## 1.1. \(\sigma\)-Algebras

≪ 1. Measures | Table of Contents | 1.2. Measures and Their Properties ≫In a metric space \((X, \rho)\), it makes sense to assign a distance between any two points. Similarly, we’d like to assign a *measure* to every subset of a set \(X\), but as mentioned before, this isn’t always possible without creating contradictions. So, we need to spend some time describing the structure of which sets can be measured and which ones can’t.

**Definition 1. \(\sigma\)-Algebras**

Let \(X\) be a set. A \(\sigma\)-algebra (of subsets of \(X\)) is a nonempty subset \(\Sigma\) of the power set of \(X\) such that:

- If \(A\in\Sigma\), the complement \(A^\complement\in\Sigma\).
- If \(\lbrace A_n\rbrace\) is a sequence in \(\Sigma\), then their union \(\bigcup_{n=0}^\infty A_n\) is also in \(\Sigma\).

These two properties can be made a little stronger:

**Proposition 2. **

Let \(\Sigma\) be a \(\sigma\)-algebra on a set \(X\). Then,

- \(\varnothing, X\in\Sigma\).
- If \(\lbrace A_n\rbrace\) is a sequence in \(\Sigma\), then their intersection \(\bigcap_{n=0}^\infty A_n\) is also in \(\Sigma\).
- \(\Sigma\) is closed under
*finite*unions and intersections.

**Proof:** We’ll first show that \(\Sigma\) is closed under countable intersections. Let \(\lbrace A_n\rbrace\) be a sequence in \(\Sigma\). Then,
\[\bigcap_{n=0}^\infty A_n=\left(\bigcup_{n=0}^\infty A_n^\complement\right)^\complement.\]
Since \(\Sigma\) is closed under countable unions and complements, this intersection is in \(\Sigma\). Next, let \(A\in\Sigma\) be any set. Define the sequence \(A_0=A\) and \(A_n=A^\complement\) for \(n\geq 1\); then \(\bigcup_{n=0}^\infty A_n=X\in\Sigma\) and \(\bigcap_{n=0}^\infty A_n=\varnothing\in\Sigma\). Finally, if \(A_0,\ldots, A_n\in\Sigma\), we can consider that
\[\bigcup_{j=0}^nA_j=\left(\bigcup_{j=0}^n A_j\right)\cup\left(\bigcup_{j=n+1}^\infty\varnothing\right)\in\Sigma,\]
as that’s a countable union in \(\Sigma\). Replace \(\varnothing\) with \(X\) to get countable intersections. \(\square\)

Clearly, for any set \(X\), \(\lbrace\varnothing, X\rbrace\) and \(\mathcal P(X)\) are both \(\sigma\)-algebras on \(X\). If \(A\subseteq X\) in general, then \(\lbrace \varnothing, A, A^\complement, X\rbrace\) is also a \(\sigma\)-algebra. However, other \(\sigma\)-algebras can be much harder to describe: what about a \(\sigma\)-algebra that contains two different subsets of \(X\)?

**Proposition 3. **

Let \(X\) be a set, and let \(S\subset\mathcal P(X)\). Let \[\mathcal G=\lbrace\Sigma\subset\mathcal P(X) : S\subset\Sigma\text{ and }\Sigma\text{ is a }\sigma\text{-algebra.}\rbrace.\] Define \(\Sigma(S)=\bigcap_{\Sigma\in\mathcal G}\Sigma\). Then \(\Sigma(S)\) is a \(\sigma\)-algebra containing \(S\), and if \(\Sigma\) is another \(\sigma\)-algebra containing \(S\), \(\Sigma(S)\subseteq\Sigma\). \(\Sigma(S)\) is called the "\(\sigma\)-algebra generated by \(S\)".

**Proof:** Let \(A\in\Sigma(S)\). Then, \(A\in\Sigma\) for all \(\Sigma\in\mathcal G\), and since each of these \(\Sigma\)’s is a \(\sigma\)-algebra, we get that \(A^\complement\in\Sigma\) for all \(\Sigma\in\mathcal G\). Hence, \(A^\complement\in\Sigma(S)\). Similarly, if \(\lbrace A_n\rbrace\) is a sequence in \(\Sigma(S)\), then \(\lbrace A_n\rbrace\) is a sequence in \(\Sigma\) for all \(\Sigma\in\mathcal G\). It then follows that \(\bigcup_{n=0}^\infty A_n\in\Sigma(S)\). As \(S\in\Sigma\) for all \(\Sigma\in\mathcal G\) by definition, we conclude that \(\Sigma(S)\) is a \(\sigma\)-algebra containing \(S\). If \(\Sigma\) is any \(\sigma\)-algebra containing \(S\), then \(\Sigma\in\mathcal G\implies \Sigma(S)\subseteq\Sigma.\ \square\)

If \(X\) is a topological space, the \(\sigma\)-algebra generated by the open sets is called the *Borel \(\sigma\)-algebra*; elements of it are called *Borel sets*. We’ll see these come up again later, as they allow us to connect measures and measurable spaces to continuous functions defined between them.

Finally, if \((X, \Sigma)\) and \((Y, \Tau)\) are both measurable spaces, we can define the *product \(\sigma\)-algebra* on \(X\times Y\) as the \(\sigma\)-algebra generated by
\[\lbrace A\times B:A\in\Sigma,\ B\in\Tau\rbrace.\]
Some remarks are in order. First of all, the Borel \(\sigma\)-algebra on \(\R\) is very large, but it isn’t quite all of \(\mathcal P(\R)\). This is because of the existence of non-measurable sets such as the Vitali set. In fact, every subset of \(\R\) with positive “length” contains a non-measurable set! However, the existence of such sets depends on the axiom of choice. We’ll include a section discussing such sets in greater detail later.

Second of all, we saw in Proposition 3 that the structure of a \(\sigma\)-algebra is preserved by intersections — even if it’s *uncountable*. This is generally far from true for unions, and it’s quite easy to come up with counterexamples involving unions of small \(\sigma\)-algebras.

To end things off awkwardly, we’ll note the somewhat unfamiliar choice of terminology: what the heck do the “\(\sigma\)” and “algebra” in “\(\sigma\)-algebra” mean?

At first, I thought the “algebra” referred to the funny vector spaces with vector multiplication, but it actually refers to a boolean algebra. If you’re a logic enjoyer, this comes from \(x\in A\lor x\in B\iff x\in A\cup B\): the union operation acts like a logical or (very loosely). Similarly, taking complements acts as a negation.

The \(\sigma\) then comes from the German word *summe*, which translates to *sum* (or in this context, union). Specifically, it seems to refer to *countable* sums. The character \(\delta\) originates from the German word *Durchschnitt*, or *intersection* in English. This MathOverflow discussion contains much more information on this.