8  Completion of Measures

Let \(\mu\) be a measure on a \(\sigma\)-algebra \(\mathcal S\) of subsets of \(X\), so that \((X, \mathcal S, \mu)\) is a measure space. If \(\mathcal S\) is not \(2^X\), it can be useful to extend \(\mu\) to a larger \(\sigma\)-algebra, especially if there are sets whose measure is somehow determined by the measures of sets already in \(\mathcal S\).

We will make such an extension \(\mu\) using the outer measure \(\mu^*\). Starting with an arbitrary set \(E\), we wish to find a set \(C \in \mathcal S\) with \(E \subset C\) and \(\mu(C)\) as small as possible.

Theorem 8.1 For any measurable space \((X, \mathcal S, \mu)\) and any \(E \subset X\), there is a \(C \in S\) with \(E \subset C\) such that \(\mu^*(E) = \mu(C)\)

If \(E \subset \bigcup_{n\ge 1}A_n\) and \(A_n \in \mathcal S\), let \(A := \bigcup_{n\ge 1}A_n\).

Then \(E \subset A\), \(A \in \mathcal S\), and as in the proof of Lemma 6.2, \(\mu(A) \le \sum_{n\ge 1}\mu(A_n)\). So \(\mu^*(E) = \inf\{\mu(A): E\subset A, A \in \mathcal S\}\).

For each \(m = 1,2,...,\) choose \(C_m \in \mathcal S\) with \(E \subset C_m\) and \(\mu(C_m) \le \mu^*(E) + 1/m\). Let \(C := \bigcap_m C_m\).

Then \(C \in \mathcal S\), since \(C = X \setminus (\bigcup_m X \setminus C_m)\), \(E \subset C\) and \(\mu^*(E) = \mu(C)\).

A set \(C\) as in Theorem 8.1 is called a measurable cover of \(E\).

If \(E \in \mathcal S\), then \(E\) is a measurable cover of itself. If \(C \in \mathcal S\), and the only subsets of \(C\) in \(\mathcal S\) are itself and \(\varnothing\), then \(C\) is a measurable cover of any nonempty subset \(E \subset C\).

If there is going to be a set \(A \in \mathcal S\) with \(A \subset E \subset C\) and \(\mu(C\setminus A) = 0\), then \(\mu^*(E\setminus A) = \mu^*(C\setminus E) = 0\), from now on we will call these sets with outer measure \(0\) “null sets”.

Definition 8.1 The collection of null sets for \(\mu\) is defined by \(\mathcal N(\mu) := \{F \subset X: \mu^*(F) = 0\}\).

Any subset of a countable union of sets in \(\mathcal N(\mu)\) is in \(\mathcal N(\mu)\).

Let \(\mathcal S \vee \mathcal N(\mu)\) be the \(\sigma\)-algebra generated by \(\mathcal S\cup \mathcal N(\mu)\).

Let \(\mathcal S^*_\mu := \{E \subset X: E \Delta B \in \mathcal N(\mu) \text{ for some }B \in \mathcal S\}\). Then \(E\) will be called almost equal to \(B\). \(\mathcal S^*_\mu\) is the collection of sets which almost equal sets in \(\mathcal S\).

For example, if \(A\subset E \subset C\) and \(\mu(C\setminus A) = 0\), then \(E \in \mathcal S^*_\mu\) and we can take \(B = A\) or \(C\) in the definition.

The two ways just defined of extending a \(\sigma\)-algebra by sets of measure \(0\) give the same result.

Proposition 8.1 For any measure space \((X, \mathcal S,\mu)\) and the collection \(\mathcal S^*_\mu\) of sets almost equal to sets in \(\mathcal S\), we have \(\mathcal S^*_\mu = \mathcal S \vee \mathcal N(\mu)\), the smallest \(\sigma\)-algebra including \(\mathcal S\) and \(\mathcal N(\mu)\).

Example 8.1 Let \(\mu\) be a measure on the set \(\mathbb Z\), with \(\mu\{k\} > 0\) for each \(k > 0\) and \(\mu\{k\} = 0\) for all \(k \le 0\). Then \(\mathcal N(\mu)\) is the ring of all subsets of \(-\mathbb N\) of nonpositive integers.

