proof of existence of the Lebesgue measure

First, let $\mathcal{C}$ be the collection of bounded open intervals of the real numbers. As this is a $\pi$ -system (http://planetmath.org/PiSystem), uniqueness of measures extended from a $\pi$ -system (http://planetmath.org/UniquenessOfMeasuresExtendedFromAPiSystem) shows that any measure defined on the $\sigma$ -algebra $\sigma(\mathcal{C})$ is uniquely determined by its values restricted to $\mathcal{C}$ . It remains to prove the existence of such a measure.

Define the length of an interval as $p((a,b))=b-a$ for $a<b$ . The Lebesgue outer measure $\mu^{*}\colon\mathcal{P}(X)\rightarrow\mathbb{R}_{+}\cup\{\infty\}$ is defined as

$\mu^{*}(A)=\inf\left\{\sum_{i=1}^{\infty}p(A_{i}):A_{i}\in\mathcal{C},\ A% \subseteq\bigcup_{i=1}^{\infty}A_{i}\right\}.$

(1)

This is indeed an outer measure (http://planetmath.org/OuterMeasure2) (see construction of outer measures) and, furthermore, for any interval of the form $(a,b)$ it agrees with the standard definition of length, $\mu^{*}((a,b))=p((a,b))=b-a$ (see proof that the outer (Lebesgue) measure of an interval is its length (http://planetmath.org/ProofThatTheOuterLebesgueMeasureOfAnIntervalIsItsLength)).

We show that intervals $(-\infty,a)$ are $\mu^{*}$ -measurable (http://planetmath.org/CaratheodorysLemma). Choosing any $\epsilon>0$ and interval $A\in\mathcal{C}$ the definition of $p$ gives

$p(A)=p(A\cap(-\infty,a))+p(A\cap(a,\infty)).$

So, choosing an arbitrary set $E\subseteq\mathbb{R}$ and a sequence $A_{i}\in\mathcal{C}$ covering $E$ ,

$\begin{split}\displaystyle\sum_{i=1}^{\infty}p(A_{i})&\displaystyle=\sum_{i=1}% ^{\infty}p(A_{i}\cap(-\infty,a))+\sum_{i=1}^{\infty}p(A_{i}\cap(a,\infty))\\ &\displaystyle\geq\mu^{*}(E\cap(-\infty,a))+\mu^{*}(E\cap(a,\infty)).\end{split}$

So, from equation (1)

$\mu^{*}(E)\geq\mu^{*}(E\cap(-\infty,a))+\mu^{*}(E\cap(a,\infty)).$

(2)

Also, choosing any $\epsilon>0$ and using the subadditivity of $\mu^{*}$ ,

$\begin{split}\displaystyle\mu^{*}(E\cap(a,\infty))&\displaystyle\geq\mu^{*}(E% \cap(a-\epsilon,\infty))-\mu^{*}(E\cap(a-\epsilon,a+\epsilon))\\ &\displaystyle\geq\mu^{*}(E\cap[a,\infty))-\mu^{*}((a-\epsilon,a+\epsilon))\\ &\displaystyle=\mu^{*}(E\cap[a,\infty))-2\epsilon.\end{split}$

As $\epsilon>0$ is arbitrary, $\mu^{*}(E\cap(a,\infty))\geq\mu^{*}(E\cap[a,\infty))$ and substituting into (2) shows that

$\mu^{*}(E)\geq\mu^{*}(E\cap(-\infty,a))+\mu^{*}(E\cap[a,\infty)).$

Consequently, intervals of the form $(-\infty,a)$ are $\mu^{*}$ -measurable. As such intervals generate the Borel $\sigma$ -algebra and, by Caratheodory’s lemma, the $\mu^{*}$ -measurable sets form a $\sigma$ -algebra on which $\mu^{*}$ is a measure, it follows that the restriction of $\mu^{*}$ to the Borel $\sigma$ -algebra is itself a measure.

Title	proof of existence of the Lebesgue measure
Canonical name	ProofOfExistenceOfTheLebesgueMeasure
Date of creation	2013-03-22 18:33:14
Last modified on	2013-03-22 18:33:14
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	10
Author	gel (22282)
Entry type	Proof
Classification	msc 26A42
Classification	msc 28A12