Proving Lebesgue measure of a set is zero 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.
 A: 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 $\lambda(A) > 0$.
