On partial absolute continuity

$$\newcommand\B{\mathscr B}\newcommand\A{\mathscr A}\newcommand\si{\sigma}$$Let $$I:=[0,1]$$, and let $$\B$$ and $$\B^2$$ denote the Borel $$\si$$-algebras over $$I$$ and $$I^2$$, respectively. Let $$\A$$ stand for the algebra generated by the set of all the product sets $$A\times B\in\B\times\B$$. Let $$L^2$$ be the Lebesgue measure on $$\B^2$$. Let $$U$$ be another measure on $$\B^2$$.

Question: Suppose that $$L^2$$ is "absolutely continuous on the algebra $$\A$$" with respect to $$U$$ -- in the sense that $$L^2(C)=0$$ whenever $$C\in\A$$ and $$U(C)=0$$. Does it then necessarily follow that $$L^2$$ is (truly) absolutely continuous with respect to $$U$$ (on the $$\si$$-algebra $$\B^2$$)?

The same question, restated: Suppose that $$U(A\times B)>0$$ whenever $$A\times B\in\B\times\B$$ and $$L^2(A\times B)>0$$. Does it then necessarily follow that $$U(C)>0$$ whenever $$C\in\B^2$$ and $$L^2(C)>0$$?

The notation $$L^2$$ for Lebesgue measure is confusing. I denote $$\lambda$$ and $$\lambda_2$$ the Lebesgue measure on $$\mathbb{R}$$ and $$\mathbb{R}^2$$, respectively.
Fix a discrete measure $$\mu$$ which gives a positive mass to every rational number in $$[0,2]$$. Call $$\nu$$ the image of $$\lambda \otimes \mu$$ by the map $$(x,y) \mapsto (x,y-x)$$.
Given Borel subsets $$A$$ and $$B$$ of $$[0,1]$$ with positive measure, we know that the function $$1_A*1_B$$ is non-negative, continuous since it is a convolution between two functions in $$L^2(\mathbb{R})$$, and has a positive integral on $$[0,2]$$. Therefore, it is positive on some non-empty open subinterval of $$[0,2]$$. Thus $$\begin{eqnarray*} \nu(A \times B) &=& \int_{\mathbb{R}}\int_{\mathbb{R}} 1_{A \times B}(x,y-x)~d\lambda(x)~ d\mu(y) \\ &=& \int_{\mathbb{R}} (1_A*1_B)(y)~d\mu(y) \\ &>&0.\end{eqnarray*}$$ Yet, $$\nu$$ is carried by $$\{(x,z) \in \mathbb{R}^2 : x+z \in \mathbb{Q}\}$$, whose $$\lambda_2$$ measure is null. Furthermore, $$\lambda_2$$ is carried by the complement of $$\{(x,z) \in \mathbb{R}^2 : x+z\in \mathbb{Q}\}$$, which has null $$\nu$$-measure. Hence, the measures $$\lambda_2$$ and $$\nu$$ are mutually singular.
The measure $$(1_{I^2})\nu$$ yields a couterexample.
• It appears that you have $(1_{I^2})\nu$ for $U$. If so, then in your example $U$ is not absolutely continuous w.r. to $L^2$. However, the question was, vice versa, about the absolute continuity of $L^2$ w.r. to $U$. Commented Nov 20, 2022 at 0:13