1

Consider $ \mathcal{X}$ that is a countable set and is a set of isolated points in $\mathbb{R}$. Let $A = \{ t \in \mathbb{R}: \mathcal{X} \cap \{ \mathcal{X}-t \} \neq \varnothing\}$. Here, $\mathcal{X}-t$ means that $\mathcal{X}-t = \{ x - t : x\in \mathcal{X} \}$.

How large can the Lebesgue measure of $A$ be? In particular, does there exists $\mathcal{X}$ such that $A$ has a positive Lebesgue measure?

Here is what I've thought so far: Since $\mathcal{X}$ is a set of isolated points, any $t$ such that $\mathcal{X}\cap \{\mathcal{X}-t \}\neq \varnothing$ is also isolated. Thus, $A$ has measure zero. But I am not sure this is correct.

KCd
  • 46,062
sabo
  • 35
  • The possibility that $\mathcal{X} \cap (\mathcal{X} - t)$ contains only isolated points won't help you since you are looking at the collection of values of $t$. – Nick F Nov 16 '22 at 22:20

1 Answers1

0

Let's enumerate the elements of $\mathcal{X}$, say $\mathcal{X} = \{x_n : n \in \mathbb{Z}^+\}$ (I'll leave the case when $\mathcal{X}$ is finite to you). If $t \in A$, then necessarily $x_n - t \in \mathcal{X}$ for some $n$, say $x_n - t = x_m$ for some $m$. Therefore we see that $t = x_n - x_m$.

The conclusion is that $A \subseteq \mathcal{X} - \mathcal{X} = \{x_n - x_m : n,m\in\mathbb{Z}^+\}$. This latter set is countable, being indexed by the countable set $(\mathbb{Z}^+)\times(\mathbb{Z}^+)$. A basic result concerning Lebesgue measure $\lambda$ is that countable sets have measure zero. Hence $\lambda(A) = 0$ by monotonicity of the measure.

We did not use the fact that the points of $\mathcal{X}$ were all isolated. Are you sure $\mathcal{X}$ had to be countable? I think there may be more interesting possibilities if we let it be uncountable. There's an exercise in Folland's Real Analysis: Modern Techniques and Their Applications which essentially states $\lambda(E - E) > 0$ for any measurable $E \subseteq \mathbb{R}$ with $\lambda(E) > 0$ (see Exercise 1.5 #31). So if your $\mathcal{X}$ were uncountable with positive measure, there is the possibility that $\lambda(A) > 0$.

Nick F
  • 1,219
  • Thank you! Yes, I know how to prove it if $\mathcal{X}$ is finite. But, one thing here I am confused about is that the latter set $\mathcal{X}-\mathcal{X}$ is countable. Is the cardinality of $\mathcal{X}-\mathcal{X}$ larger than that of natural numbers? I think it is. If it is, the set is not countable by definition. Could you give some hints about how the set becomes countable? – sabo Nov 16 '22 at 23:07
  • There's a surjective function $f : (\mathbb{Z}^+)\times(\mathbb{Z}^+) \to \mathcal{X} - \mathcal{X}$ given by $f(n,m) = x_n - x_m$. So $\mathcal{X} - \mathcal{X}$ is at most as large as $(\mathbb{Z}^+)\times(\mathbb{Z}^+)$, which is countable. (See here) – Nick F Nov 17 '22 at 00:08
  • I think I didn't know the definition of cardinality. I learned it from your comment. Thanks a lot! – sabo Nov 18 '22 at 03:27
  • To your credit, cardinality gets weird in the infinite case. – Nick F Nov 18 '22 at 04:42