Maximal element w.r.t. abolute continuity of measures

Question

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.

Answer 1

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.