Borel sigma algebras of subsets of $\mathbb{R}$

Question

I'm trying to prove a result that seems to be used routinely, but the proof is skipped over.

Let $\mathbb{R}$ have the Euclidean metric, and this defines a topology. Let $A \subset \mathbb{R}$ be equipped with the subspace topology. Then,

1. $\mathcal{B}(A) = \{ B \cap A : B \in \mathcal{B}(\mathbb{R})\}$

2. If $A \in \mathcal{B}(\mathbb{R})$, then $\mathcal{B}(A) = \{ B \in \mathcal{B}(\mathbb{R}) : B \subset A\}$.

The proof for the $\subset$ direction of 1. is simple; note that $\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\}$ is a sigma algebra, which contains all the open sets of $A$. There are also other questions on stackexchange which address this direction of the proof. However, I am stuck on the $\supset$ direction. Does anyone have a nice proof for this?

(2. is an immediate consequence of 1., as far as I can see)

Answer 1

You have already proved that $\mathcal{B}(A) \subseteq \{ B \cap A : B \in \mathcal{B}(\mathbb{R})\} $, and your prooof is correct: $\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\}$ is a sigma algebra, which contains all the open sets of $A$. So $\mathcal{B}(A) \subseteq \{ B \cap A : B \in \mathcal{B}(\mathbb{R})\} $.

Let us prove, in a simple way, that $$\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\} \subseteq \mathcal{B}(A)$$

Proof: Let $\tau$ be the collections of open sets in $\mathbb{R}$.

Let $$K = \{ E \cup F : E \in \mathcal{B}(A) \textrm{ and } F \in \mathcal{B}(A^c) \}$$

(note that $E$ and $F$ are disjoint). It is easy to see that $K$ is a $\sigma$-algebra. For all $O \in \tau$, $O =(O\cap A) \cup (O\cap A^c)$, $(O\cap A)\in\mathcal{B}(A)$ and $(O\cap A^c)\in\mathcal{B}(A^c)$. So $O \in K$. It follows that $\mathcal{B}(\mathbb{R}) \subseteq K$. So we have

$$\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\} \subseteq \{ G \cap A : G \in K\}$$

But $\{ G \cap A : G \in K\}=\mathcal{B}(A)$. So we have: $$\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\} \subseteq \mathcal{B}(A)$$

Remark: A second way to do this proof (which is just a variant) is:

Proof 2: Let $\tau$ be the collections of open sets in $\mathbb{R}$.

Let $$K = \{ G \subseteq \mathbb{R} : G\cap A \in \mathcal{B}(A) \}$$

It is easy to see that $K$ is a $\sigma$-algebra. For all $O \in \tau$, $(O\cap A)\in\mathcal{B}(A)$. So $O \in K$. It follows that $\mathcal{B}(\mathbb{R}) \subseteq K$.

$$\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\} \subseteq \{ G \cap A : G \in K\}$$

But $\{G \cap A : G \in K\} \subseteq \mathcal{B}(A)$. So we have: $$\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\} \subseteq \mathcal{B}(A)$$

Answer 2

The family $\mathcal S=\{X\subseteq\mathbb R:X\cap A\in\mathcal B(A)\}$ is a $\sigma$-algebra of subsets of $\mathbb R$, and it contains all open subsets of $\mathbb R$ (because their intersections with $A$ are open in the subspace topology of $A$). So $\mathcal S$ contains all of the Borel subsets of $\mathbb R$.

Answer 3

This question permits a very elegant and general solution, which I would like to try and present. We begin by introducing a very elementary and equally important construction. For arbitrary map $f \colon A \to B$ and subsets $X \subseteq A$ respectively $Y \subseteq B$ we denote by $f[X]$ respectively $f^{-1}[Y]$ the direct respectively inverse images through $f$ of the subsets in question. We proceed to introduce the direct and inverse image maps as follows: $$\begin{align} \hat{f} \colon \mathscr{P}(A) &\to \mathscr{P}(B)\\ \hat{f}(X)&=f[X]\\ \check{f} \colon \mathscr{P}(B) &\to \mathscr{P}(A)\\ \check{f}(Y)&=f^{-1}[Y]. \end{align}$$ We also introduce another highly important notion: given fixed set $A$, we say $\mathscr{K}$ is a closure system on $A$ if $A \in \mathscr{K} \subseteq \mathscr{P}(A)$ such that for every nonempty $\varnothing \neq \mathscr{M} \subseteq \mathscr{K}$ we have that $\bigcap \mathscr{M} \in \mathscr{K}$. In natural language, $\mathscr{K}$ is a collection of subsets of $A$ containing $A$ in particular and closed with respect to arbitrary intersections. Given arbitrary subset $X \subseteq A$ we introduce the notation: $$[X]_{\mathscr{K}}\colon=\bigcap_{\substack{M \in \mathscr{K}\\M \supseteq X}}M$$ for the $\mathscr{K}$-substructure generated by $X$, which is easily seen to be the minimum (with respect to inclusion) of the subset $\{M \in \mathscr{K} \mid M \supseteq X\}$.

We proceed to adopt the formal (and usual) definition according to which $\mathscr{M}$ is a $\sigma$-algebra on $A$ if $\varnothing \neq \mathscr{M} \subseteq \mathscr{P}(A)$ satisfies the axioms: $$\begin{align} &\left(\forall X \right)\left(X \in \mathscr{M}^{\mathbb{N}} \Rightarrow \bigcup_{n \in \mathbb{N}} X_n \in \mathscr{M}\right)\\ &(\forall X)(X \in \mathscr{M} \Rightarrow A \setminus X \in \mathscr{M}). \end{align}$$ For fixed set $A$ we denote by $\mathscr{Alg}^{\sigma}(A)$ the collection of all $\sigma$-algebras on $A$ and remark that it is a closure system on the powerset $\mathscr{P}(A)$.

