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.

`$\mu(X_2)\leq\mu(X_1)$`

? With the conclusion as stated,`$X_2\subseteq X_1$`

, there are easy counterexamples. Let $X$ be the unit cube. Let`$\gamma_1$`

consist of leftward translations by distances ranging from 0 to $1/2$, and let`$\gamma_2$`

be similar but with rightward translations. $\endgroup$ – Andreas Blass Apr 25 '11 at 15:08