7.8: Lebesgue Measure
We shall now consider the most important example of a measure in \(E^{n},\) due to Lebesgue. This measure generalizes the notion of volume and assigns "volumes" to a large set family, the "Lebesgue measurable" sets, so that "volume" becomes a complete topological measure. For "bodies" in \(E^{3},\) this measure agrees with our intuitive idea of "volume."
We start with the volume function \(v : \mathcal{C} \rightarrow E^{1}\) ("Lebesgue premeasure") on the semiring \(\mathcal{C}\) of all intervals in \(E^{n}\) (§1). As we saw in §§5 and 6, this premeasure induces an outer measure \(m^{*}\) on all subsets of \(E^{n};\) and \(m^{*},\) in turn, induces a measure \(m\) on the \(\sigma\)-field \(\mathcal{M}^{*}\) of \(m^{*}\)-measurable sets. These sets are, by definition, the Lebesgue-measurable (briefly \(L\)-measurable) sets; \(m^{*}\) and \(m\) so defined are the (\(n\)-dimensional) Lebesgue outer measure and Lebesgue measure.
Lebesgue premeasure \(v\) is \(\sigma\)-additive on \(\mathcal{C},\) the intervals in \(E^{n}\). Hence the latter are Lebesgue measurable \(\left(\mathcal{C} \subseteq \mathcal{M}^{*}\right),\) and the volume of each interval equals its Lebesgue measure:
\[v=m^{*}=m \text { on } \mathcal{C}.\]
This follows by Corollary 1 in §2 and Theorem 2 of §6
Note 1 . As \(\mathcal{M}^{*}\) is a (\(\sigma\)-field §6), it is closed under countable unions, countable intersections, and differences. Thus
\[\mathcal{C} \subseteq \mathcal{M}^{*} \text { implies } \mathcal{C}_{\sigma} \subseteq \mathcal{M}^{*};\]
i.e., any countable union of intervals is \(L\)-measurable. Also, \(E^{n} \in \mathcal{M}^{*}\).
Any countable set \(A \subset E^{n}\) is \(L\)-measurable, with \(m A=0\).
- Proof
-
The proof is as in Corollary 6 of §2.
The Lebesgue measure of \(E^{n}\) is \(\infty\).
- Proof
-
Prove as in Corollary 5 of §2.
(a) Let
\[R=\left\{\text {rationals in } E^{1}\right\}.\]
Then \(R\) is countable (Corollary 3 of Chapter 1, §9); so \(m R=0\) by Corollary 1. Similarly for \(R^{n}\) (rational points in \(E^{n})\).
(b) The measure of an interval with endpoints \(a, b\) in \(E^{1}\) is its length, \(b-a.\)
Let
\[R_{o}=\{\text { all rationals in }[a, b]\};\]
so \(m R_{o}=0.\) As \([a, b]\) and \(R_{o}\) are in \(\mathcal{M}^{*}\) (a \(\sigma\)-field), so is
\[[a, b]-R_{o},\]
the irrationals in \([a, b].\) By Lemma 1 in §4, if \(b>a,\) then
\[m\left([a, b]-R_{o}\right)=m([a, b])-m R_{o}=m([a, b])=b-a>0=m R_{o}.\]
This shows again that the irrationals form a "larger" set than the rationals (cf. Theorem 3 of Chapter 1, §9).
(c) There are uncountable sets of measure zero (see Problems 8 and 10 below).
Lebesgue measure in \(E^{n}\) is complete, topological, and totally \(\sigma\)-finite. That is,
(i) all null sets (subsets of sets of measure zero) are \(L\)-measurable;
(ii) so are all open sets \(\left(\mathcal{M}^{*} \supseteq \mathcal{G}\right),\) hence all Borel sets \(\left(\mathcal{M}^{*} \supseteq \mathcal{B}\right);\) in particular, \(\mathcal{M}^{*} \supseteq \mathcal{F}, \mathcal{M}^{*} \supseteq \mathcal{G}_{\delta}, \mathcal{M}^{*} \supseteq \mathcal{F}_{\sigma}, \mathcal{M}^{*} \supseteq \mathcal{F}_{\sigma \delta},\) etc.;
(iii) each \(A \in \mathcal{M}^{*}\) is a countable union of disjoint sets of finite measure.
