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.

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    $\begingroup$ Did you mean to ask whether $\mu(X_2)\leq\mu(X_1)$? 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
  • $\begingroup$ Yes, thanks. I did mean to ask $\mu(X_2) \le \mu(X_1)$. $\endgroup$ – Sai Apr 25 '11 at 15:20

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.

  • $\begingroup$ Thanks Andreas. I am not quite sure if I understand the example since the intersection of $X$ with the transformation $(1,1)$ produces a measure zero set. But I think I understand the rationale, the hypothesis requires one $q$ for every $p$ globally, and is a a bit weak. So, what happens if the hypothesis is refined to 'For each $p \in \gamma_1$ there exists a distinct $q \in \gamma_2$ in a neighborhood of $p$ such that $\mu(X \cap pX) \ge \mu(X \cap qX)$' (such a neighborhood can be constructed for example by picking $q$ in a ball around $p$ in $\gamma_1^c$) $\endgroup$ – Sai Apr 25 '11 at 16:31
  • $\begingroup$ Thanks for catching a typo; $f(1)$ should be 0, not 1. ($x-x^2$ now works.) I edited the answer to correct this. As for requiring $q$ to be close to the corresponding $p$, I would expect that once you specify how close you want it, there will still be counterexamples similar to the one I described. Just take $f$ very small. $\endgroup$ – Andreas Blass Apr 25 '11 at 16:50

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