Measure Spaces

Author

John Robin Inston

Published

September 25, 2026

1 Concept of measure

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

  1. for disjoint \(E_{1},E_{2},\dots\subseteq X\) we have \(\mu\left(\bigcup_{j=1}^\infty E_{j}\right)=\sum_{j=1}^\infty\mu(E_{j})\) (countable additivity);
  2. for every interval \([a,b]\) we have \(\mu([a,b])=b-a\); and
  3. for \(E\subseteq\mathbb{R}\) and \(c\in\mathbb{R}\) we have \(\mu(E+c)=\mu(E)\) (translation invariance).

The Vitalli Non-Measurability Theorem shows that no such function exists. The closest generalisation of mass is countable additivity, which gives rise to the Lebesgue measure, but not every set has a measure (for example a Vitali set).

There are two standard ways out:

  1. weaken the requirements, which leads to outer measures (Outer Measure); or
  2. restrict the domain to a family of well-behaved sets, which leads to measures and measure spaces.

For the second route the family of sets should be non-empty and closed under complements and countable unions, so that it preserves the properties we want the measure to have. Such a family is a σ-algebra, a special kind of algebra.

2 Algebras and σ-algebras

A collection \(\mathcal{A}\subseteq2^X\) is an algebra of subsets of \(X\) if

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

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

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

Proof.

  1. Since \(\mathcal{A}\) is non-empty there is some \(E\in\mathcal{A}\), and \(E^c\in\mathcal{A}\). Then \(X=E\cup E^c\in\mathcal{A}\) and \(\emptyset=X^c\in\mathcal{A}\).

  2. By De Morgan’s laws, \(\bigcap_{i=1}^nE_i=\big(\bigcup_{i=1}^nE_i^c\big)^c\in\mathcal{A}\).

An algebra \(\mathcal{F}\subseteq2^X\) that is also closed under countable unions, \[ \{E_i\}_{i=1}^\infty\subseteq\mathcal{F}\implies\bigcup_{i=1}^\infty E_i\in\mathcal{F}, \] is called a σ-algebra.

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

Proof. For \(\{E_i\}_{i=1}^\infty\subseteq\mathcal{F}\), De Morgan’s laws give \[ \bigcap_{i=1}^\infty E_i=\left(\bigcup_{i=1}^\infty E_i^c\right)^c\in\mathcal{F}, \] by closure under complements and countable unions.

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

For any set \(X\), both \(2^X\) and the trivial σ-algebra \(\{\emptyset,X\}\) are σ-algebras. For \(X=\{1,2,3\}\), the collection \(\{\emptyset,\{1\},\{2,3\},X\}\) is a σ-algebra.

2.1 Generated σ-algebras

Given any non-empty collection \(\mathcal{C}\) of σ-algebras on \(X\), \[ \bigcap\mathcal{C}:=\{E\subseteq X:E\in\mathcal{A}\text{ for all }\mathcal{A}\in\mathcal{C}\} \] is a σ-algebra on \(X\).

Proof.

  1. \(X\in\mathcal{A}\) for every \(\mathcal{A}\in\mathcal{C}\), so \(X\in\bigcap\mathcal{C}\).
  2. If \(E\in\bigcap\mathcal{C}\) then \(E\in\mathcal{A}\) and so \(E^c\in\mathcal{A}\) for every \(\mathcal{A}\in\mathcal{C}\), hence \(E^c\in\bigcap\mathcal{C}\).
  3. If \(\{E_n\}\subseteq\bigcap\mathcal{C}\) then \(\bigcup_nE_n\in\mathcal{A}\) for every \(\mathcal{A}\in\mathcal{C}\), hence \(\bigcup_nE_n\in\bigcap\mathcal{C}\).

