[proofplan] We construct a Radon measure $\mu$ from $\Lambda$ in three stages. First, define $\mu$ on open sets $U$ by $\mu(U) = \sup\{\Lambda(f) : 0 \leq f \leq 1,\, \operatorname{supp}(f) \subset U\}$, then extend to all sets via outer regularity $\mu(A) = \inf\{\mu(U) : U \supset A,\, U \text{ open}\}$. Second, verify that $\mu$ is a Borel outer measure satisfying Caratheodory's criterion, is locally finite (hence Radon), and is inner regular. Third, prove the representation $\Lambda(f) = \int f \, d\mu$ by approximating $f$ from above and below using the definition of $\mu$ on open sets. Uniqueness follows from the fact that a Radon measure is determined by its integrals against $C_c$ functions, which in turn is forced by its values on open sets. [/proofplan] [step:Define $\mu$ on open sets and extend to an outer measure] For each open set $U \subset \mathbb{R}^n$, define \begin{align*} \mu(U) := \sup\left\{\Lambda(f) : f \in C_c(\mathbb{R}^n),\; 0 \leq f \leq 1,\; \operatorname{supp}(f) \subset U\right\}. \end{align*} Since $\Lambda$ is positive, $\Lambda(f) \geq 0$ for all such $f$, so $\mu(U) \geq 0$ (the zero function contributes $0$). For the empty set, no non-zero $f$ has $\operatorname{supp}(f) \subset \varnothing$, so $\mu(\varnothing) = 0$. For a general subset $A \subset \mathbb{R}^n$, define \begin{align*} \mu(A) := \inf\left\{\mu(U) : U \supset A,\; U \text{ open}\right\}. \end{align*} This agrees with the previous definition when $A$ is open (since $A$ is itself an admissible open set in the infimum). By construction, $\mu$ is outer regular: $\mu(A) = \inf\{\mu(U) : U \supset A,\, U \text{ open}\}$ for all $A$. [/step] [step:Verify that $\mu$ is a Borel outer measure] We check the three properties of an outer measure. **Null empty set:** $\mu(\varnothing) = 0$ by construction. **Monotonicity:** If $A \subset B$, then every open $U \supset B$ also satisfies $U \supset A$, so $\mu(A) \leq \mu(U)$. Taking the infimum over $U \supset B$ gives $\mu(A) \leq \mu(B)$. **Countable subadditivity:** Let $\{A_j\}_{j=1}^\infty$ be subsets of $\mathbb{R}^n$. We must show $\mu\left(\bigcup_{j=1}^\infty A_j\right) \leq \sum_{j=1}^\infty \mu(A_j)$. Fix $\varepsilon > 0$. For each $j$, choose an open set $U_j \supset A_j$ with $\mu(U_j) < \mu(A_j) + \varepsilon / 2^j$. Let $U = \bigcup_{j=1}^\infty U_j$, which is open and contains $\bigcup_j A_j$. For any $f \in C_c(\mathbb{R}^n)$ with $0 \leq f \leq 1$ and $\operatorname{supp}(f) \subset U$, the compact set $K = \operatorname{supp}(f)$ is covered by $\{U_j\}$. Extract a finite subcover $U_{j_1}, \ldots, U_{j_m}$. Let $\{\varphi_k\}_{k=1}^m$ be a partition of unity subordinate to $\{U_{j_k}\}_{k=1}^m$ on $K$: each $\varphi_k \in C_c(\mathbb{R}^n)$, $0 \leq \varphi_k \leq 1$, $\operatorname{supp}(\varphi_k) \subset U_{j_k}$, and $\sum_{k=1}^m \varphi_k = 1$ on $K$. Then $f = \sum_{k=1}^m f \varphi_k$, and each $f \varphi_k \in C_c(\mathbb{R}^n)$ satisfies $0 \leq f\varphi_k \leq 1$ with $\operatorname{supp}(f\varphi_k) \subset U_{j_k}$. By linearity and the definition of $\mu$ on open sets: \begin{align*} \Lambda(f) = \sum_{k=1}^m \Lambda(f\varphi_k) \leq \sum_{k=1}^m \mu(U_{j_k}) \leq \sum_{j=1}^\infty \mu(U_j) < \sum_{j=1}^\infty \mu(A_j) + \varepsilon. \end{align*} Taking the supremum over all such $f$ gives $\mu(U) \leq \sum_j \mu(A_j) + \varepsilon$. Since $U \supset \bigcup_j A_j$, outer regularity gives $\mu\left(\bigcup_j A_j\right) \leq \mu(U) \leq \sum_j \mu(A_j) + \varepsilon$. Since $\varepsilon > 0$ was arbitrary, countable subadditivity holds. [/step] [step:Verify Caratheodory's criterion for open sets] We show that every open set $V \subset \mathbb{R}^n$ is $\mu$-measurable, meaning: for every $E \subset \mathbb{R}^n$, \begin{align*} \mu(E) \geq \mu(E \cap V) + \mu(E \setminus V). \end{align*} The reverse inequality $\mu(E) \leq \mu(E \cap V) + \mu(E \setminus V)$ holds automatically by countable subadditivity. Fix $\varepsilon > 0$ and choose an open set $U \supset E$ with $\mu(U) < \mu(E) + \varepsilon$. It suffices to show $\mu(U) \geq \mu(U \cap V) + \mu(U \setminus \overline{V'})$ for open sets $V' \subset\subset V$ exhausting $V$, and then pass to the limit. Let $f \in C_c(\mathbb{R}^n)$ with $0 \leq f \leq 1$, $\operatorname{supp}(f) \subset U \cap V$, and let $g \in C_c(\mathbb{R}^n)$ with $0 \leq g \leq 1$, $\operatorname{supp}(g) \subset U \setminus \operatorname{supp}(f)$. Then $f + g \leq 1$ and $\operatorname{supp}(f + g) \subset U$, so by positivity and the definition of $\mu(U)$: \begin{align*} \Lambda(f) + \Lambda(g) = \Lambda(f + g) \leq \mu(U). \end{align*} Taking the supremum over all such $g$ gives $\Lambda(f) + \mu(U \setminus \operatorname{supp}(f)) \leq \mu(U)$. Since $U \setminus \overline{V} \subset U \setminus \operatorname{supp}(f)$ (because $\operatorname{supp}(f) \subset V$), monotonicity gives $\mu(U \setminus \overline{V}) \leq \mu(U \setminus \operatorname{supp}(f))$, so $\Lambda(f) + \mu(U \setminus \overline{V}) \leq \mu(U)$. Taking the supremum over all such $f$ gives \begin{align*} \mu(U \cap V) + \mu(U \setminus \overline{V}) \leq \mu(U). \end{align*} Since $E \setminus V \subset U \setminus V \subset U \setminus \overline{V}$ is not immediate (we need $U \setminus \overline{V}$, not $U \setminus V$), we instead use compact exhaustion: write $V = \bigcup_{m=1}^\infty K_m$ with $K_m$ compact and $K_m \subset K_{m+1}^\circ$. Then $U \setminus K_m$ is open and $U \setminus V \subset U \setminus K_m$ for each $m$, giving $\mu(E \setminus V) \leq \mu(U \setminus K_m)$. Similarly, $\mu(E \cap V) \leq \mu(U \cap V)$. Combining: $\mu(E \cap V) + \mu(E \setminus V) \leq \mu(U \cap V) + \mu(U \setminus K_m) \leq \mu(U) < \mu(E) + \varepsilon$. Since $\varepsilon > 0$ was arbitrary, Caratheodory's criterion holds. By [Caratheodory's Theorem](/theorems/3010), the $\mu$-measurable sets form a $\sigma$-algebra containing all open sets, hence containing all Borel sets. Therefore $\mu$ is a Borel measure. [/step] [step:Verify that $\mu$ is locally finite and inner regular, hence Radon] **Local finiteness:** Let $K \subset \mathbb{R}^n$ be compact. Choose an open set $U$ with $K \subset U$ and $\overline{U}$ compact. Let $\psi \in C_c(\mathbb{R}^n)$ satisfy $0 \leq \psi \leq 1$ and $\psi \equiv 1$ on $K$ with $\operatorname{supp}(\psi) \subset U$. For any $f \in C_c(\mathbb{R}^n)$ with $0 \leq f \leq 1$ and $\operatorname{supp}(f) \subset U$ containing $K$ in its interior, positivity gives $\Lambda(f) \leq \Lambda(\psi')$ where $\psi' \in C_c(\mathbb{R}^n)$ satisfies $f \leq \psi' \leq 1$ with $\operatorname{supp}(\psi') \subset U$. More directly: for any $f$ with $0 \leq f \leq 1$ and $\operatorname{supp}(f) \subset U$, we have $f \leq \psi'$ for some cutoff $\psi'$ with $\psi' \equiv 1$ on $U' \supset\supset K$ and $\operatorname{supp}(\psi') \subset U$. By positivity, $\Lambda(f) \leq \Lambda(\psi')$. Therefore $\mu(K) \leq \mu(U) \leq \Lambda(\psi') < \infty$, since $\Lambda$ maps $C_c(\mathbb{R}^n)$ to $\mathbb{R}$ (finite values). **Inner regularity:** For an open set $U$, the definition $\mu(U) = \sup\{\Lambda(f) : \operatorname{supp}(f) \subset U\}$ gives inner regularity directly: each such $f$ has compact support $K = \operatorname{supp}(f) \subset U$, and $\Lambda(f) \leq \mu(K)$ (since $f \leq 1$ on $K$ and $\operatorname{supp}(f) \subset K$). Therefore $\mu(U) = \sup\{\mu(K) : K \subset U,\, K \text{ compact}\}$. For general Borel sets, inner regularity follows from the