Let \(\mathcal S\) be the \(\sigma\)-algebra of all sets \(A \subset \mathbb Z\) such that for some set \(B\) of positive integers, either \(A = B\) or \(A = -\mathbb N \cup B\). Then the two equal \(\sigma\)-algebras in Proposition 8.1 both equal the \(\sigma\)-algebra \(2^{\mathbb Z}\).

Given a measure \(\mu\), we will show that \(\mu\) can be extended to a measure \(\bar\mu\) so that whenever two sets are almost equal for \(\mu\), and \(\mu\) is defined for at least one of them, then \(\bar\mu\) will be defined and equal for both.

Proposition 8.2 For any measure space \((X, \mathcal S, \mu)\), if \(A \Delta B \in \mathcal N(\mu)\) and \(B \in \mathcal S\), let \(\bar\mu(A) := \mu(B)\) for any \(A \in \mathcal S \vee \mathcal N(\mu)\). Then, \(\bar\mu\) is a well-defined measure on the \(\sigma\)-algebra \(\mathcal S \vee \mathcal N(\mu)\) which equals \(\mu\) on \(\mathcal S\).

The measure \(\bar\mu\) is called the completion of \(\mu\).

Let \(\tilde\mu\) be the restrition of \(\mu^*\) to \(\mathcal M(\mu^*)\), which is a measure and extends \(\mu\) by Lemma 6.2 and Lemma 6.4. This extension is usually also called \(\mu\). We need to check that outer measure \(\mu^*\) is not changed by the extension.

Proposition 8.3 For any countably additive nonnegative function \(\mu\) on an algebra \(\mathcal A\) in a set \(X\), \(\tilde\mu^* = \mu^*\) on \(2^X\).

Since \(\tilde\mu = \mu\) on \(\mathcal A\) by Lemma 6.2, clearly \(\tilde\mu^* \le \mu^*\).

Conversely, for any \(A \subset X\), by Theorem 8.1 we can take \(C \in \mathcal M(\mu^*)\) with \(A \subset C\) and \(\tilde\mu^*(A) = \tilde\mu(C)\).

Then \(\mu^*(A)\le\mu^*(C)=\tilde\mu(C)=\tilde\mu(A)\)

Definition 8.2 A measure space \((X, \mathcal S, \mu)\) is called

  • finite iff \(\mu(X) < \infty\)
  • \(\sigma\)-finite iff there is a sequence of sets \(E_n\in\mathcal S\) with \(\mu(E_n) < \infty\) and \(X =\bigcup_n E_n\).

Theorem 8.2 For any \(\sigma\)-finite measure space \((X, \mathcal S, \mu)\), \(\mathcal S\vee \mathcal N(\mu)=\mathcal M(\mu^*)\)

Let \(X = \bigcup_n E_n\), \(\mu(E_n)<\infty\), and \(E \in \mathcal M(\mu^*)\).

To show that \(E \in S \vee \mathcal N(\mu)\), it is enough to show that \(E\cap E_n \in \mathcal S \vee \mathcal N(\mu)\) for all \(n\). Thus we can assume \(\mu\) is finite.

Let \(A\) be a measurable cover of \(E\), by Theorem 8.1. Then \(\mu^*(E)=\mu(A) = \mu^*(A)=\mu^*(E) + \mu^*(A\setminus E)\), so \(\mu^*(A\setminus E)=0\) and \(E = A \setminus(A\setminus E)\in \mathcal S \vee \mathcal N(\mu)\), so \(\mathcal M(\mu^*) \subset \mathcal S \vee \mathcal N(\mu)\).

The converse inclusion follows from Lemma 6.3, Lemma 6.4, and Proposition 6.1.

We will show a counterexample to see why \(\sigma\)-finiteness is needed.

Example 8.2 Let \(\mathcal S := \{\varnothing, X\}\), \(\mu(\varnothing)=0\) and \(\mu(X)=\infty\). Then \(\mu^*(E)=\infty\) whenever \(\varnothing \ne E \subset X\).

Thus all subsets of \(X\) are in \(\mathcal M(\mu^*)\), while \(\mathcal S \vee \mathcal N(\mu) = \{\varnothing, X\}\), a smaller \(\sigma\)-algebra if \(X\) has more than one point.