$\newcommand\B{\mathscr B}\newcommand\A{\mathscr A}\newcommand\si{\sigma}$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 $\B^2$.



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> **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 $\B^2$)?
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> **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 $L^2(C)>0$?
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On partial absolute continuity
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.


