Abstract:
如果一个实数的 Borel 集同时是实数集的子环，那么该集合要么是一个零维的集合（其 Hausdorff 维数为 0），要么就是整个实数集本身。我们将首先介绍若干代数与分析在交汇处的现象，并介绍几何测度论中的 Marstrand 投影定理。最后，作为该定理的应用，我们将对潜在的 Borel 环进行分类。
Jinghua Xi Nov. 17, 2024
Definition: A set $E\sub \R$ is said to be a Borel ring if $E$ is a Borel set of $\R$ and $E$ is a subring of $\R$.
Main Theorem: If $E$ is a Borel ring with $\dim_HE>0$, then $E=\R$.
In other word, there is no Borel ring with dimension strictly between $0$ and $1$.
If $G$ is a subgroup of $\R$ under addition and $G$ has interior, then $G=\R$.
The proof is motivated by the idea that the interior of G will “grow” to cover the entire real line.
Since $G$ has non-empty interior, it contains an open interval $I$. Because $G$ is closed under subtraction, the difference set $I-I$ is a subset of $G$: $I - I = \{ x - y \mid x, y \in I \} \subset G.$ The set $I-I$ is an open interval containing $0$, so $G$ contains an open interval centered at the origin, say $(-\epsilon, \epsilon)$.
Since $G$ is closed under addition and subtraction, it contains all integer multiples of $(-\epsilon, \epsilon)$ . As $n$ ranges over all integers, the union of these intervals expands indefinitely:$\bigcup_{n \in \mathbb{Z}} (-n\epsilon, n\epsilon) = \mathbb{R}.$ Therefore, $G = \mathbb{R}$ .
If $f:\R\to \R$ is additive, i.e., $f(x+y)=f(x)+f(y)$, then it necessarily follows that $f(q)=q\cdot f(1)$ for all $q\in \mathbb Q$. This naturally leads to the question: Under what conditions is $f$ linear, i.e., $f(x)=x\cdot f(1)$ for all $x\in \R$. A trivial condition we can enforce is continuity. However, a weaker condition—measurability—will also suffice. In other words, if f is both additive and measurable, then $f$ is linear.
The idea is straightforward: measurability is closely related to continuity. It suffices to prove continuity at the origin, as global continuity then follows from additivity.
Since $f$ is measurable on $[0, 1]$, Lusin’s Theorem guarantees the existence of a closed set $K\subset [0, 1]$ where $f|_K$ is continuous and $m(K) > \frac{2}{3}$. Indeed, $K$ is compact, so $f|_K$ is uniformly continuous. In other words, for any $\epsilon > 0$ , there exists $\delta \in(0, \frac{1}{3})$ such that for all $x, y \in K$ with $|x - y| < \delta$ , we have $|f(x) - f(y)| < \epsilon$.
For any $|h| < \delta$, I claim there exist $x, y \in K$ with $|x - y| = h$. Assuming the claim holds, additivity implies $|f(h)| = |f(x) - f(y)| < \epsilon$ due to the uniform continuity of $f$ on $K$. This establishes the proposition.
Proof of the Claim:
Suppose, for contradiction, that there exists $h$ with $|h| < \delta < \frac{1}{3}$ such that there are no $x, y \in K$ with $|x - y| = h$ . In other words, $(h + K) \cap K = \emptyset$. This implies $m\left( (h + K) \cup K \right) = m(h + K) + m(K) = 2m(K) > \frac{4}{3}$. However, since $h < \frac{2}{3}$ , we also have $m\left( (h + K) \cup K \right) \leq m(h+[0,1]\cup [0,1] ) < 1/3+1=\frac{4}{3}$, which is a contradiction.
Marstrand’s Projection Theorem: Let $A$ be a Borel set in $\R^n$ with $\dim _HA >2$. Then for $\sigma^{n-1}$-a.e. $e\in S^{n-1}$, it follows that $P_e(A)$ has interior, where $P_e(x)=x\cdot e$.
The theorem states that if we have a Borel set $A \subset \mathbb{R}^k$ with a sufficiently large Hausdorff dimension (specifically, $\dim_H A > 2$ ), then there exists a linear functional $\varphi: \mathbb{R}^k \to \mathbb{R}$ such that the image $\varphi(A)$ has interior. Furthermore, if $A$ possesses a group structure, the image $\varphi(A)$ inherits this structure when considering $\varphi$ as a group homomorphism. According to Phenomenon 1, this implies that $\varphi(A) = \mathbb{R}$ .
The primary challenge, however, lies in finding a Borel set $A$ that simultaneously has a large Hausdorff dimension and a group structure.
A Borel ring with positive Hausdorff dimension serves as an excellent candidate for our purposes. However, a slight modification is necessary to address the dimension-related challenges. Specifically, the positive dimension allows us to employ the Cartesian product to effectively accumulate dimensions. This is evident from the observation that