Sigma-Algebras and Measurable Spaces

Let $X$ be a set. A collection $\mathcal{A} \subseteq \mathcal{P}(X)$ is an algebra of sets if:

1. $X \in \mathcal{A}$.
2. $A \in \mathcal{A}$ implies $A^c \in \mathcal{A}$ (closed under complements).
3. $A, B \in \mathcal{A}$ implies $A \cup B \in \mathcal{A}$ (closed under finite unions).

From these axioms it follows that $\varnothing \in \mathcal{A}$, $\mathcal{A}$ is closed under finite intersections ($A \cap B = (A^c \cup B^c)^c$), and under set difference ($A \setminus B = A \cap B^c$).

An algebra $\mathcal{F}$ is called a $\sigma$-algebra (or sigma-algebra) if it is also closed under countable unions: if $\{A_n\}_{n=1}^{\infty} \subseteq \mathcal{F}$, then $\bigcup_{n=1}^{\infty} A_n \in \mathcal{F}$.

The pair $(X, \mathcal{F})$ is called a measurable space.

Proposition 1.1. A $\sigma$-algebra is also closed under countable intersections and countable complements:

$\bigcap_{n=1}^{\infty} A_n = \left(\bigcup_{n=1}^{\infty} A_n^c\right)^c$

If $\mathcal{C} \subseteq \mathcal{P}(X)$ is any collection of subsets of $X$, the $\sigma$-algebra generated by $\mathcal{C}$, denoted $\sigma(\mathcal{C})$, is the smallest $\sigma$-algebra containing $\mathcal{C}$. It equals the intersection of all $\sigma$-algebras containing $\mathcal{C}$:

$\sigma(\mathcal{C}) = \bigcap\{\mathcal{F} : \mathcal{C} \subseteq \mathcal{F},\ \mathcal{F} \text{ is a } \sigma\text{-algebra}\}$

Definition. Let $X$ be a topological space with topology $\tau$. The Borel $\sigma$-algebra $\mathcal{B}(X)$ is $\sigma(\tau)$, the $\sigma$-algebra generated by the open sets. Elements of $\mathcal{B}(X)$ are called Borel sets.

Proposition 1.2. In $\mathbb{R}^n$, $\mathcal{B}(\mathbb{R}^n) = \sigma(\mathcal{O}) = \sigma(\mathcal{C}) = \sigma(\mathcal{K})$, where $\mathcal{O}$ is the collection of open sets, $\mathcal{C}$ is the collection of closed sets, and $\mathcal{K}$ is the collection of compact sets.

Proposition 1.3. $\mathcal{B}(\mathbb{R})$ contains all intervals: $(a, b)$, $[a, b]$, $(a, b]$, $[a, b)$ for $a, b \in \mathbb{R} \cup \{-\infty, +\infty\}$.

Example 1. For any set $X$, $\{\varnothing, X\}$ and $\mathcal{P}(X)$ are $\sigma$-algebras (the trivial and discrete $\sigma$-algebras).

Example 2. The countable-cocountable $\sigma$-algebra on $X$: $\mathcal{F} = \{A \subseteq X : A \text{ is countable or } A^c \text{ is countable}\}$.

Example 3. On $\mathbb{R}$, the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ is generated by intervals of the form $(a, \infty)$ with $a \in \mathbb{R}$.

• Confusing algebras with $\sigma$-algebras. An algebra is closed only under finite unions; a $\sigma$-algebra requires countable unions. The collection of finite and cofinite subsets of $\mathbb{N}$ is an algebra but not a $\sigma$-algebra.
• Thinking the Borel $\sigma$-algebra contains all subsets of $\mathbb{R}$. It does not. The existence of non-Lebesgue-measurable sets (using the axiom of choice) shows $\mathcal{P}(\mathbb{R})$ is strictly larger than $\mathcal{B}(\mathbb{R})$.
• Assuming generated $\sigma$-algebras are easy to describe explicitly. The Borel $\sigma$-algebra is enormous; there is no constructive way to list all its elements.
• Forgetting that $\sigma(\mathcal{C})$ is the intersection of all $\sigma$-algebras containing $\mathcal{C}$. This definition is useful but non-constructive; it does not tell us what the elements look like.

Problem 1. Show that the countable-cocountable $\sigma$-algebra on an uncountable set $X$ is not generated by a countable collection of subsets.

Solution. Suppose $\mathcal{C}$ is a countable collection generating $\mathcal{F}$. Then each $A \in \mathcal{F}$ is obtained from $\mathcal{C}$ by countably many set operations, so every non-cocountable set in $\mathcal{F}$ must be countable. But $\mathcal{C}$ itself is countable, so the $\sigma$-algebra it generates is at most the size of the continuum, while the cocountable $\sigma$-algebra on an uncountable $X$ contains uncountably many cocountable sets (each complement of a singleton). Contradiction. $\blacksquare$

Problem 2. Prove that if $\mathcal{F}$ is a $\sigma$-algebra and $\{A_n\} \subseteq \mathcal{F}$, then $\limsup_{n\to\infty} A_n = \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k \in \mathcal{F}$.

Solution. For each $n$, $\bigcup_{k=n}^\infty A_k \in \mathcal{F}$ (countable union). The intersection of these sets is then also in $\mathcal{F}$ (countable intersection). So $\limsup A_n \in \mathcal{F}$. Similarly $\liminf A_n \in \mathcal{F}$. $\blacksquare$

• Probability theory: Kolmogorov’s axioms define a probability space $(\Omega, \mathcal{F}, P)$ where $\mathcal{F}$ is a $\sigma$-algebra of events. Filtrations (increasing families of $\sigma$-algebras) model information flow in stochastic processes.
• Statistics: Sufficient statistics are defined in terms of conditional expectations, which require $\sigma$-algebras generated by the statistic. The Lehmann-Scheffé theorem uses $\sigma$-algebras for minimal sufficient statistics.
• Ergodic theory: Invariant $\sigma$-algebras capture the long-term behaviour of dynamical systems. The ergodic theorem relates time averages to space averages via conditional expectation on the invariant $\sigma$-algebra.
• Economics: In financial mathematics, the $\sigma$-algebra generated by asset prices models the information available to traders, and martingale pricing uses filtrations.