Measure Spaces

Author

John Robin Inston

Published

September 25, 2026

1 Concept of Measure

For general set \(X\subseteq \mathbb{R}\) we wish to construct a function \(\mu:2^{X}\to[0,\infty]\) that assigns each subset \(E\subset X\) a number \(\mu(E)\in[0,\infty]\), some measure of \(E\) where \(\mu\) satisfies:

  1. For disjoint \(E_{1}, E_{2}, \dots \in X\) we have \(\mu\left( \bigcup_{j=1}^\infty E_{j} \right)=\sum_{j=1}^\infty \mu(E_{j})\) (countably additive);
  2. For interval \([a,b]\) we have \(\mu([a,b])=b-a\); and
  3. For set \(E\subset \mathbb{R}\) and constant \(c\in \mathbb{R}\) we have \(\mu(E+c)=\mu(E)\) (translation invariance).

The Vitalli Non-Measurability Theorem demonstrates that no such function exists. Indeed the closest mathematical generalization of mass is σ-additivity which gives rise to the Lebesgue measure, however not all sets have such a measure (for example, a Vitali set).

To overcome this problem we turn to some standard approaches for mathematicians, either:

  1. Weaken the criteria; or
  2. Restrict the domain of the function to the family of well-behaved sets.

Both options are valid and we will see that (1) leads us to the concept of outer-measure, and (2) leads to the concept of measure spaces and measures.

Let us consider how we might restrict the domain of the function and what kind of family of subsets we should restrict to. The answer is a non-empty collection of subsets that is both closed under compliments and closed under countable unions (i.e. sets that preserve the properties we wish our measure to have) also known as a \(\sigma\)-algebra, a specific type of algebra.

2 Algebra & σ-Algebra

An algebra is a non-empty collection of subsets of a set \(X\) that is closed under finite unions and compliments.

A collection \(\mathcal{A}\subseteq 2^X\) is an algebra of subsets of \(X\) if it satisfies the following properties:

  1. \(\mathcal{A}\) is non-empty.
  2. Closed under finite unions. \(\{E_i\}_{i=1}^n\subseteq\mathcal{A}\implies\bigcup_{i=1}^nE_i \in\mathcal{A}\).
  3. Closed under compliments. \(E\in\mathcal{A}\implies E^c\in\mathcal{A}\).

The following results follow immediately from the definition of an algebra.

If \(\mathcal{A}\) is an algebra then:

  1. \(\emptyset\in\mathcal{A}\), \(X\in\mathcal{A}\).
  2. \(\{E_i\}_{i=1}^n\subseteq\mathcal{A}\implies \bigcap_{i=1}^nE_i\in\mathcal{A}\).

Proof. For (1) since \(\mathcal{A}\) is non-empty there exists \(E\in\mathcal{A}\) and also \(E^c\in\mathcal{A}\) by closure under compliments. Then \(\emptyset=E\cap E^c\in\mathcal{A}\) by closure under finite unions. Additionally \(X=\emptyset^c\in\mathcal{A}\) by closure under compliments. For (2) we have that

A algebra is closed under countable unions when

\[ \{E_i\} _{i=1}^\infty\subseteq\mathcal{A}\implies\bigcup_{i=1}^\infty E_i \in\mathcal{A}. \]

An algebra \(\mathcal{F}\subseteq 2^X\) that is closed under countable unions (a property known as σ-additivity) is called a σ-algebra.

The following results follow immediately from the definition of a \(\sigma\)-algebra and will be used frequently in the development of measure theory.

A \(\sigma\)-algebra is also closed under countable intersections

A \(\sigma\)-algebra \(\mathcal{F}\) is closed under countable intersections.

Proof. Assume that \(\mathcal{A}\) is closed under countable disjoint unions and let \(\{E_i\}_{i=1}^\infty\subseteq\mathcal{F}\). We can write

\[ \bigcap_{i=1}^\infty E_i = \left( \bigcup_{i=1}^\infty E_i^c \right)^c\in \mathcal{F}, \]

where the first equality follows from DeMorgan’s law and the remainder of the proof follows from closure under countable unions and compliments of \(\mathcal{F}\).

A double \((X, \mathcal{F})\) consisting of a set \(X\) and a σ-algebra \(\mathcal{F}\) of subsets of \(X\) is called a measurable space. Any set \(E\in\mathcal{F}\) is called a measurable set.

In measure theory we will spend a lot of time working with σ-algebras and thus we spend some time deriving the most important results about them.

