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.

Theorem $\PageIndex{1}$

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, $E^{n} \in \mathcal{M}^{*}$.

Corollary $\PageIndex{1}$

Any countable set $A \subset E^{n}$ is $L$-measurable, with $m A=0$.

Proof

The proof is as in Corollary 6 of §2.

Corollary $\PageIndex{2}$

The Lebesgue measure of $E^{n}$ is $\infty$.

Proof

Prove as in Corollary 5 of §2.

Examples

(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).

Theorem $\PageIndex{2}$

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.)

Theorem $\PageIndex{3}$

(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.$\quad \square$