1.1 Foundations of σ-Algebras and Measure Spaces

In this chapter, we will establish the most fundamental framework of measure theory. Unlike the classical Riemann integral, which relies on partitions of intervals, the Lebesgue integral (and abstract measure theory) is built upon partitions of sets. Therefore, we first need to define which sets are "measurable", which leads to the concept of σ-algebras.

1. Algebras and σ-Algebras of Sets

In order for addition of measures to be meaningful, the family of measurable sets must be closed under basic set operations.

Definition 1.1 (Algebra)

Let \(X\) be a nonempty set and \(\mathcal{A} \subset 2^X\) be a family of subsets of \(X\). \(\mathcal{A}\) is called an algebra if it satisfies the following conditions:

For any \(E_1, E_2 \in \mathcal{A}\), we have \(E_1 \cup E_2 \in \mathcal{A}\).
For any \(E \in \mathcal{A}\), its complement \(E^c \in \mathcal{A}\).

From the definition it follows easily that an algebra is also closed under finite intersections and necessarily contains the whole set \(X\) and the empty set \(\emptyset\).

Definition 1.2 (σ-Algebra)

Let \(\mathcal{M}\) be an algebra. If \(\mathcal{M}\) is additionally closed under countable unions, i.e., for any sequence of sets \(\{E_k\}_{k=1}^\infty \subset \mathcal{M}\),

\[ \bigcup_{k=1}^\infty E_k \in \mathcal{M}, \]

then \(\mathcal{M}\) is called a σ-algebra.

By De Morgan's laws and closure under complements, a σ-algebra is necessarily closed under countable intersections as well.

Theorem 1.1 (Generated σ-Algebra and Borel σ-Algebra)

Generated σ-algebra: Given any family \(\mathcal{E} \subset 2^X\) of subsets of \(X\), the intersection of all σ-algebras containing \(\mathcal{E}\) is still a σ-algebra, denoted by \(\mathcal{M}(\mathcal{E})\). It is called the smallest σ-algebra generated by \(\mathcal{E}\).
Borel σ-algebra: If \(X\) is a topological space, let \(\mathcal{T}\) be the family of all open sets in \(X\). The σ-algebra generated by \(\mathcal{T}\) is called the Borel σ-algebra, denoted by \(\mathcal{B}_X = \mathcal{M}(\mathcal{T})\).

In the Euclidean space \(\mathbb{R}^n\), the Borel σ-algebra can be generated not only by open sets but also by open intervals \((a,b)\), closed intervals \([a,b]\), half-open intervals \([a,b)\), and even families of rays of the form \((-\infty, a)\) or \([a, +\infty)\).

Product σ-Algebra and Elementary Family

Definition 1.3 (Product σ-Algebra)

Let \(\{(X_\alpha, \mathcal{M}_\alpha)\}_{\alpha \in A}\) be a family of measurable spaces. Define the product space \(X = \prod_{\alpha \in A} X_\alpha\), and let the projection maps be \(\pi_\alpha: X \rightarrow X_\alpha\).

Then the product σ-algebra on \(X\) is defined as the smallest σ-algebra that makes all projection maps \(\pi_\alpha\) measurable:

\[ \bigotimes_{\alpha \in A} \mathcal{M}_\alpha = \mathcal{M} \left( \left\{ \pi_\alpha^{-1}(E_\alpha) \mid E_\alpha \in \mathcal{M}_\alpha, \alpha \in A \right\} \right). \]

Definition 1.4 (Elementary Family)

A family \(\mathcal{E} \subset 2^X\) is called an elementary family if it satisfies:

\(\emptyset \in \mathcal{E}\).
If \(E, F \in \mathcal{E}\), then \(E \cap F \in \mathcal{E}\).
If \(E \in \mathcal{E}\), then \(E^c\) can be expressed as a finite union of pairwise disjoint sets from \(\mathcal{E}\).

2. Measures and Measure Spaces

Having the concept of "measurable sets" provided by a σ-algebra, we can now define the abstract notion of "length", "area", or "volume", i.e., a measure.

Definition 1.5 (Measure)

Let \((X, \mathcal{M})\) be a measurable space. A set function \(\mu: \mathcal{M} \rightarrow [0, +\infty]\) is called a measure on \(\mathcal{M}\) if it satisfies the following conditions:

Non-negativity: The values of the measure lie in \([0, +\infty]\).
Empty set zero: \(\mu(\emptyset) = 0\).
Countable additivity: For any sequence of pairwise disjoint sets \(\{E_k\}_{k=1}^\infty\) in \(\mathcal{M}\) (i.e., \(E_i \cap E_j = \emptyset\) for \(i \ne j\)), we have

\[ \mu\left(\bigcup_{k=1}^\infty E_k\right) = \sum_{k=1}^\infty \mu(E_k). \]

If we are given a space \(X\), a σ-algebra \(\mathcal{M}\) on it, and a measure \(\mu\), the triple \((X, \mathcal{M}, \mu)\) is called a measure space.

Some common simple examples of measures:

Counting measure: For any set \(E\), \(\mu(E)\) equals the number of elements in the set.
Dirac delta measure: Fix a point \(x_0\) in the space. Then \(\mu(E)=1\) if \(x_0 \in E\) and \(\mu(E)=0\) if \(x_0 \notin E\).