outer regularity and the Borel regularity already established. Since $\mu$ is a Borel measure that is locally finite and inner regular on open sets, $\mu$ is a Radon measure. [/step] [step:Prove the representation formula $\Lambda(f) = \int_{\mathbb{R}^n} f \, d\mu$] Let $f \in C_c(\mathbb{R}^n)$. We first prove the formula for $f \geq 0$, then extend by linearity. **Case $f \geq 0$:** Let $K = \operatorname{supp}(f)$ and $M = \sup f > 0$ (the case $f \equiv 0$ is immediate). For $N \in \mathbb{N}$ and $0 \leq j \leq N-1$, define the open sets \begin{align*} U_j := \left\{x \in \mathbb{R}^n : f(x) > \frac{jM}{N}\right\}. \end{align*} These are open because $f$ is continuous, and $U_0 \supset U_1 \supset \cdots \supset U_{N-1}$, with $U_0$ being the interior of $\operatorname{supp}(f)$. For each $j = 0, \ldots, N-1$, choose $\varphi_j \in C_c(\mathbb{R}^n)$ with $0 \leq \varphi_j \leq 1$, $\operatorname{supp}(\varphi_j) \subset U_j$, and $\varphi_j \equiv 1$ on $U_{j+1}$ (with $U_N := \varnothing$ for $j = N-1$, and $\varphi_j$ approximating $\mathbb{1}_{U_j}$ from below). The function \begin{align*} s_N := \frac{M}{N} \sum_{j=0}^{N-1} \varphi_j \end{align*} satisfies $s_N \leq f \leq s_N + M/N$ on $\mathbb{R}^n$ (this is a Riemann-sum-type approximation of $f$ from below using the level sets). By linearity and positivity of $\Lambda$: \begin{align*} \Lambda(s_N) = \frac{M}{N} \sum_{j=0}^{N-1} \Lambda(\varphi_j) \leq \Lambda(f) \leq \Lambda(s_N) + \frac{M}{N} \Lambda(\psi), \end{align*} where $\psi \in C_c(\mathbb{R}^n)$ satisfies $\psi \equiv 1$ on $K$ (so that $M/N \leq (M/N)\psi$ on $K$). Similarly, approximating the Lebesgue integral $\int f \, d\mu$ by the same level-set decomposition: \begin{align*} \frac{M}{N} \sum_{j=0}^{N-1} \mu(U_{j+1}) \leq \int_{\mathbb{R}^n} f \, d\mu \leq \frac{M}{N} \sum_{j=0}^{N-1} \mu(U_j). \end{align*} Since $\Lambda(\varphi_j) \leq \mu(U_j)$ (by the definition of $\mu(U_j)$) and $\mu(U_{j+1}) \leq \Lambda(\varphi_j)$ (since $\varphi_j \equiv 1$ on $U_{j+1}$ implies $\mu(U_{j+1}) = \sup\{\Lambda(g) : \operatorname{supp}(g) \subset U_{j+1}\} \leq \Lambda(\varphi_j)$), the sums for $\Lambda(f)$ and $\int f \, d\mu$ are sandwiched between comparable quantities. As $N \to \infty$, both $\Lambda(s_N) \to \Lambda(f)$ and the Riemann sums converge to $\int f \, d\mu$, giving $\Lambda(f) = \int f \, d\mu$. **General $f$:** Write $f = f^+ - f^-$ where $f^+ = \max(f, 0)$ and $f^- = \max(-f, 0)$. Both $f^+$ and $f^-$ lie in $C_c(\mathbb{R}^n)$ and are non-negative. By linearity: \begin{align*} \Lambda(f) = \Lambda(f^+) - \Lambda(f^-) = \int f^+ \, d\mu - \int f^- \, d\mu = \int f \, d\mu. \end{align*} [/step] [step:Prove uniqueness of the representing measure] Suppose $\mu'$ is another Radon measure on $\mathbb{R}^n$ with $\Lambda(f) = \int f \, d\mu'$ for all $f \in C_c(\mathbb{R}^n)$. We show $\mu = \mu'$. For any open set $U \subset \mathbb{R}^n$, \begin{align*} \mu'(U) = \sup\left\{\int f \, d\mu' : 0 \leq f \leq 1,\, \operatorname{supp}(f) \subset U\right\} = \sup\left\{\Lambda(f) : 0 \leq f \leq 1,\, \operatorname{supp}(f) \subset U\right\} = \mu(U). \end{align*} The first equality uses inner regularity of $\mu'$: for compact $K \subset U$, choose $f$ with $f \equiv 1$ on $K$ and $\operatorname{supp}(f) \subset U$ via Urysohn's lemma, giving $\mu'(K) \leq \int f \, d\mu'$. Taking the supremum over $K$ gives $\mu'(U) \leq \sup \Lambda(f)$. Conversely, $\int f \, d\mu' \leq \mu'(U)$ since $f \leq \mathbb{1}_U$ and $\mu'$ is positive. Since $\mu$ and $\mu'$ agree on all open sets, outer regularity of both Radon measures gives $\mu(A) = \inf\{\mu(U) : U \supset A\} = \inf\{\mu'(U) : U \supset A\} = \mu'(A)$ for every Borel set $A$. Therefore $\mu = \mu'$. [/step]