Given an arbitrary map $f \colon A \to B$ we signal the following two important constructions:

1. Given a $\sigma$-algebra $\mathscr{M} \in \mathscr{Alg}^{\sigma}(A)$ it holds that $\mathscr{N}\colon=\check{f}^{-1}\left[\mathscr{M}\right] \in \mathscr{Alg}^{\sigma}(B)$. We call this $\sigma$-algebra on $B$ the direct image of $\mathscr{M}$ through $f$ and denote it by $f^{*}\left[\mathscr{M}\right]$. We also remark that it is by definition constituted of those $Y \subseteq B$ such that $f^{-1}[Y] \in \mathscr{M}$.
2. Given a $\sigma$-algebra $\mathscr{N} \in \mathscr{Alg}^{\sigma}(B)$ it holds that $\mathscr{M}\colon=\check{f}\left[\mathscr{N}\right] \in \mathscr{Alg}^{\sigma}(A)$. We call this $\sigma$-algebra on $A$ the inverse image of $\mathscr{N}$ through $f$ and denote it by $f_{*}\left[\mathscr{N}\right]$.

We make the important observation that given any two $\sigma$-algebras $\mathscr{M} \in \mathscr{Alg}^{\sigma}(A)$ and $\mathscr{N} \in \mathscr{Alg}^{\sigma}(B)$ we have the following dualisation relation: $f^{*}\left[\mathscr{M}\right] \supseteq \mathscr{N} \Leftrightarrow \mathscr{M} \supseteq f_{*}\left[\mathscr{N}\right]$.

Let us also note that in the particular case when $B \in \mathscr{M} \in \mathscr{Alg}^{\sigma}(A)$, if we denote by $\mathrm{i}^B_A \colon B \to A$ the respective inclusion map we have the following simple description: $\left(\mathrm{i}^B_A\right)_{*}\left[\mathscr{M}\right]=\{X \in \mathscr{M} \mid X \subseteq B\}$.

With these preparations in place, we can now state a very general:

Proposition. Let $f \colon A \to B$ be an arbitrary map and $\mathscr{Y} \subseteq \mathscr{P}(B)$ be an arbitrary collection of subsets of $B$. Then the relation $f_{*}\left[\left[\mathscr{Y}\right]_{\mathscr{Alg}^{\sigma}(B)}\right]=\left[\check{f}\left[\mathscr{Y}\right]\right]_{\mathscr{Alg}^{\sigma}(A)}$ holds.

Proof. For ease of notation let us write $\mathscr{X}\colon=\check{f}\left[\mathscr{Y}\right] \subseteq \mathscr{P}(A)$, $\mathscr{N}\colon=\left[\mathscr{Y}\right]_{\mathscr{Alg}^{\sigma}(B)}$, $\mathscr{M}\colon=\left[\mathscr{X}\right]_{\mathscr{Alg}^{\sigma}(A)}$ respectively $\mathscr{M'}\colon=f_{*}\left[\mathscr{N}\right]$. Our objective thus becomes to show that $\mathscr{M}=\mathscr{M'}$.

On the one hand we have by definition that $\mathscr{Y} \subseteq \mathscr{N}$ whence $\mathscr{X}=\check{f}\left[\mathscr{Y}\right] \subseteq \check{f}\left[\mathscr{N}\right]=\mathscr{M'}$. Since $\mathscr{M'}$ is a $\sigma$-algebra on $A$ including $\mathscr{X}$ we infer that $\mathscr{M} \subseteq \mathscr{M'}$.

On the other hand, since by construction $\mathscr{M} \supseteq \mathscr{X}=\check{f}\left[\mathscr{Y}\right]$ we have $f^{*}\left[\mathscr{M}\right]=\check{f}^{-1}\left[\mathscr{M}\right] \supseteq \check{f}^{-1}\left[\check{f}\left[\mathscr{Y}\right]\right] \supseteq \mathscr{Y}$. Since $f^{*}\left[\mathscr{M}\right]$ is a $\sigma$-algebra on $B$ including $\mathscr{Y}$ we infer that $f^{*}\left[\mathscr{M}\right] \supseteq \mathscr{N}$ which by virtue of the dualisation relation mentioned above leads to $\mathscr{M} \supseteq \mathscr{M'}$, thus bringing our argument to an end. $\Box$

Given a fixed topology $\mathscr{T}$ on a set $X$ let us abbreviate by $\mathscr{B}\left(\mathscr{T}\right)\colon=\left[\mathscr{T}\right]_{\mathscr{Alg}^{\sigma}(X)}$ the Borel $\sigma$-algebra associated to the space $(X, \mathscr{T})$. If we adopt notations similar to the ones above for initial and final topologies induced by arbitrary maps, the general proposition above applies in the following particular form to Borel $\sigma$-algebras:

Corollary. Let $(Y, \mathscr{T})$ be an arbitrary space and $f \colon X \to Y$ an arbitrary map. We then have the relation $f_{*}\left[\mathscr{B}\left(\mathscr{T}\right)\right]=\mathscr{B}\left(f_{*}\left[\mathscr{T}\right]\right)$.

The result you are interested in is a special case of the above corollary, when the map $f$ is an inclusion map.