Basic Properties of Measures

From the axiomatic definition of a measure, the following important analytic properties follow naturally.

Theorem 1.2 (Basic Properties of Measures)

Let \((X, \mathcal{M}, \mu)\) be a measure space. Then \(\mu\) satisfies:

Monotonicity: If \(E, F \in \mathcal{M}\) and \(E \subset F\), then \(\mu(E) \le \mu(F)\). Moreover, if \(\mu(E) < \infty\), then \(\mu(F \setminus E) = \mu(F) - \mu(E)\).
Subadditivity: For any sequence of sets \(\{E_j\}_{j=1}^\infty \subset \mathcal{M}\) (not necessarily pairwise disjoint), we always have

\[ \mu\left(\bigcup_{j=1}^\infty E_j\right) \le \sum_{j=1}^\infty \mu(E_j). \]

Continuity from below: For an increasing sequence of sets \(E_1 \subset E_2 \subset \cdots\), we have

\[ \mu\left(\bigcup_{j=1}^\infty E_j\right) = \lim_{j \rightarrow \infty} \mu(E_j). \]

Continuity from above: For a decreasing sequence of sets \(E_1 \supset E_2 \supset \cdots\) with \(\mu(E_1) < \infty\), we have

\[ \mu\left(\bigcap_{j=1}^\infty E_j\right) = \lim_{j \rightarrow \infty} \mu(E_j). \]

3. Null Sets and Completion of Measure Spaces

In later discussions of real analysis (e.g., the concept of "almost everywhere"), null sets play a central role. However, a general measure space does not guarantee that all subsets of a null set are measurable, which leads to the notion of completion.

Definition 1.6 (Null Set and Complete Measure Space)

Null set: If \(E \in \mathcal{M}\) satisfies \(\mu(E) = 0\), then \(E\) is called a \(\mu\)-null set.
Complete measure space: A measure space \((X, \mathcal{M}, \mu)\) is called complete if every subset of a null set belongs to \(\mathcal{M}\) (and therefore its measure is automatically zero).

In a general measure space, subsets of a null set are not necessarily measurable. However, we can "complete" any measure space by artificially adding these subsets to the original σ-algebra.

Theorem 1.3 (Completion of a Measure Space)

For any measure space \((X, \mathcal{M}, \mu)\), we can construct its completion \((X, \overline{\mathcal{M}}, \overline{\mu})\), where the new σ-algebra is defined as

\[ \overline{\mathcal{M}} = \{E \cup F \mid E \in \mathcal{M}, F \subset N \text{ and } N \in \mathcal{M}, \mu(N)=0\}. \]

On this extended σ-algebra, define \(\overline{\mu}(E \cup F) = \mu(E)\). Then \(\overline{\mathcal{M}}\) is a σ-algebra and \(\overline{\mu}\) is the unique complete measure on it extending \(\mu\).

Proof of the Completion Theorem

To prove that \((X, \overline{\mathcal{M}}, \overline{\mu})\) is a complete measure space, we need to verify two things: \(\overline{\mathcal{M}}\) is a σ-algebra, and \(\overline{\mu}\) is a well-defined complete measure on it.

Step 1: Verify that \(\overline{\mathcal{M}}\) is a σ-algebra

For closure under countable unions: Let \(A_j = E_j \cup F_j \in \overline{\mathcal{M}}\), where \(E_j \in \mathcal{M}\), \(F_j \subset N_j\) and \(\mu(N_j) = 0\).

\[ \bigcup_{j=1}^\infty A_j = \left(\bigcup_{j=1}^\infty E_j\right) \cup \left(\bigcup_{j=1}^\infty F_j\right). \]

Since \(\mathcal{M}\) is a σ-algebra, \(\bigcup_{j=1}^\infty E_j \in \mathcal{M}\). Also, \(\bigcup_{j=1}^\infty F_j \subset \bigcup_{j=1}^\infty N_j\). By subadditivity of the measure:

\[ \mu\left(\bigcup_{j=1}^\infty N_j\right) \le \sum_{j=1}^\infty \mu(N_j) = 0, \]

so the union again has the form \(E \cup F\), proving closure under countable unions.

Next, verify closure under complements. Let \(A = E \cup F \in \overline{\mathcal{M}}\). For convenience we may assume \(E \cap N = \emptyset\). Since \(A^c = (E \cup F)^c = E^c \cap F^c\) and \(F \subset N\), we can rewrite it as

\[ A^c = (E \cup N)^c \cup (N \setminus F) \cap E^c. \]

Here \((E \cup N)^c \in \mathcal{M}\), while \((N \setminus F) \cap E^c \subset N\) is a subset of a null set, showing that \(A^c \in \overline{\mathcal{M}}\). Hence \(\overline{\mathcal{M}}\) is a σ-algebra.

Step 2: Verify completeness

Suppose there exists a subset \(S \subset A \in \overline{\mathcal{M}}\) with \(\overline{\mu}(A) = \mu(E) = 0\). Because \(A = E \cup F \subset E \cup N\), we have \(S \subset E \cup N\). Since \(\mu(E \cup N) \le \mu(E) + \mu(N) = 0\), this shows that \(S\) is contained in a null set of \(\mathcal{M}\). By definition, \(S = \emptyset \cup S \in \overline{\mathcal{M}}\). This proves the completeness of the completed measure space.