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: How do we define an example of $A$, such that:
- $\lambda(A\cap B)>0$ for all $B$
- $\lambda(A\cap B)\neq\lambda(B)$ for all $B$?
For a potential answer, see this reddit post:
Find a set $A\subseteq\mathbb{R}$, where $0<\lambda(A\cap I)<\lambda(I)$ for every interval $I$, then take its cross product with $\mathbb{R}$ to get the set desired.
Therefore, construct a subset of $[0,1]$ with this property (then copy and paste this set to get the set $A\subseteq\mathbb{R}$):
We do this by constructing a strange map from $[0,1]\to\mathbb{R}$. Take a real number $x\in[0,1]$, expand that number in binary as $0.b_0 b_1 b_2 {\cdot\cdot\cdot}$ and map the value to the series $\sum_{n=1}^{\infty}(2(b_n)-1)/n$. It's possible using Khintchine's inequality to show the sum converges for a.e. $x\in[0,1]$. Thus, our desired set $A$ will just consist of those $x$ for which the sum is positive.
The fact this set works is a little bit annoying to prove, but relies on Khintchine's inequality and the divergence of the Harmonic series. Essentially, we want to show that for any initial seqeuence $b_0,b_1,\cdot\cdot\cdot,b_n$ of digits there is a positive probability that the final sum is positive and a positive probability that the final sum is negative.
It seems the answer is correct; however, the answer states "its a little bit annoying to prove". I wonder if there's a simpler version which requires less effort to prove.
