# 7.8: Lebesgue Measure

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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$$

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