Define
$$f(x) = (2x) \mod 1$$ to be the doubling map on $X = (0, 1)$ and the Lebesgue measure mod zero partition $P = \{(0, 1/2), (1/2, 1)\}$. Here, given some finite measure space $(X, \mathcal{F}, \mu)$, a collection $P$ of elements of $\mathcal{F}$ is said to be a measure $\mu$ mod zero partition of $X$ if there exists a full measure subset $X'\subset X$ such that elements of the set $\{A\cap X'\mid A\in P\}$ are pairwise disjoint and their union is equal to $X'$. I am trying to understand when a strict inequality may occur in the inequality of the following statement:
Note that given any $n\geq 0$, if $x,y$ are points in $X$ such that $f^n(x), f^n(y)$ are in the same element of $P$, $\forall n\geq 0:\forall x,y\in X:\exists A_n\in P:f^n(x),f^n(y)\in A$, then $$|x-y|\leq 2^{-n}|f^n(x) - f^n(y)|$$
One can verify that e.g.
$$f^{-1}\left[(0, 1/2)\right] = (0,1/4)\cup (1/2, 3/4),$$
$$f^{-2}\left[(0, 1/2)\right] = (0,1/8)\cup (1/2, 5/8)\cup (1/4, 3/8)\cup (3/4, 7/8)$$
so there is definitely a geometric shrinking in the distance between any such $x$ and $y$. If we take e.g. $x = 3/4 + \epsilon_1, y = 3/4 + \epsilon_2$ with $0 < \epsilon_2 < \epsilon_1 \ll 1/2$, then
$|f^2(x) - f^2(y)| = 4|\epsilon_1 - \epsilon_2|$, which would lead to an equality in the aforementioned statement. But what about a strict inequality, when can it occur?
While this statement sounds tempting I am currently unable to explicitly state
Since for any $x,y\in X,n\geq 0$ we have that $0\leq |f^n(x) - f^n(y)|\leq 1$, it follows that given some $n\geq 0$, the possible points $