Definitions of Algebra and Measure

24 Feb 2025

1. Vitali’s Theorem

Definition. A measure $\mu$ on a set $X$ satisfies the following conditions:

1. $\mu(\varnothing) = 0$
2. For any countable collection of pairwise disjoint sets $\lbrace A_n \rbrace$, $\mu\left( \bigcup A_n \right) = \sum \mu(A_n)$

Unfortunately, when $X = \mathbb{R}$, the domain of measure cannot range over all subsets of the real numbers.

Vitali’s Theorem. There does not exist a measure $\mu$ on $\mathbb{R}$ that satisfies all of the following conditions:

1. It is not identically zero.
2. It is invariant under translation. That is, for any $r \in \mathbb{R}$, $\mu(A + r) = \mu(A)$.
3. $\operatorname{dom} \mu = \mathcal{P}(\mathbb{R})$

Proof. Suppose such a measure $\mu$ exists. Define the following equivalence relation on $\mathbb{R}$:

\[x \sim y \iff x - y \in \mathbb{Q}\]

By the axiom of choice, there exists a choice function $\iota$ for $[0, 1]/{\sim}$. Let $V = \operatorname{im} \iota$. We call $V$ a Vitali set. For example, $V = \lbrace 0.1, \pi - 3, \sqrt{2} - 1, \dots \rbrace$. We shall show that $\mu$ cannot be defined on $V$.

By the definition of $V$, for $q \in \mathbb{Q}$, $V$ and $V + q$ are disjoint. Moreover, $[0, 1] \subset \bigcup_{q \in \mathbb{Q}} (V + q) \subset [-1, 3]$. Therefore,

\[1 \leq \sum_{q \in \mathbb{Q}}\mu(V + q) \leq 4.\]

However, since $\mu(V + q) = \mu(V)$, if $\mu(V) = 0$, then the left inequality fails, and if $\mu(V) > 0$, then the right inequality fails. This is a contradiction. ■

2. $\sigma$-algebra

Therefore, to construct a proper measure, it is necessary to appropriately restrict the domain of the measure. To this end, we introduce the concept of an algebra.

Definition. An algebra $\mathcal{A}_0$ on a set $X$ is a collection of sets satisfying the following conditions:

1. $\varnothing, X \in \mathcal{A}_0$
2. $A \in \mathcal{A}_0 \implies A^c \in \mathcal{A}_0$
3. $A, B \in \mathcal{A}_0 \implies A \cup B \in \mathcal{A}_0$

Remark. By axiom 2 and De Morgan’s laws, axiom 3 also implies that $A \cap B \in \mathcal{A}_0$.

Strengthening the third condition of an algebra to countable unions yields the definition of a $\sigma$-algebra. That is,

Definition. A $\sigma$-algebra $\mathcal{A}$ on a set $X$ is an algebra satisfying the following condition:

1. $\lbrace A_n \rbrace _{n \in \mathbb{N}} \subset \mathcal{A}_0 \implies \bigcup_{n \in \mathbb{N}} A_n \in \mathcal{A}$

Theorem. If $\lbrace \mathcal{A}_i \rbrace_{i \in I}$ is a collection of $\sigma$-algebras on $X$, then $\bigcap_{i \in I}\mathcal{A}_i$ is also a $\sigma$-algebra on $X$.

Proof. This can be proved easily from the definition of a $\sigma$-algebra. However, to spice things up, we provide a slightly unconventional proof. By the Łoś-Tarski theorem, a first-order theory is preserved under intersections if and only if every sentence of the first-order theory is a $\Pi_1$ sentence. Since all three axioms of a $\sigma$-algebra are $\Pi_1$ sentences, $\sigma$-algebras are preserved under intersections. ■

Corollary. Let $\mathcal{C}$ be a collection of subsets of $X$. There exists the smallest $\sigma$-algebra containing $\mathcal{C}$. Such $\sigma$-algebra is denoted $\sigma(\mathcal{C})$.

Proof. Let $\mathcal{S} = \lbrace \mathcal{A} : \mathcal{A} \text{ is a $\sigma$-algebra containing } \mathcal{C} \rbrace$. Since $\mathcal{P}(X) \in \mathcal{S}$, $\mathcal{S}$ is non-empty. Then $\sigma(\mathcal{C}) = \bigcap_{\mathcal{A} \in \mathcal{S}} \mathcal{A}$.

As a representative example of a $\sigma$-algebra, let us examine the Borel $\sigma$-algebra.

Definition. Let $\mathcal{G}$ be the collection of open sets in $\mathbb{R}$. We call $\sigma(\mathcal{G})$ the Borel $\sigma$-algebra and denote it by $\mathcal{B}$.

The Borel $\sigma$-algebra can also be understood through the Borel hierarchy. Let $\Sigma_1$ be the collection of open sets and $\Pi_1$ be the collection of closed sets. We define:

• $\Delta_n = \Sigma_n \cap \Pi_n$
• $\Sigma_{n + 1} = \lbrace S : \exists \lbrace P_n \rbrace _{n \in \mathbb{N}} \subset \Pi_n \; S = \bigcup_{n \in \mathbb{N}} P_n \rbrace$
• $\Pi_n = \lbrace P : \exists S \in \Sigma_n \; P = S^c \rbrace$

That is, $\Sigma_2 = F_\sigma$ and $\Pi_2 = G_\delta$. Note that when we think of union as $\exists$ and intersection as $\forall$, the form of the definition is similar to the arithmetic hierarchy.

Theorem. $\mathcal{B} = \Sigma_{\omega_1} = \Pi_{\omega_1} = \Delta_{\omega_1}$

Proof. Omitted. However, this can be understood intuitively. Since $\mathcal{B}$ must be closed under countable intersections, countable unions, and complements, we have $\Sigma_\alpha, \Pi_\alpha, \Delta_\alpha \subset \mathcal{B}$ for all countable ordinals $\alpha$. From this fact, we obtain the theorem by transfinite induction. ■

There is a known method for defining a measure from an algebra and a premeasure on some set. This method is called the Carathéodory method. Since premeasures are very easy to construct, using this method allows us to construct measures very easily. The Carathéodory method will be discussed in the next post.