7.5: Nonnegative Set Functions. Premeasures. Outer Measures
We now concentrate on nonnegative set functions
\[m : \mathcal{M} \rightarrow[0, \infty]\]
(we mostly denote them by \(m\) or \(\mu\)). Such functions have the advantage that
\[\sum_{n=1}^{\infty} m X_{n}\]
exists and is permutable (Theorem 2 in §2) for any sets \(X_{n} \in \mathcal{M},\) since \(m X_{n} \geq\) \(0.\) Several important notions apply to such functions (only). They "mimic" §§1 and 2.
A set function
\[m : \mathcal{M} \rightarrow[0, \infty]\]
is said to be
(i) monotone (on \(\mathcal{M}\)) iff
\[m X \leq m Y\]
whenever
\[X \subseteq Y \text { and } X, Y \in \mathcal{M};\]
(ii) (finitely) subadditive (on \(\mathcal{M}\)) iff for any finite union
\[\bigcup_{k=1}^{n} Y_{k},\]
we have
\[m X \leq \sum_{k=1}^{m} m Y_{k}\]
whenever \(X, Y_{k} \in \mathcal{M}\) and
\[X \subseteq \bigcup_{k=1}^{n} Y_{k} \text { (disjoint or not);}\]
(iii) \(\sigma\)-subadditive (on \(\mathcal{M}\)) iff (1) holds for countable unions, too.
Recall that \(\left\{Y_{k}\right\}\) is called a covering of \(X\) iff
\[X \subseteq \bigcup_{k} Y_{k}.\]
We call it an \(\mathcal{M}\)-covering of \(X\) if all \(Y_{k}\) are \(\mathcal{M}\)-sets. We now obtain the following corollary.
Subadditivity implies monotonicity.
Take \(n=1\) in formula (1).
If \(m : \mathcal{C} \rightarrow[0, \infty]\) is additive (\(\sigma\)-additive) on \(\mathcal{C}\), a semiring, then \(m\) is also subadditive (\(\sigma\)-subadditive, respectively), hence monotone, on \(\mathcal{C}\).
- Proof
-
The proof is a mere repetition of the argument used in Lemma 1 in §1.
Taking \(n=1\) in formula (ii) there, we obtain finite subadditivity.
For \(\sigma\)-subadditivity, one only has to use countable unions instead of finite ones.
Note 1. The converse fails: subadditivity does not imply additivity.
Note 2. Of course, Corollary 2 applies to rings, too (see Corollary 1 in §3).
A premeasures is a set function
\[\mu : \mathcal{C} \rightarrow[0, \infty]\]
such that
\[\emptyset \in \mathcal{C} \text { and } \mu \emptyset=0.\]
(\(\mathcal{C}\) may, but need not, be a semiring.)
A premeasure space is a triple
\[(S, \mathcal{C}, \mu),\]
where \(\mathcal{C}\) is a family of subsets of \(S\) (briefly, \(\mathcal{C} \subseteq 2^{S})\) and
\[\mu : \mathcal{C} \rightarrow[0, \infty]\]
is a premeasure. In this case, \(\mathcal{C}\)-sets are also called basic sets.
If
\[A \subseteq \bigcup_{n} B_{n},\]
with \(B_{n} \in \mathcal{C},\) the sequence \(\left\{B_{n}\right\}\) is called a basic covering of \(A,\) and
\[\sum_{n} \mu B_{n}\]
is a basic covering value of \(A;\left\{B_{n}\right\}\) may be finite or infinite.
(a) The volume function \(v\) on \(\mathcal{C}\) (= intervals in \(E^{n}\)) is a premeasure, as \(v \geq 0\) and \(v \emptyset=0.\) (\(E^{n}, \mathcal{C}, v\)) is the Lebesgue premeasure space.
(b) The LS set function \(s_{\alpha}\) is a premeasure if \(\alpha \uparrow\) (see Problem 7 in §4). We call it the \(\alpha\)-induced Lebesgue-Stieltjes \((L S)\) premeasure in \(E^{1}\).
