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.