Given \(\mathcal{E}\subseteq2^X\) there is a smallest σ-algebra containing \(\mathcal{E}\), denoted \(\sigma(\mathcal{E})\) and called the σ-algebra generated by \(\mathcal{E}\) (see Generated Sigma Algebra). Explicitly, \[ \sigma(\mathcal{E})=\bigcap\{\mathcal{F}:\mathcal{F}\text{ is a σ-algebra on }X,~\mathcal{E}\subseteq\mathcal{F}\}. \]

Proof. The family is non-empty since \(2^X\) belongs to it, so the intersection is a σ-algebra by the previous theorem. It contains \(\mathcal{E}\), and it is contained in every σ-algebra containing \(\mathcal{E}\), so it is the smallest.

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

Proof. \(\sigma(\mathcal{A})\) is a σ-algebra containing \(\mathcal{E}\), and \(\sigma(\mathcal{E})\) is contained in every such σ-algebra.

The most important example is the Borel σ-algebra of \(\mathbb{R}\), generated by the open sets; see Borel σ-Algebra and Measure. Sigma-algebras on product spaces are treated in Product σ-Algebra.

3 Measures

On a measurable space we define a measure, a function assigning each measurable set a size.

Given a measurable space \((X,\mathcal{F})\), a measure \(\mu:\mathcal{F}\to[0,+\infty]\) satisfies

  1. \(\mu(\emptyset)=0\); and
  2. (countable additivity) for disjoint \(\{E_i\}_{i=1}^\infty\subseteq\mathcal{F}\), \(\mu\left(\bigcup_{i=1}^\infty E_i\right)=\sum_{i=1}^\infty\mu(E_i)\).

The triple \((X,\mathcal{F},\mu)\) is a measure space.

A measure with \(\mu(X)=1\) is a probability measure and gives a probability space.

For \((X,2^X)\) fix \(x_{0}\in X\) and define \[ \delta_{x_{0}}(A)=\begin{cases}1&x_{0}\in A,\\0&\text{otherwise,}\end{cases} \] the Dirac measure at \(x_0\).

For a measure space \((X,\mathcal{F},\mu)\) and \(A,B,A_i\in\mathcal{F}\):

  1. Monotonicity: \(A\subseteq B\implies\mu(A)\leq\mu(B)\);
  2. \(A\subseteq B\) and \(\mu(A)<\infty\implies\mu(B\setminus A)=\mu(B)-\mu(A)\);
  3. Countable subadditivity: \(\mu\left(\bigcup_{i}A_{i}\right)\leq\sum_{i}\mu(A_{i})\);
  4. Continuity from below: \(A_{i}\subseteq A_{i+1}\) for all \(i\) implies \(\mu\left(\bigcup_{i}A_{i}\right)=\lim_{i\to\infty}\mu(A_{i})\);
  5. Continuity from above: \(A_{i}\supseteq A_{i+1}\) for all \(i\) and \(\mu(A_1)<\infty\) imply \(\mu\left(\bigcap_{i}A_{i}\right)=\lim_{i\to\infty}\mu(A_{i})\).

Proof. (1), (2) Write \(B=A\cup(B\setminus A)\), a disjoint union, so \(\mu(B)=\mu(A)+\mu(B\setminus A)\geq\mu(A)\), and subtract if \(\mu(A)<\infty\).

  1. Let \(B_{1}=A_{1}\) and \(B_{n}=A_{n}\setminus\bigcup_{i<n}A_{i}\). The \(B_n\) are disjoint, \(B_n\subseteq A_n\) and \(\bigcup_nB_n=\bigcup_nA_n\), so \[ \mu\left(\bigcup_{n}A_{n}\right)=\sum_{n}\mu(B_{n})\leq\sum_{n}\mu(A_{n}). \]

  2. Let \(B_{1}=A_{1}\) and \(B_{i}=A_{i}\setminus A_{i-1}\) for \(i\geq2\). These are disjoint with \(\bigcup_{i\le n}B_i=A_n\) and \(\bigcup_iB_i=\bigcup_iA_i\), so \[ \mu(A_{n})=\sum_{i=1}^n\mu(B_{i})\xrightarrow[n\to\infty]{}\sum_{i=1}^\infty\mu(B_{i}) =\mu\left(\bigcup_{i}A_{i}\right). \]

