counterexample checked!

@JonMarkPerry If $k=1$, then a position $j$ is $2b$ iff $a_j=0$. Since an interval only has one zero at the end, then the element before is $1g1b$. for instance, if the interval ends with $(\ldots,1,0]$ and this 1 is $a_j$, then $U_{1,j}(1)=1$ and $U_{2,j}(1)=0$ (because you start the counting from the position $j$ until the next $0$, which is $a_{j+1}$.

The definition of good intervals includes at least two elements 2g before the first 1g1b, while the neutral intervals have only one element 2g and then 1g1b.

If we take an infinite graph of finite valency d, I believe is $\omega$-stable, $\omega$-categorical and its geometry is trivial. Doesit interpretable in the pure set? How? (btw, I also answered by mistake the question, does anyone know how to take it down?)

Also, how exactly the last inequality help me to find the two indices satisfying $\mu(A_i\cap A_j)\geq \epsilon^2$? Couldn't it be possible that the indices witnessing the maximum value of $\mu(A_i\cap A_j)$ also depend on N, and the measure of the intersection keep being always strictly less than $\epsilon^2$?

@AnthonyQuas No, it is a lemma that I needed. I could prove it assuming the bound to be $\epsilon^t$ for any $t>2$ using inclusion-exclusion (which was enough to my purposes), and I was just curious about this particular case.