$\newcommand\B{\mathscr B}\newcommand\A{\mathscr A}\newcommand\si{\sigma}$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 $L^2(C)>0$?