Example of the good set principle

Question

Can anyone help me with the following?

For $A \subset \mathbb{R}^d$ and $\mathbb{a}> 1$, let $aA:=\{ax|x\in A\}$.
If $A$ is a Borel set in $\mathbb{R}^d$, then $\mathbb{a}A$ is also a Borel set.

Our lecturer gave us this as an example, but without a proof. He only gave us the hint to use the good set principle. Can someone show me how to proof this? Or at least explain to me how to use the principle in general? I saw a question similar to this one, but it didn't help me. It'd be great, if anyone could help me.

Answer 1

In an abstract setup, the good-set principle is used as follows: Suppose you want to show that the each member of a $\sigma$-algebra $\mathscr{F}$ has some property $p$. Define $\mathscr{A} := \{A \in \mathscr{F} : A \ \text{has property} \ p\}$ (ie., $\mathscr{A}$ is the class of 'good' sets having property $p$). We will show that:

• $\mathscr{A}$ is a $\sigma$-algebra.
• $\mathscr{A}$ contains a class $\mathscr{C}$, such that, $\sigma(\mathscr{C}) = \mathscr{F}$, ie., the $\sigma$-algebra generated by $\mathscr{C}$ is $\mathscr{F}$.

This would mean that $\mathscr{C} \subset \mathscr{A}$ which implies $\sigma(\mathscr{C}) \subset \sigma(\mathscr{A})$. But, $\sigma(\mathscr{C}) = \mathscr{F}$ and $\mathscr{A}$ is already a $\sigma$-algebra, so $\sigma(\mathscr{A}) = \mathscr{A}$. Thus, $\mathscr{F} \subset \mathscr{A}$, and from definition of $\mathscr{A}$, we have, $\mathscr{A} \subset \mathscr{F}$. Hence, $\mathscr{A} = \mathscr{F}$ or in other words, every set in $\mathscr{F}$ is a good set.

Now, I will extend @martini's answer.

Let $\mathscr{B}$ denote the Borel $\sigma$-algebra over $\mathbb{R}^d$. In our case, a set $A$ in $\mathscr{B}$ is good if $aA$ is Borel. Equivalently, the class of good sets is, $\mathscr{A} = \{A \in \mathscr{B}: aA \in \mathscr{B}\}$.

Claim 1: $\mathscr{A}$ forms a $\sigma$-algebra.

Proof.

• It is obvious that $\mathbb{R}^d \in \mathscr{A}$.
• Let $A \in \mathscr{A}$, that is, $aA$ is Borel. Then $a(A^c) = a(\mathbb{R}^d \setminus A) = a\mathbb{R}^d \setminus aA = \mathbb{R}^d \setminus aA = (aA)^c$ which is Borel since $aA$ is Borel. Thus, $A^c \in \mathscr{A}$.
• Let $\mathscr{C}$ be a countable class of good sets, then, $a(\bigcup_{A \in \mathscr{C}} A) = \bigcup_{A \in \mathscr{C}} aA$, which is a Borel set since each of $aA$ is Borel.

Hence, $\mathscr{A}$ is a $\sigma$-algebra.

Claim 2: $\mathscr{A}$ contains all the open rectangles (ie. open rectangles are good).

Proof. We know that open rectangles are Borel. Let $A = (x_1, y_1)\times(x_2, y_2)\times \dots \times (x_d, y_d)$ be an open rectangle. Then, $aA = (ax_1, ay_1)\times(ax_2, ay_2)\times \dots \times (ax_d, ay_d)$ which is again an open rectangle, and hence Borel. Hence, $\mathscr{A}$ contains all the open rectangles.

Now, we know that the $\sigma$-algebra generated by the open rectangles is indeed the Borel $\sigma$-algebra $\mathscr{B}$. Hence, by good-set principle, it follows, $\mathscr{A} = \mathscr{B}$.

Answer 2

The basic idea of the good sets principle is the following: Consider the collection of the good sets, that it the sets that have the property you want them to have: $$ \mathcal A:= \{A \subseteq \mathbf R^d : A \text{ is Borel}, aA \text{ is Borel} \} $$ We want to prove that $\mathcal A$ is the Borel $\sigma$-Algebra. To prove that, show two things: (a) $\mathcal A$ is a $\sigma$-Algebra, (b) $\mathcal A$ contains a generator of the Borel $\sigma$-Algebra, for example, all open sets.

Answer 3

Let $\varphi(x) = ax: \mathbb{R}^d \to \mathbb{R}^d$. Then $\varphi$ is a homeomorphism.

Since $\varphi$ preserves the open sets, it also preserves the $\sigma$-algebra generated by the open sets.