# 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. 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  ProofOfExistenceOfTheLebesgueMeasure 2013-03-22 18:33:14 2013-03-22 18:33:14 gel (22282) gel (22282) 10 gel (22282) Proof msc 26A42 msc 28A12