- Proof
-
(i) This follows by Theorem 1 in §6.
(ii) By Lemma 2 in §2, each open set is in \(\mathcal{C}_{\sigma},\) hence in \(\mathcal{M}^{*}\) (Note 1). Thus \(\mathcal{M}^{*} \supseteq \mathcal{G}.\) But by definition, the Borel field \(\mathcal{B}\) is the least \(\sigma\)-ring \(\supseteq \mathcal{G}.\) Hence \(\mathcal{M}^{*} \supseteq \mathcal{B}^{*}\).
(iii) As \(E^{n}\) is open, it is a countable union of disjoint half-open intervals,
\[E^{n}=\bigcup_{k=1}^{\infty} A_{k} \text { (disjoint),}\]
with \(m A_{k}<\infty\) (Lemma 2 §2). Hence
\[\left(\forall A \subseteq E^{n}\right) \quad A \subseteq \bigcup A_{k};\]
so
\[A=\bigcup_{k}\left(A \cap A_{k}\right) \text { (disjoint).}\]
If, further, \(A \in \mathcal{M}^{*},\) then \(A \cap A_{k} \in \mathcal{M}^{*},\) and
\[m\left(A \cap A_{k}\right) \leq m A_{k}<\infty. \text{ (Why?)} \quad \square\]
Note 2. More generally, a \(\sigma\)-finite set \(A \in \mathcal{M}\) in a measure space \((S, \mathcal{M}, \mu)\) is a countable union of disjoint sets of finite measure (Corollary 1 of §1).
Note 3. Not all \(L\)-measurable sets are Borel sets. On the other hand, not all sets in \(E^{n}\) are \(L\)-measurable (see Problems 6 and 9 below.)
(a) Lebesgue outer measure \(m^{*}\) in \(E^{n}\) is \(\mathcal{G}\)-regular; that is,
\[\left(\forall A \subseteq E^{n}\right) \quad m^{*} A=\inf \{m X | A \subseteq X \in \mathcal{G}\}\]
(\(\mathcal{G}=\) open sets in \(E^{n}\)).
(b) Lebesgue measure \(m\) is strongly regular (Definition 5 and Theorems 1 and 2, all in §7).
- Proof
-
By definition, \(m^{*} A\) is the glb of all basing covering values of \(A.\) Thus given \(\varepsilon>0,\) there is a basic covering \(\left\{B_{k}\right\} \subseteq \mathcal{C}\) of nonempty sets \(B_{k}\) such that
\[A \subseteq \bigcup B_{k} \text { and } m^{*} A+\frac{1}{2} \varepsilon \geq \sum_{k} v B_{k}.\]
(Why? What if \(m^{*} A=\infty\)?)
Now, by Lemma 1 in §2, fix for each \(B_{k}\) an open interval \(C_{k} \supseteq B_{k}\) such that
\[v C_{k}-\frac{\varepsilon}{2^{k+1}}<v B_{k}.\]
Then (2) yields
\[m^{*} A+\frac{1}{2} \varepsilon \geq \sum_{k}\left(v C_{k}-\frac{\varepsilon}{2^{k+1}}\right)=\sum_{k} v C_{k}-\frac{1}{2} \varepsilon;\]
so by \(\sigma\)-subadditivity,
\[m \bigcup_{k} C_{k} \leq \sum_{k} m C_{k}=\sum_{k} v C_{k} \leq m^{*} A+\varepsilon.\]
Let
\[X=\bigcup_{k} C_{k}.\]
Then \(X\) is open (as the \(C_{k}\) are). Also, \(A \subseteq X,\) and by (3),
\[m X \leq m^{*} A+\varepsilon.\]
Thus, indeed, \(m^{*} A\) is the \(g l b\) of all \(m X, A \subseteq X \in \mathcal{G},\) proving (a).
In particular, if \(A \in \mathcal{M}^{*},\) (1) shows that \(m\) is regular (for \(m^{*} A=m A). Also, by Theorem 2, \(m\) is \(\sigma\)-finite, and \(E^{n} \in \mathcal{M}^{*};\) so (b) follows by Theorem 1 in §7.\(\quad \square\)