# Some clarification needed on what it means for a $\sigma$-algebra to be generated, Borel $\sigma$-algebra

Here is where my real analysis textbook explains what it means for something to generate a $$\sigma$$-algebra, and subsequently what a Borel $$\sigma$$-algebra is. I couldn't really follow what it was saying.

Lemma 2.7. If $$A_\alpha$$ is a $$\sigma$$-algebra for each $$\alpha$$ in some non-empty index set $$I$$, then $$\cap_{\alpha \in I} \mathcal{A}_\alpha$$ is a $$\sigma$$-algebra.

Proof. This follows immediately from the definition.$$\tag*{\square}$$If we have a collection $$\mathcal{C}$$ of subsets of $$X$$, define$$\sigma(\mathcal{C}) = \cap\{\mathcal{A}_\alpha : \mathcal{A}_\alpha \text{ is a }\sigma\text{-algebra, }\mathcal{C} \subset \mathcal{A}_\alpha\},$$the intersection of all $$\sigma$$-algebras containing $$\mathcal{C}$$. Since there is at least one $$\sigma$$-algebra containing $$\mathcal{C}$$, namely, the one consisting of all subsets of $$X$$, we are never taking the intersection over an empty class of $$\sigma$$-algebras. In view of Lemma 2.7, $$\sigma(\mathcal{C})$$ is a $$\sigma$$-algebra. We call $$\sigma(\mathcal{C})$$ the $$\sigma$$-algebra generated by the collection $$\mathcal{C}$$, or say that $$\mathcal{C}$$ generates the $$\sigma$$-algebra $$\sigma(\mathcal{C})$$. it is clear that if $$\mathcal{C}_1 \subset \mathcal{C}_2$$, then $$\sigma(\mathcal{C}_1) \subset \sigma(\mathcal{C}_2)$$. since $$\sigma(\mathcal{C})$$ is a $$\sigma$$-algebra, then $$\sigma(\sigma(\mathcal{C})) = \sigma(\mathcal{C})$$.

If $$X$$ has some additional structure, say, it is a metric space, then we can talk about open sets. If $$\mathcal{G}$$ is the collection of open subsets of $$X$$, then we call $$\sigma(\mathcal{G})$$ the Borel $$\sigma$$-algebra on $$X$$, and this is often denoted $$\mathcal{B}$$. Elements of $$\mathcal{B}$$ are called Borel sets and are said to be Borel measurable. We will see later that when $$X$$ is the real line $$\mathcal{B}$$ is not equal to the collection of all subsets of $$X$$.

My questions are as follows.

1. I get quite confused/muddled when we consider a set of $$\sigma$$-algebras, i.e. a set of sets of subsets of a set, as per somewhere in the above. Can anyone tell me how they their thinking clear when playing with these things?
2. Could anybody explain to me how they think about $$\sigma$$-algebras generated by a collection?
3. Why are Borel $$\sigma$$-algebras important, i.e. why should we care specifically about the $$\sigma$$-algebra generated by the collection of open subsets of $$X$$? .
• For 3: We can use it to develop,(among other things) Lebesgue measure and Lebesgue integration, which extends the family of real integrable functions to a much wider class than classical integration. It has applications in many subjects, including quantum mechanics. Aug 24, 2016 at 18:11

This an explanation of why the definition in your textbook is equivalent to point $$(2)$$ in the answer by Alexis Olson; it’s much too long for a comment. It does require a little familiarity with transfinite recursion; in particular, you need to know that $$\omega_1$$ is the first uncountable ordinal. I’ve added a couple of analogous closure constructions in hopes of making the idea a bit clearer.

Let $$\mathscr{A}$$ be a collection of subsets of $$X$$. We can define $$\sigma(\mathscr{A})$$ from the outside in, as the intersection of all $$\sigma$$-algebras containing $$\mathscr{A}$$, or from the inside out, by adding to $$\mathscr{A}$$ the bare minimum collection of subsets of $$X$$ needed to expand $$\mathscr{A}$$ to a $$\sigma$$-algebra.

To do the latter, let $$\mathscr{A}_0=\mathscr{A}$$. Given $$\mathscr{A}_\alpha$$ for some ordinal $$\alpha<\omega_1$$, let $$\mathscr{C}_\alpha=\{X\setminus A:A\in\mathscr{A}_\alpha\}$$, the set of complements of members of $$\mathscr{A}_\alpha$$. Then let

$$\mathscr{A}_{\alpha+1}=\left\{\bigcup\mathscr{S}:\mathscr{S}\subseteq\mathscr{C}_\alpha\text{ and }\mathscr{S}\text{ is countable}\right\}\;,$$

the family of all unions of countable subcollections of $$\mathscr{C}_\alpha$$.

Example. If $$X$$ is a topological space, and $$\mathscr{A}$$ is the family of all open sets in $$X$$, then $$\mathscr{C}_0$$ is the family of all closed sets in $$X$$. $$\mathscr{A}_1$$ is the family of all unions of countably many closed sets, so it’s the family of all $$F_\sigma$$-sets in $$X$$. $$\mathscr{C}_1$$ is the family of all complements of $$F_\sigma$$-sets in $$X$$, so it’s the family of all $$G_\delta$$-subsets of $$X$$. (The complement of a union of countably many closed sets is the intersection of countably many open sets.)

If $$\alpha$$ is a limit ordinal, let

$$\mathscr{A}_\alpha=\bigcup_{\xi<\alpha}\mathscr{A}_\xi\;,$$

and keep going.

