Is there a more non-uniform set with positive measure in any rectangle of the 2-d plane, where the measures don't equal the area of the rectangles?

Question

(If you don't need the motivation, skip it.)

Motivation: I want to find a set $A\subseteq\mathbb{R}^2$ which is more non-uniform and difficult to "meaningfully average" (with this paper) than a set with a positive Lebesgue measure in any rectangle of the plane, where the measures don’t equal the area of the rectangles. I need such a set to test, with the same paper, that set $A\subseteq\mathbb{R}^2$ can be "meaningfully averaged".

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Suppose $A\subseteq\mathbb{R}^{2}$ is Borel and $B$ is a rectangle of $\mathbb{R}^2$. In addition, suppose the Lebesgue measure on the Borel $\sigma$-algebra is $\lambda(\cdot)\;\!$:

Question: Is there an example of $A$, such that:

1. $\lambda(A\cap B)>0$ for all $B$

2. $\lambda(A\cap B)\neq\lambda(B)$ for all $B$

3. For all rectangles $\mathscr{B}\subseteq B$

   a. $\lambda(B\setminus\mathscr{B})>\lambda(\mathscr{B})\Rightarrow\lambda(A\cap(B\setminus\mathscr{B}))<\lambda(A\cap\mathscr{B})$

   b. $\lambda(B\setminus\mathscr{B})<\lambda(\mathscr{B})\Rightarrow\lambda(A\cap(B\setminus\mathscr{B}))>\lambda(A\cap\mathscr{B})$

   c. $\lambda(B\setminus\mathscr{B})=\lambda(\mathscr{B})\Rightarrow\lambda(A\cap(B\setminus\mathscr{B}))\neq\lambda(A\cap\mathscr{B})$?

If so, how do we define such a set? If not, how do we modify the question so an example of $A$ exists?

Edit: According to a user on reddit,

Let's just use $\mathbb{R}$ so "rectangles" are just intervals.

Your conditions in 3 imply $\lambda(A\cap(2,3)) > \lambda(A\cap(0,2))$, and condition 1 implies $\lambda(A \cap (0,2)) >\lambda(A\cap(1,2))$. So $\lambda(A \cap (2,3)) > \lambda(A\cap(1,2))$. Likewise, $\lambda(A\cap(1,2)) > \lambda(A \cap(2,4))$ and $\lambda(A\cap(2,4)) > \lambda(A\cap(2,3))$, so $\lambda(A\cap(1,2)) > \lambda(A\cap(2,3))$, which is a contradiction.

Does this apply to $\mathbb{R}^2$? Are the conditions in this question nonsensical?

Answer 1

The conditions seem to be nonsensical. Construct rectangles $B\supseteq\mathscr B_1\supseteq\mathscr B_2\supseteq\mathscr B_3\supseteq\cdots $ with $\lambda(B\setminus\mathscr B_1)\gt\lambda(\mathscr B_1)$ and $\lambda(\mathscr B_n)\lt\frac1n$. If $A$ is a measurable set with $\lambda(A\cap B)\gt0$ then $\lim_{n\to\infty}\lambda(A\cap\mathscr B_n)=0$ while $\lim_{n\to\infty}\lambda(A\cap(B\setminus\mathscr B_n))=\lambda(A\cap B)\gt0$, so $\lambda(A\cap(B\setminus\mathscr B_n))\gt\lambda(A\cap\mathscr B_n)$ for some $n$, while $\lambda(B\setminus\mathscr B_n)\ge\lambda(B\setminus\mathscr B_1)\gt\lambda(\mathscr B_1)\ge\lambda(\mathscr B_n)$.