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 $L^2(C)>0$?

 A: 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.
The answer is no.
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.