Proposition. $$\mathscr{A}_{\omega_1}$$ is a $$\sigma$$-algebra.

Proof. If $$A\in\mathscr{A}_{\omega_1}$$, then there is an $$\alpha<\omega_1$$ such that $$A\in\mathscr{A}_\alpha$$. Clearly $$X\setminus A\in\mathscr{C}_\alpha$$, and $$\{X\setminus A\}$$ is a countable subset of $$\mathscr{C}_\alpha$$, so $$\bigcup\{X\setminus A\}=X\setminus A\in\mathscr{A}_{\alpha+1}\subseteq\mathscr{A}_{\omega_1}$$. Thus, $$\mathscr{A}_{\omega_1}$$ is closed under complementation.

Let $$\mathscr{S}$$ be a countable subset of $$\mathscr{A}_{\omega_1}$$. For each $$S\in\mathscr{S}$$ there is an $$\alpha(S)<\omega_1$$ such that $$S\in\mathscr{A}_{\alpha(S)}$$. Then $$\{\alpha(S):S\in\mathscr{S}\}$$ is a countable set of ordinals less than $$\omega_1$$, so there is a limit ordinal $$\beta<\omega_1$$ such that $$\alpha(S)<\beta$$ for each $$S\in\mathscr{S}$$. By definition $$\mathscr{A}_{\alpha(S)}\subseteq\mathscr{A}_\beta$$ for each $$S\in\mathscr{S}$$, so $$\mathscr{S}\subseteq\mathscr{A}_\beta$$, and hence $$\{X\setminus S:S\in\mathscr{S}\}\subseteq\mathscr{C}_\beta$$.

We showed in the first paragraph that $$\mathscr{C}_\beta\subseteq\mathscr{A}_{\beta+1}$$, so $$\{X\setminus S:S\in\mathscr{S}\}\subseteq\mathscr{A}_{\beta+1}$$, and therefore, taking complements again, $$\mathscr{S}\subseteq\mathscr{C}_{\beta+1}$$. That is, $$\mathscr{S}$$ is a countable subset of $$\mathscr{C}_{\beta+1}$$, so by definition $$\bigcup\mathscr{S}\in\mathscr{A}_{\beta+2}\subseteq\mathscr{A}_{\omega_1}$$, and $$\mathscr{A}_{\omega_1}$$ is therefore closed under taking countable unions. Thus, $$\mathscr{A}_{\omega_1}$$ is a $$\sigma$$-algebra containing $$\mathscr{S}$$.

At every stage of the recursive construction of $$\mathscr{A}_{\omega_1}$$ we added only sets that have to be in any $$\sigma$$-algebra containing $$\mathscr{S}$$: we added only complements of sets that we already had and countable unions of sets that we already had. We just kept adding these required sets until we got to a point at which we could prove that no further additions were required. We could keep going, letting $$\mathscr{C}_{\omega_1}=\{X\setminus A:A\in\mathscr{A}_{\omega_1}\}$$ and so on, but all of the families $$\mathscr{C}_\alpha$$ and $$\mathscr{A}_\alpha$$ for $$\alpha\ge\omega_1$$ are just $$\mathscr{A}_{\omega_1}$$ all over again, since that family is already a $$\sigma$$-algebra: taking complements and finite unions gets us nothing new at this point.

Analogies. In a topological space $$X$$ we can define the closure of a set $$A$$ from the outside in, as the intersection of all closed sets containing $$A$$, or from the inside out, as $$A\cup A'$$, where $$A'$$ is the set of accumulation points of $$A$$. The outside-in approach works because the intersection of closed sets is closed. The inside-out approach works because we don’t add to $$A$$ anything that doesn’t have to be in every closed set containing $$A$$, and we do add enough to get a closed set.

In a group $$G$$ we can define the subgroup $$\langle X\rangle$$ generated by a subset $$X$$ of $$G$$ from the outside in, as the intersection of all subgroups of $$G$$ containing $$X$$, or from the inside out, by recursively closing $$X$$ under the group operation and under taking inverses in a manner similar to the construction of $$\mathscr{A}_{\omega_1}$$ above; the main difference is that only $$\omega$$ steps are required instead of $$\omega_1$$, so that we can actually write down the generated group at one go, as in this answer.

1. The intersection does have a few levels. $\mathcal{A}_I = \bigcap_{\alpha \in I} \mathcal{A}_\alpha$ is an intersection of $\sigma$-algebras and each $\sigma$-algebra is a collection of sets. For a set $S$ to be in the $\sigma$-algebra $\mathcal{A}_I$, it must be in every $\mathcal{A}_\alpha$ for $\alpha \in I$. In general, you have to keep careful track of notation. Write out the objects at each level.

2. A $\sigma$-algebra generated by a collection of sets $\{S_\beta\}_{\beta \in J}$ (for some indexing set $J$) is the collection of all sets that can be generated by a countable amount of union, complement, and intersection operations on the sets $S_\beta$.

3. The Borel $\sigma$-algebra is important in measure theory. Since a measure must satisfy $\sigma$-additivity (a.k.a. countable additivity), if a measure is defined on the open sets, then it has to be defined on the entire Borel $\sigma$-algebra.

• thanks for the answer! Is it possible you could elaborate on 3 a bit, and why is what you wrote for 2 equivalent to what was written in my textbook?
– user362105
Aug 24, 2016 at 7:57