[proofplan] We apply the [Caratheodory Extension Theorem](/theorems/522) to extend the length function from a ring of half-open intervals to $\mathcal{B}(\mathbb{R})$, then invoke the [Uniqueness of Measure Extension](/theorems/506) to show the extension is unique. The main technical work is verifying that the length function is countably additive on the ring, which requires a compactness argument. [/proofplan] [step:Define the ring of finite unions of half-open intervals and the length function] Let $\mathcal{A}$ be the collection of all finite disjoint unions of half-open intervals $(a, b]$ (including $\varnothing$). This is a ring: finite unions and set differences of such intervals can always be expressed as finite disjoint unions of half-open intervals. Define \begin{align*} \mu_0\!\left(\bigsqcup_{k=1}^n (a_k, b_k]\right) = \sum_{k=1}^n (b_k - a_k). \end{align*} [claim:Mu Zero Is Well Defined And Finitely Additive] $\mu_0$ is well-defined and finitely additive on $\mathcal{A}$. [/claim] [proof] If a set $A \in \mathcal{A}$ admits two disjoint representations, summing lengths over a common refinement shows both sums are equal. Finite additivity is immediate from the definition. [/proof] [/step] [step:Verify countable additivity of $\mu_0$ on $\mathcal{A}$ using a compactness argument] [claim:Mu Zero Is Countably Additive On The Ring] If $(A_n)_{n=1}^\infty$ is a sequence of pairwise disjoint sets in $\mathcal{A}$ with $\bigcup_n A_n = A \in \mathcal{A}$, then $\mu_0(A) = \sum_{n=1}^\infty \mu_0(A_n)$. [/claim] [proof] By finite additivity, $\mu_0(A) \geq \sum_{n=1}^\infty \mu_0(A_n)$. For the reverse, it suffices to treat $A = (a, b]$. Fix $\varepsilon > 0$. Shrink $A$ to the compact set $K = [a + \varepsilon, b] \subseteq (a, b]$. Expand each component $(c, d]$ of $A_n$ to $(c - \varepsilon_n, d + \varepsilon_n)$, choosing $\varepsilon_n > 0$ so that the total enlargement is at most $\varepsilon/2^n$. Call the enlarged open sets $U_n$. Then $K \subseteq \bigcup_n U_n$. Since $K$ is compact, finitely many $U_n$ cover $K$, say $K \subseteq \bigcup_{n=1}^N U_n$. Therefore \begin{align*} b - a - \varepsilon \leq \mu_0(K) \leq \sum_{n=1}^N \mu_0(U_n) \leq \sum_{n=1}^\infty \mu_0(A_n) + \varepsilon. \end{align*} Since $\varepsilon$ was arbitrary, $\mu_0(A) \leq \sum_n \mu_0(A_n)$. [/proof] [/step] [step:Apply the Caratheodory Extension Theorem to obtain Lebesgue measure] Since $\mu_0$ is a countably additive set function on the ring $\mathcal{A}$ with $\mu_0(\varnothing) = 0$, the [Caratheodory Extension Theorem](/theorems/522) provides a measure $\lambda$ on $\sigma(\mathcal{A})$ extending $\mu_0$. Since the half-open intervals $(a, b]$ with rational endpoints generate $\mathcal{B}(\mathbb{R})$, we have $\sigma(\mathcal{A}) = \mathcal{B}(\mathbb{R})$. [/step] [step:Prove uniqueness via the Dynkin $\pi$-system lemma] The half-open intervals $\{(a, b] : a < b\}$ form a $\pi$-system generating $\mathcal{B}(\mathbb{R})$. The sequence $E_n = (-n, n] \in \mathcal{A}$ satisfies $E_n \uparrow \mathbb{R}$ and $\lambda(E_n) = 2n < \infty$, so $\lambda$ is $\sigma$-finite on this $\pi$-system. By the [Uniqueness of Measure Extension](/theorems/506), any Borel measure agreeing with $\lambda$ on all half-open intervals must equal $\lambda$ on $\mathcal{B}(\mathbb{R})$. [/step] [step:Verify $\sigma$-finiteness and translation invariance] $\sigma$-finiteness was established in the previous step. For translation invariance: the measure $\mu_t(A) = \lambda(A - t)$ satisfies $\mu_t((a, b]) = \lambda((a - t, b - t]) = b - a = \lambda((a, b])$ for all intervals, so $\mu_t = \lambda$ on $\mathcal{B}(\mathbb{R})$ by uniqueness. [/step]