  3. Let \(B_{i}=A_{1}\setminus A_{i}\). Then \(B_{i}\subseteq B_{i+1}\) and \(\bigcup_iB_i=A_{1}\setminus\bigcap_iA_i\). Since \(\mu(A_1)<\infty\), \(\mu(B_i)=\mu(A_1)-\mu(A_i)\), and by (4) \[ \mu(A_1)-\mu\left(\bigcap_iA_i\right)=\mu\left(\bigcup_iB_i\right) =\lim_{i\to\infty}\big(\mu(A_1)-\mu(A_i)\big). \] Rearranging gives the result.

For a measure space \((X,\mathcal{F},\mu)\):

  1. \(\mu\) is finite if \(\mu(X)<\infty\);
  2. \(\mu\) is σ-finite if there are \(\{E_{i}\}_{i=1}^\infty\subseteq\mathcal{F}\) with \(\bigcup_iE_{i}=X\) and \(\mu(E_{i})<\infty\) for all \(i\);
  3. \(E\) is a null set if \(E\in\mathcal{F}\) and \(\mu(E)=0\);
  4. a property holds (\(\mu\)-)almost everywhere if the set where it fails is a null set;
  5. \(\mu\) is complete if every subset of a null set is measurable (and so has measure zero).

Let \((X,\mathcal{F},\mu)\) be a measure space, \(\mathcal{N}=\{N\in\mathcal{F}:\mu(N)=0\}\) and \[ \overline{\mathcal{F}}=\{E\cup F:E\in\mathcal{F},~F\subseteq N\text{ for some }N\in\mathcal{N}\}. \] Then \(\overline{\mathcal{F}}\) is a σ-algebra, the completion of \(\mathcal{F}\) with respect to \(\mu\), and \(\mu\) has a unique extension \(\overline\mu\) to a complete measure on \(\overline{\mathcal{F}}\), given by \(\overline\mu(E\cup F)=\mu(E)\). The space \((X,\overline{\mathcal{F}},\overline\mu)\) is the completion of \((X,\mathcal{F},\mu)\).

Proof. Since \(\mathcal{F}\) and \(\mathcal{N}\) are closed under countable unions, so is \(\overline{\mathcal{F}}\), and \(\emptyset\in\overline{\mathcal{F}}\). For complements, let \(E\cup F\in\overline{\mathcal{F}}\) with \(F\subseteq N\). Then \[ (E\cup F)^c=E^c\cap F^c=(E^c\cap N^c)\cup\big(E^c\cap(N\setminus F)\big), \] where the first set is in \(\mathcal{F}\) and the second is a subset of \(N\).

If \(E_1\cup F_1=E_2\cup F_2\) with \(F_j\subseteq N_j\), then \(E_1\subseteq E_2\cup N_2\), so \(\mu(E_1)\leq\mu(E_2)\), and symmetrically, so \(\overline\mu\) is well defined. Countable additivity is inherited from \(\mu\), and every subset of a \(\overline\mu\)-null set is in \(\overline{\mathcal{F}}\). Uniqueness follows since any complete extension must give subsets of null sets measure zero.

4 Measures on the real line

One naturally asks whether there is a measure \(\mu\) on the measurable space \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), where \(\mathcal{B}_{\mathbb{R}}\) is the Borel σ-algebra (see Borel σ-Algebra and Measure), with \(\mu((a,b))=b-a\) and translation invariant. Constructing it requires outer measures, which are defined on all subsets, and the Carathéodory Extension Theorem.

5 References

  • G. B. Folland, Real Analysis: Modern Techniques and Their Applications, 2nd ed., Wiley, 1999, Chapter 1.
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