Given any nonempty collection \(\mathcal{C}\) of \(\sigma\)-algebras on \(X\), then

\[ \bigcap \mathcal{C} \coloneqq \{E\subseteq X: E\in \mathcal{A}, \forall \mathcal{A}\in\mathcal{C}\}, \]

is a \(\sigma\)-algebra on \(X\).

Proof. To show this holds

Given a set \(\mathcal{E}\subseteq 2^X\) there exists a smallest \(\sigma\)-algebra containing \(\mathcal{E}\), denoted \(\sigma(\mathcal{E})\), and referred to as the \(\sigma\)-algebra generated by \(\mathcal{E}\).

Proof. To show

By the smallest we mean that, all other \(\sigma\)-algebra \(\mathcal{F}\) containing \(\mathcal{E}\) satisfy \(\sigma(\mathcal{E})\subseteq\mathcal{F}\). You can also think of \(\sigma(\mathcal{E})\) as the intersection of all other \(\sigma\)-algebra that contain \(\mathcal{E}\).

If \(\mathcal{E}\subset\sigma(\mathcal{A})\) then \(\sigma(\mathcal{E})\subset\sigma(\mathcal{A}).\)

Proof. Assume that \(\mathcal{E}\subset \sigma(\mathcal{A})\), then

3 Measure

Upon a measurable space we are able to define a measure, a function assigning elements of \(\mathcal{F}\) a measure of size.

Given a measurable space \((X,\mathcal{F})\) a measure \(\mu:\mathcal{F}\to[0,+\infty]\) is a mapping frm the collection of measurable sets to \([0,+\infty]\) such that: 1. \(\mu(\emptyset)=0\) 2. \(\{ E_{i} \}_{i=1}^\infty \subseteq \mathcal{F}~\&~\text{disjoint}\implies \mu(\cup_{i=1}^\infty E_{i})=\sum_{i=1}^\infty \mu(E_{i})\).

We call the triple \((X, \mathcal{F}, \mu)\) a measure space.

For \((X,\mathcal{F})=(X, 2^X)\) we fix \(x_{0}\in X\) and define

\[ \mu(A)=\delta_{x_{0}}(A)=\begin{cases} 1&x_{0}\in A \\ 0& \text{o.w.} \end{cases} \]

known as the Dirac measure.

For any measure space \((X,\mathcal{F},\mu)\) and \(A,B\in \mathcal{F}\), \(\{ A_{i} \}_{i=1}^n\subseteq \mathcal{F}\) we have the following properties: 1. \(A \subseteq B\implies \mu(A)\leq \mu(B)\) (Monotonicity) 2. \(A \subseteq B\implies \mu(B\setminus A)=\mu(B)-\mu(A)\) 3. \(\mu(\cup_{i=1}^\infty A_{i})\leq \sum_{i=1}^\infty \mu(A_{i})\) (Countable subadditivity) 4. \(A_{i}\subseteq A_{i+1},~\forall i\implies \mu(\cup_{i=1}^\infty A_{i})=\lim_{_{i \to \infty}}\mu(A_{i})\) (Continuity from below) 5. \(A_{i}\supseteq A_{i+1},~\forall i\) where \(\mu(A_1)<+\infty\implies \mu(\cap_{i=1}^\infty A_{i})=\lim_{_{i \to \infty}}\mu(A_{i})\) (Continuity from above)

Proof.

