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

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.

• 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. Commented Feb 21 at 14:54
• @ChristianRemling What do you mean by "implies the same for the marginals"? Commented Feb 21 at 15:11
• 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). Commented Feb 21 at 15:15

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.