Given a closed bounded set $X \subset \mathbb{R}^3$ and two curves $\gamma\_1$ and $\gamma\_2$ in the group of orientation preserving isometries of $\mathbb{R}^3$. Define the sets $X\_1$ and $X\_2$ as the infinite intersections $X\_1 = \bigcap\_{p \in \gamma\_1} pX$ and $X\_2 = \bigcap\_{q \in \gamma\_2} qX$ where $kX$ represents the set $X$ transformed by $k$. If each of the $pX$ and $qX$ are measurable sets with measure $\mu$ defined on $\mathbb{R}^3$, and for each $p \in \gamma\_1$ there exists a distinct $q \in \gamma\_2$ such that $\mu(X \cap pX) \ge \mu(X \cap qX)$, can I conclude that $\mu(X\_2) \le \mu(X\_1)$ ? How so/ Why not? (We can assume the identity element of the isometry group is in both $\gamma\_1$ and $\gamma\_2$ if needed)


EDIT: Changed conclusion from $X\_2 \subseteq X\_1$ to $\mu(X\_2) \le \mu(X\_1)$ per Andreas Blass's comment.



Measure of infinite intersection of sets 
Here's a counterexample in $\mathbb{R}^2$; to get one in $\mathbb{R}^3$ just take the product with $[0,1]$. Let $X=[0,2]\times[0,1]$. Let $f:[0,1]\to[0,1]$ be a continuous function with $f(0)=0$, $f(1)=0$, $f(x)>0$ for $0<x<1$, and $(2-x)(1-f(x))$ strictly decreasing. (If my arithmetic is good, $f(x)=x-x^2$ works, but even if my arithmetic is bad, some $f$ works.) Let $\gamma\_1$ consists of the translations by the vectors $(t,f(t))$ for $t\in[0,1]$, and let $\gamma\_2$ consist of the translations by $(t,0)$. In both cases, the measure of the intersection of $X$ and its image under the $t$ translation begins at $\mu(X)=2$ for $t=0$ and decreases monotonically and continuously to $\mu([1,2]\times[0,1])=1$ at $t=1$. So the hypothesis in the question (that for each $p$ there is an appropriate $q$) is satisfied (in fact with equality of the two measures). But $X\_2=[1,2]\times[0,1]$ while $X\_1$ is a proper subset of that, missing some points near the bottom edge because $\gamma\_1$ lifted the rectangle $X$ a little ways above the $x$-axis during the motion.


