If a joint density factorizes on a square, does this imply that the marginal random variables are locally independent?

Question

Let $Z=(X,Y) : \Omega\rightarrow\mathbb{R}^2$ be a Borel-measurable random vector and $U\subset\mathbb{R}$ be open. Suppose that $Z$ is absolutely continuous with Lebesgue density $\zeta$.

I was wondering about the following question: If we assume that $\zeta$ factorizes on the square $U\times U$, i.e. that the restriction

$$\tag{1}q:=\left.\zeta\right|_{U\times U} \quad\text{is such that}\quad q\equiv q(x,y)=q_1(x)q_2(y) \quad\text{for some } \ q_1, q_2 : U\rightarrow\mathbb{R},$$ does this imply that $X$ and $Y$ are independent on $U$ in the sense that

$$\tag{2}\mathbb{P}(X^{-1}(A)\cap Y^{-1}(B)) = \mathbb{P}_X(A)\cdot\mathbb{P}_Y(B)\quad\forall\, A, B\subseteq U \ (\mathrm{Borel})?$$

Clearly this holds for both the special cases $U=\mathbb{R}$ and $q_i = \zeta_i$ ($i=1,2$, with $\zeta_i$ the marginal densities of $\zeta$), however I couldn't show it in the above generality; do you have a counterexample?

Answer 1

The answer is no. E.g., let $U=(0,1)$ and $$\zeta=\frac12\times 1_{(0,1)^2}+\frac12\times 1_{(1,2)^2};$$ that is, the joint distribution of $(X,Y)$ is the half-and-half mixture of the uniform distributions on the the squares $(0,1)^2$ and $(1,2)^2$. Then $$P_{X,Y}(U\times U)=\frac12\ne\frac12\times\frac12=P_X(U)P_Y(U).$$

Answer 2

No. For example suppose $Z=(X,Y)$ distributes uniformly in the triangle $\{(x,y):x,y\geq 0, x+y\leq 1\}$, and $U=(0,\frac 1 2)$. Let $A=(0,\frac 1 2)$ and $B=(0,\frac 1 {10})$. Then $P(Z\in A\times B)=\frac 1 {10}$, and $P(X\in A)P(Y\in B)=\frac 3 4 \times \frac{19}{100}$.