      1. Assume there exists \(A\subseteq B\) then we can partition \(B=A\cup(B\setminus A)\) and so \[ \mu(B)=\mu(A)+\mu(B\setminus A)\geq \mu(A). \]
  2. Similarly, we can define \(A_{1}=B_{1}\) and \(B_{n}=A_{n}\setminus \cup_{i=1}^{n-1}A_{i}\) then we have that \[ \mu(\cup_{i=1}^\infty A_{i})=\mu(\cup_{i=1}^\infty B_{i})=\sum_{i=1}^\infty \mu(B_{i})=\sum_{i=1}^\infty \mu(A_{i})-\mu(\cup_{i=1}^{n-1}A_{i})\leq \sum_{i=1}^\infty \mu(A_{i}). \]
  3. Defining \(B_{1}=A_{1}\) and \(B_{i}=A_{i}\setminus A_{i-1}\) for \(i \geq 2\). By definition \(\cup_{i=1}^n B_{i}=A_{n}\) hence \[ \cup_{i=1}^\infty B_{i}=\cup_{n=1}^\infty A_{n}\implies \mu(A_{n})=\mu(\cup_{i=1}^nB_{i})=\sum_{i=1}^n\mu(B_{i}). \] Consequently taking the limit as \(n \to \infty\) we find \[ \lim_{n \to \infty}\mu(A_{n})=\sum_{i=1}^\infty \mu(B_{i})=\mu\left( \bigcup_{i=1}^\infty B_{i} \right) =\mu\left(\bigcup_{i=1}^n A_{i}\right) \]
  4. Define \(B_{i}=A_{1}\setminus A_{i}\). By constraint \(B_{i}\subseteq B_{i+1}\) and so \[ \begin{align} \bigcup_{i=1}^\infty B_{i} & =\bigcup_{i=1}^\infty(A_{1}\setminus A_{i})=\bigcup_{i=1}^\infty(A_{i}\cap A_{i}^c)=A_{1}\cap\left( \bigcup_{i=1}^\infty A_{i}^c \right) \\ & = A_{1}\cap \left( \bigcap_{i=1}^\infty A_{i} \right)^c=A_{1}\setminus \left( \bigcap_{i=1}^\infty A_{i} \right). \end{align} \] Therefore we have that \[ \begin{align} \mu(A_{1}) & =\mu\left( A_{1}\cap\left( \bigcap_{i=1}^\infty A_{i} \right) \right)+\mu\left( A_{1}\setminus \bigcap_{i=1}^\infty A_{i} \right) =\mu\left( \bigcap_{i=1}^\infty A_{i} \right)+\lim_{i \to \infty} \mu(B_{i}) \\ & = \mu\left( \bigcap_{i=1}^\infty A_{i} \right)+\lim_{i \to \infty}\mu(A_{i})-\mu(A_{i} ) = \mu\left( \bigcap_{i=1}^\infty A_{i} \right)+\mu(A_{1})-\lim_{i \to \infty}\mu(A_{i}), \end{align} \]

and rearranging gives the result.

Given a measure space \((X,\mathcal{F}, \mu)\) we define the following terminology: 1. the measure is finite if \(\mu(X)<+\infty\) 2. the measure is \(\sigma\)-finite if \(\exists \{ E_{i} \}_{i=1}^\infty \subseteq \mathcal{F}\) such that \(\cup_{i=1}^\infty E_{i}=X\) and \(\mu(E_{i})< \infty\) for all \(i \in \mathbb{N}\) 3. \(E \in S^X\) is a null set (of \(\mu\)) if \(E\in \mathcal{F}\) and \(\mu(E)=0\) 4. a property holds (\(\mu\)-) almost everywhere if the set of points where it fails is a null set 5. a measure whose domain includes all subsets of null sets is called complete.

The next theorem defines a complete measure space, a measure space in which every subset of every null set is measurable (with measure zero).

Suppose that \((X,\mathcal{F}, \mu)\) is a measure space. Let \(\mathcal{N}=\{ N \in \mathcal{F}:\mu(N)=0 \}\) and \(\overline{\mathcal{F}}=\{ E \cup F:E \in\mathcal{F},~F\subset N~\text{for some }N \in\mathcal{N} \}\). Then \(\overline{\mathcal{F}}\) is a \(\sigma\)-algebra known as the completion of \(\mathcal{F}\) with respect to \(\mu\), and there is a unique extension \(\overline\mu\) of \(\mu\) to a complete measure on \(\overline{\mathcal{F}}\), known as the completion of \(\mu\). The resulting measure space \((X,\overline{F}, \overline{\mu})\) is called a complete measure space.

Proof. Firstly, we show that \(\overline{\mathcal{F}}\) is a \(\sigma\)-algebra. Note that since \(\mathcal{F}\) and \(\mathcal{N}\) are both closed under countable unions, so to is \(\overline{\mathcal{F}}\) and further, since \(\emptyset \in \mathcal{F},\mathcal{N}\) then we have that \(\emptyset \in\overline{\mathcal{F}}\), hence we need only show that \(\overline{\mathcal{F}}\) is closed under compliments.

Naturally you might ask whether there exists a measure \(\mu\) on the measurable real space \((\mathbb{R}, \mathcal{B}_{\mathbb{R}})\) (i.e. the couple of the real numbers and the [[borel-sigma-algebra]]) where \(\mu((a,b))=b-a\) and \(\mu\) is translation invariant. To prove existence we must first consider outer-measures, functions similar to measures but that are defined on all collections of subsets.

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