Pi:\mu\ll\nu\}. $$ Assuming that $\mathcal{A}\ne \emptyset$, does this set contain a maximal element (here $\nu\le \nu'$ iff $\nu'\ll \nu$)? The answer is trivial when $\mu\in\Pi$, and I think that it might be true when $\mu$ is abolutely continuous w.r.t. the product of its marginals.

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  • $\begingroup$ It's certainly true when $\mu\ll\mu_1\otimes\mu_2$ because $\mu\ll\nu_1\otimes\nu_2$ implies the same for the marginals, so if $\nu_j=\mu_j$ works, it's the best you can do. $\endgroup$ Commented Feb 21 at 14:54
  • $\begingroup$ @ChristianRemling What do you mean by "implies the same for the marginals"? $\endgroup$
    – d.k.o.
    Commented Feb 21 at 15:11
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    $\begingroup$ If $\mu\ll \nu_1\otimes\nu_2$, then also $\mu_1\ll \nu_1$ etc., with $\mu_1(A)=\mu(A\times X_2)$ (I'm perhaps assuming that the measures are finite, but I don't think that matters here). $\endgroup$ Commented Feb 21 at 15:15

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Yes, this works. I'll assume (for convenience) that $n=2$ and the measures are finite. First of all, as already observed in my comment above, if $\mu\ll\nu$, then the marginals $\mu_1(A)=\mu(A\times X_2)$ etc. also satisfy $\mu_j\ll\nu_j$.

Now if $d\mu=f\, d(\nu_1\otimes\nu_2)$ for some product measure, then we can simply optimize the marginals by letting \begin{align*} N_1 &=\{ x\in X_1 : \int_{X_2} f(x,y)\, d\nu_2(y)=0\} \\ N_2 &=\{ y\in X_2 : \int_{X_1} f(x,y)\, d\nu_1(x)=0\} . \end{align*} Then $N_j$ is well defined modulo $\nu_j$ null sets. In particular, we can unambiguously set $\sigma_j=\chi_{N^c_j}\nu_j$. Then still $\mu\ll\sigma_1\otimes\sigma_2$, and this new product measure is minimal.

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  • $\begingroup$ I'm a bit confused by "Yes, this works." :) $\endgroup$
    – d.k.o.
    Commented Feb 21 at 16:02
  • $\begingroup$ @d.k.o.: I could have said just "yes", in answer to the question you asked above. In other words, the statement is true (and proved in the rest of the answer). $\endgroup$ Commented Feb 21 at 18:34

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