Let $\Omega \subset \mathbb{R}^2$ be a bounded region. The random variable $(X,Y) \sim U(\Omega)$. (Uniform distribution)
If $\Omega = [a,b]\times [c,d] $ is a rectangle, I know that $X$ and $Y$ are independent, thus not correlated.
My question is: When $\Omega$ is not a rectangle, is $X$ necessarily correlated with $Y$ ? Thanks!
Let $m_2(\Omega)$ denote the 2-dimensional Lebesgue measure of $\Omega$, which is just its area. First let's check when $X,Y$ are independent, in which case they are also uncorrelated. Take any $(x_0,y_0) \in \Omega$, then $$f_X(x_0) = \int_{-\infty }^{\infty }f_{XY}(x_0,y)dy=\frac{m_1((x=x_0)\cap\Omega)}{m_2(\Omega)}$$
$$f_Y(y_0) = \int_{-\infty }^{\infty }f_{XY}(x,y_0)dx=\frac{m_1((y=y_0)\cap\Omega)}{m_2(\Omega)}$$ Then $X,Y$ are independent iff for all $(x_0,y_0)\in \Omega$, $$\frac{m_1((x=x_0)\cap\Omega)\cdot m_1((y=y_0)\cap\Omega)}{m_2(\Omega)^2}=f_X(x_0)f_Y(y_0) =f_{XY}(x_0,y_0)=\frac1{m_2(\Omega)}$$ $$\iff m_1((x=x_0)\cap\Omega)\cdot m_1((y=y_0)\cap\Omega)=m_2(\Omega)$$ This means $m_1((x=x_0)\cap\Omega)\cdot m_1((y=y_0)\cap\Omega)$ is constant, certainly a rectangle satisfies this requirement. You say $\Omega$ is a bounded region on $\mathbb R^2$, and I take that to mean a bounded open connected subset. So the question is whether there exists a bounded open connected subset $\Omega$ of $\mathbb R^2$ which is not a rectangle, such that $m_1((x=x_0)\cap\Omega)\cdot m_1((y=y_0)\cap\Omega)$ is constant for all $(x_0,y_0) \in \Omega$.
I'm not very good at geometry but intuitively I think it's impossible. Since $\Omega$ is bounded, $y_0$ is also bounded, which means that as $x_0$ moves along the $x$-axis, the measure of $y_0$'s such that $(x_0,y_0)\in\Omega$, or $m_1((x=x_0)\cap\Omega)$(does this have a name in geometry?), changes abruptly from $0$ to some positive number (otherwise $m_1((y=y_0)\cap\Omega)\to\infty$ and $x_0$ is unbounded). And I can imagine this scenario only when $\Omega$ is a rectangle. Perhaps someone else good at geometry can help me finish my argument.
It depends on what you call not being a rectangle. For example $\Omega=([0,1]\times[0,1])\cup\{(2,3)\}$ is not a rectangle but you would still have independence.
Let $\widetilde\Omega=\{x\mid\exists y\in\mathbb R,\ (x,y)\in\Omega\}$, that is $\widetilde\Omega$ is the projection of $\Omega$ on the $x$-axis, for instance for $\Omega=[a,b]\times[c,d]$ you would have $\widetilde\Omega=[a,b]$. One way to say that $\Omega$ cannot be described as a rectangle would be for instance that there exist two measurable sets $A,B\subset\widetilde\Omega$ with same positive Lebesgue measure such that $\Omega\cap(A\times\mathbb R)$ and $\Omega\cap(B\times\mathbb R)$ do not have the same Lebesgue measure.
With that definition, then $(X,Y)$ are obviously not independent, because the distribution of $Y$ given $X\in A$ is not the same as that of $Y$ given $X\in B$.
However you could still have uncorrelated $X$ and $Y$, in the sense of covariance 0. Examples are given by any $\Omega$ such that either the $x$-axis or the $y$-axis is an axis of symmetry.
