Proof of Existence of Lebesgue Measure on the Real Line

Throughout, $\lambda^{*}$ is the Lebesgue outer measure and sums of sequences in $[0,\infty]$ are as in Measure, Measure Space, and Probability Measure.

Claim 1. $\lambda^{*}(\varnothing)=0$: cover by the degenerate intervals $(0,1/m)$-shrunk covers, e.g. for any $\varepsilon>0$ the single interval $(0,\varepsilon)$ (padded with copies of $(0,0)=\varnothing$, i.e. intervals with $a_m=b_m$) covers $\varnothing$ with total length $\varepsilon$. Monotonicity: any covering sequence for $B$ also covers $A\subseteq B$, so the infimum for $A$ is over a larger set of admissible values. Countable subadditivity: let $(A_m)$ be subsets of $\mathbb{R}$; if some $\lambda^{*}(A_m)=\infty$ there is nothing to prove, so assume all finite. Given $\varepsilon>0$, choose for each $m$ a covering sequence of open intervals $\bigl(I_{m,j}\bigr)_{j\in\mathbb{N}}$ of $A_m$ with total length at most $\lambda^{*}(A_m)+\varepsilon 2^{-m}$. The doubly indexed family $(I_{m,j})$ is countable, hence can be enumerated as a single sequence (a bijection between $\mathbb{N}\times\mathbb{N}$ and $\mathbb{N}$, e.g. the diagonal enumeration); it covers $\bigcup_m A_m$, and its total length — which is independent of the enumeration, since for series of nonnegative terms every rearrangement has the same supremum of finite partial sums — is at most $\sum_m\lambda^{*}(A_m)+\varepsilon$. Letting $\varepsilon\to 0$ gives subadditivity. Hence $\lambda^{*}$ is an outer measure.

Claim 2. By Caratheodory Extension Theorem, the Carathéodory measurable sets form a $\sigma$-algebra $\mathcal{M}$; since the Borel $\sigma$-algebra is generated by the open sets, and every open subset of $\mathbb{R}$ is a countable union of open intervals with rational endpoints (each point of an open set lies in such an interval inside the set, and there are countably many of them), while each open interval is obtained from rays $(a,\infty)$ by countable intersections, unions, and complements, it suffices to show that every ray $E=(a,\infty)$ lies in $\mathcal{M}$. Let $A\subseteq\mathbb{R}$ with $\lambda^{*}(A)<\infty$ (otherwise the splitting inequality is trivial) and let $\varepsilon>0$; choose a covering sequence $\bigl((a_m,b_m)\bigr)_m$ of $A$ with $\sum_m(b_m-a_m)\le\lambda^{*}(A)+\varepsilon$. For each $m$, the sets $(a_m,b_m)\cap(a,\infty)$ and $(a_m,b_m)\cap(-\infty,a]$ are contained, respectively, in the open intervals

whose lengths sum to at most $(b_m-a_m)+\varepsilon 2^{-m}$. The sequences $(I_m')$ and $(I_m'')$ cover $A\cap E$ and $A\setminus E$ respectively, so

Letting $\varepsilon\to 0$ and using the automatic reverse inequality (subadditivity) shows $E\in\mathcal{M}$. Hence $\mathcal{B}(\mathbb{R})\subseteq\mathcal{M}$.

Claim 3. Immediate from claims 1 and 2 and Caratheodory Extension Theorem: the restriction $\lambda$ of $\lambda^{*}$ to $\mathcal{B}(\mathbb{R})$ is a measure.

Claim 4. First, $\lambda^{*}([a,b])\le b-a$: for $\varepsilon>0$ the single interval $(a-\varepsilon,b+\varepsilon)$ covers $[a,b]$. Conversely, let $\bigl((a_m,b_m)\bigr)_m$ be any covering sequence of $[a,b]$. The closed interval $[a,b]$ is compact by Closed Interval $[a,b]$ is Compact in $\mathbb{R}$, so the open cover $\{(a_m,b_m)\}$ admits a finite subcover, using Compact Subset Criterion via Open Covers in the Ambient Space and Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$ to pass between subspace and ambient open covers. For a finite cover of $[a,b]$ by open intervals, an induction on the number of intervals shows the total length exceeds $b-a$: choose an interval $(a_{m_1},b_{m_1})$ containing $a$; if $b_{m_1}>b$ we are done since $b_{m_1}-a_{m_1}>b-a$; otherwise apply the inductive hypothesis to $[b_{m_1},b]$, covered by the remaining intervals. Hence $\sum_m(b_m-a_m)\ge b-a$, so $\lambda^{*}([a,b])=b-a$. For the open interval: $\lambda^{*}((a,b))\le b-a$ by the cover $(a,b)$ itself, and $\lambda^{*}((a,b))\ge\lambda^{*}([a+\varepsilon,b-\varepsilon])=b-a-2\varepsilon$ for small $\varepsilon>0$ by monotonicity; so $\lambda^{*}((a,b))=b-a$. Half-open intervals are squeezed between the open and closed ones by monotonicity. All these sets are Borel, so the values are values of $\lambda$.

Claim 5. $\mathbb{R}=\bigcup_{m\in\mathbb{N}}[-m,m]$ with $\lambda([-m,m])=2m<\infty$, so $\lambda$ is $\sigma$-finite in the sense of Measure, Measure Space, and Probability Measure. $\blacksquare$