We now develop a method for constructing \(\sigma\)-subadditive premeasures. (This is a first step toward achieving \(\sigma\)-additivity; see §4.)
For any premeasure space \((S, \mathcal{C}, \mu),\) we define the \(\mu\)-induced outer measure \(m^{*}\) on \(2^{S}\) (= all subsets of \(S\)) by setting, for each \(A \subseteq S\),
\[m^{*} A=\inf \left\{\sum_{n} \mu B_{n} | A \subseteq \bigcup_{n} B_{n}, B_{n} \in \mathcal{C}\right\},\]
i.e., \(m^{*} A\) (called the outer measure of \(A\)) is the glb of all basic covering values of \(A.\)
If \(\mu=v, m^{*}\) is called the Lebesgue outer measure in \(E^{n}\).
Note 3. If \(A\) has no basic coverings, we set \(m^{*} A=\infty.\) More generally, we make the convention that inf \(\emptyset=+\infty\).
Note 4. By the properties of the glb, we have
\[(\forall A \subseteq S) \quad 0 \leq m^{*} A.\]
If \(A \in \mathcal{C},\) then \(\{A\}\) is a basic covering; so
\[m^{*} A \leq \mu A.\]
In particular, \(m^{*} \emptyset=\mu \emptyset=0\).
The set function \(m^{*}\) so defined is \(\sigma\)-subadditive on \(2^{S}\).
- Proof
-
Given
\[A \subseteq \bigcup_{n} A_{n} \subset S,\]
we must show that
\[m^{*} A \leq \sum_{n} m^{*} A_{n}.\]
This is trivial if \(m^{*} A_{n}=\infty\) for some \(n.\) Thus assume
\[(\forall n) \quad m^{*} A_{n}<\infty\]
and fix \(\varepsilon>0\).
By Note 3, each \(A_{n}\) has a basic covering
\[\left\{B_{n k}\right\}, \quad k=1,2, \ldots\]
(otherwise, \(m^{*} A_{n}=\infty.\)) By properties of the glb, we can choose the \(B_{n k}\) so that
\[(\forall n) \quad \sum_{k} \mu B_{n k}<m^{*} A_{n}+\frac{\varepsilon}{2^{n}}.\]
(Explain from (2)). The sets \(B_{n k}\) (for all \(n\) and all \(k )\) form a countable basic covering of all \(A_{n},\) hence of \(A.\) Thus by Definition 3,
\[m^{*} A \leq \sum_{n}\left(\sum_{k} \mu B_{n k}\right) \leq \sum_{n}\left(m^{*} A_{n}+\frac{\varepsilon}{2^{n}}\right) \leq \sum^{n} m^{*} A_{n}+\varepsilon.\]
As \(\varepsilon\) is arbitrary, we can let \(\varepsilon \rightarrow 0\) to obtain the desired result.\(\quad \square\)
Note 5. In view of Theorem 1, we now generalize the notion of an outer measure in \(S\) to mean any \(\sigma\)-subadditive premeasure defined on all of \(2^{S}\).
By Note 4, \(m^{*} \leq \mu\) on \(\mathcal{C},\) not \(m^{*}=\mu\) in general. However, we obtain the following result.
With \(m^{*}\) as in Definition 3, we have \(m^{*}=\mu\) on \(\mathcal{C}\) iff \(\mu\) is \(\sigma\)-subadditive on \(\mathcal{C}.\) Hence, in this case, \(m^{*}\) is an extension of \(\mu.\)
- Proof
-
Suppose \(\mu\) is \(\sigma\)-subadditive and fix any \(A \in \mathcal{C}.\) By Note 4,
\[m^{*} A \leq \mu A.\]
We shall show that
\[\mu A \leq m^{*} A,\]
too, and hence \(\mu A=m^{*} A\).
Now, as \(A \in \mathcal{C}, A\) surely has basic coverings, e.g., \(\{A\}.\) Take any basic covering:
\[A \subseteq \bigcup_{n} B_{n}, \quad B_{n} \in \mathcal{C}.\]
As \(\mu\) is \(\sigma\)-subadditive,
\[\mu A \leq \sum_{n} \mu B_{n}.\]
Thus \(\mu A\) does not exceed any basic covering values of \(A;\) so it cannot exceed their glb, \(m^{*} A.\) Hence \(\mu=m^{*},\) indeed.
