Measure of infinite intersection of sets  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.
 A: 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.