Conversely, if \(\mu=m^{*}\) on \(\mathcal{C},\) then the \(\sigma\)-subadditivity of \(m^{*}\) (Theorem 1) implies that of \(\mu\) (on \(\mathcal{C}\)). Thus all is proved.\(\quad \square\)
Note 6. If, in (2), we allow only finite basic coverings, then the \(\mu\)-induced set function is called the \(\mu\)-induced outer content, \(c^{*}.\) It is only finitely subadditive, in general.
In particular, if \(\mu=v\) (Lebesgue premeasure), we speak of the Jordan outer content in \(E^{n}.\) (It is superseded by Lebesgue theory but still occurs in courses on Riemann integration.)
We add two more definitions related to the notion of coverings.
A set function \(s : \mathcal{M} \rightarrow E\left(\mathcal{M} \subseteq 2^{S}\right)\) is called \(\sigma\)-finite iff every \(X \in \mathcal{M}\) can be covered by a sequence of \(\mathcal{M}\)-sets \(X_{n},\) with
\[\left|s X_{n}\right|<\infty \quad(\forall n).\]
Any set \(A \subseteq S\) which can be so covered is said to be \(\sigma\)-finite with respect to \(s\) (briefly, (\(s\)) \(\sigma\)-finite).
If the whole space \(S\) can be so covered, we say that \(s\) is totally \(\sigma\)-finite.
For example, the Lebesgue premeasure \(v\) on \(E^{n}\) is totally \(\sigma\)-finite.
A set function \(s : \mathcal{M} \rightarrow E^{*}\) is said to be regular with respect to a set family \(\mathcal{A}\) (briefly, \(\mathcal{A}\)-regular) iff for each \(A \in \mathcal{M}\),
\[s A=\inf \{s X | A \subseteq X, X \in \mathcal{A}\};\]
that is, \(s A\) is the glb of all \(s X,\) with \(A \subseteq X\) and \(X \in \mathcal{A}\).
These notions are important for our later work. At present, we prove only one theorem involving Definitions 3 and 5.
For any premeasure space \((S, \mathcal{C}, \mu),\) the \(\mu\)-induced outer measure \(m^{*}\) is \(\mathcal{A}\)-regular whenever
\[\mathcal{C}_{\sigma} \subseteq \mathcal{A} \subseteq 2^{S}.\]
Thus in this case,
\[(\forall A \subseteq S) \quad m^{*} A=\inf \left\{m^{*} X | A \subseteq X, X \in \mathcal{A}\right\}.\]
- Proof
-
As \(m^{*}\) is monotone, \(m^{*} A\) is surely a lower bound of
\[\left\{m^{*} X | A \subseteq X, X \in \mathcal{A}\right\}.\]
We must show that there is no greater lower bound.
This is trivial if \(m^{*} A=\infty\).
Thus let \(m^{*} A<\infty;\) so \(A\) has basic coverings (Note 3). Now fix any \(\varepsilon>0\).
By formula (2), there is a basic covering \(\left\{B_{n}\right\} \subseteq \mathcal{C}\) such that
\[A \subseteq \bigcup_{n} B_{n}\]
and
\[m^{*} A+\varepsilon>\sum_{n} \mu B_{n} \geq \sum_{n} m^{*} B_{n} \geq m^{*} \bigcup_{n} B_{n}.\]
(\(m^{*}\) is \(\sigma\)-subadditive!)
Let
\[X=\bigcup_{n} B_{n}.\]
Then \(X\) is in \(\mathcal{C}_{\sigma},\) hence in \(\mathcal{A},\) and \(A \subseteq X.\) Also,
\[m^{*} A+\varepsilon>m^{*} X.\]
Thus \(m^{*} A+\varepsilon\) is not a lower bound of
\[\left\{m^{*} X | A \subseteq X, X \in \mathcal{A}\right\}.\]
This proves (4).\(\quad \square\)