[proofplan] We decompose the bounded linear functional $\Lambda$ into positive and negative parts using a lattice-theoretic construction on $C_0(\mathbb{R}^n)$, apply the [Riesz Representation Theorem for Positive Functionals](/theorems/3036) to each part to obtain Radon measures $\mu^+$ and $\mu^-$, and set $\mu = \mu^+ - \mu^-$. The norm identity $\|\Lambda\| = |\mu|(\mathbb{R}^n)$ is established by proving both inequalities: the upper bound via the triangle inequality for integration, and the lower bound by approximating the sign function of the Jordan decomposition with continuous test functions. Uniqueness follows from the density of $C_c(\mathbb{R}^n)$ in $C_0(\mathbb{R}^n)$ and the fact that a Radon measure is determined by its integrals against continuous compactly supported functions. [/proofplan] [step:Construct the positive part $\Lambda^+$ of $\Lambda$] For each non-negative $f \in C_0(\mathbb{R}^n)$ with $f \geq 0$, define \begin{align*} \Lambda^+: C_0(\mathbb{R}^n) &\to \mathbb{R}, \\ \Lambda^+(f) &= \sup\{\Lambda(g) : g \in C_0(\mathbb{R}^n),\ 0 \leq g \leq f\} \quad \text{for } f \geq 0. \end{align*} We verify that $\Lambda^+$ is well-defined and finite on non-negative functions. Since $\Lambda$ is bounded with operator norm $\|\Lambda\|$, for any $g$ with $0 \leq g \leq f$ we have $\Lambda(g) \leq |\Lambda(g)| \leq \|\Lambda\| \cdot \|g\|_\infty \leq \|\Lambda\| \cdot \|f\|_\infty$. Therefore $\Lambda^+(f) \leq \|\Lambda\| \cdot \|f\|_\infty < \infty$. Moreover $\Lambda^+(f) \geq \Lambda(0) = 0$, so $\Lambda^+(f) \in [0, \infty)$. Extend $\Lambda^+$ to all of $C_0(\mathbb{R}^n)$ by writing $f = f^+ - f^-$ (where $f^+ = \max\{f, 0\}$ and $f^- = \max\{-f, 0\}$) and setting $\Lambda^+(f) = \Lambda^+(f^+) - \Lambda^+(f^-)$. One verifies that $\Lambda^+$ is linear and positive on $C_0(\mathbb{R}^n)$: additivity $\Lambda^+(f_1 + f_2) = \Lambda^+(f_1) + \Lambda^+(f_2)$ for non-negative $f_1, f_2$ follows because $0 \leq g \leq f_1 + f_2$ can be decomposed as $g = g_1 + g_2$ with $0 \leq g_1 \leq f_1$ and $0 \leq g_2 \leq f_2$ (taking $g_1 = \min(g, f_1)$ and $g_2 = g - g_1$), and conversely any such pair gives an admissible $g$. [/step] [step:Define $\Lambda^- = \Lambda^+ - \Lambda$ and verify it is a positive linear functional] Set $\Lambda^- = \Lambda^+ - \Lambda$. Since $\Lambda^+$ and $\Lambda$ are both linear, $\Lambda^-$ is linear. To verify positivity: for $f \geq 0$, \begin{align*} \Lambda^-(f) = \Lambda^+(f) - \Lambda(f) = \sup\{\Lambda(g) : 0 \leq g \leq f\} - \Lambda(f). \end{align*} Taking $g = f$ in the supremum shows $\Lambda^+(f) \geq \Lambda(f)$, so $\Lambda^-(f) \geq 0$. Hence $\Lambda^-$ is a positive linear functional on $C_0(\mathbb{R}^n)$. Moreover, $\Lambda = \Lambda^+ - \Lambda^-$ by construction. [/step] [step:Apply the positive Riesz theorem to obtain $\mu^+$ and $\mu^-$] Both $\Lambda^+$ and $\Lambda^-$ are positive linear functionals on $C_0(\mathbb{R}^n)$. Since $C_c(\mathbb{R}^n) \subset C_0(\mathbb{R}^n)$, the restrictions $\Lambda^+|_{C_c(\mathbb{R}^n)}$ and $\Lambda^-|_{C_c(\mathbb{R}^n)}$ are positive linear functionals on $C_c(\mathbb{R}^n)$. By the [Riesz Representation Theorem for Positive Functionals](/theorems/3036), there exist unique Radon measures $\mu^+$ and $\mu^-$ on $\mathbb{R}^n$ such that \begin{align*} \Lambda^+(f) = \int_{\mathbb{R}^n} f \, d\mu^+(x), \qquad \Lambda^-(f) = \int_{\mathbb{R}^n} f \, d\mu^-(x) \end{align*} for all $f \in C_c(\mathbb{R}^n)$. The equality extends to all $f \in C_0(\mathbb{R}^n)$ by a density argument: $C_c(\mathbb{R}^n)$ is dense in $C_0(\mathbb{R}^n)$ under the supremum norm, and both sides are continuous with respect to this norm. We verify finite total variation. For any $f \in C_0(\mathbb{R}^n)$ with $\|f\|_\infty \leq 1$, we have $\Lambda^+(f) \leq \Lambda^+(|f|) \leq \|\Lambda\| \cdot \|f\|_\infty \leq \|\Lambda\|$. Taking $f = \varphi_k$ where $\varphi_k \in C_c(\mathbb{R}^n)$ with $0 \leq \varphi_k \leq 1$ and $\varphi_k \nearrow 1$ pointwise, the Monotone Convergence Theorem gives $\mu^+(\mathbb{R}^n) = \lim_{k \to \infty} \int \varphi_k \, d\mu^+ = \lim_{k \to \infty} \Lambda^+(\varphi_k) \leq \|\Lambda\|$. Similarly $\mu^-(\mathbb{R}^n) \leq \|\Lambda\|$. [/step] [step:Define $\mu = \mu^+ - \mu^-$ and verify the representation] Define the signed Radon measure $\mu = \mu^+ - \mu^-$. The total variation measure satisfies $|\mu| \leq \mu^+ + \mu^-$ (with equality when $\mu^+$ and $\mu^-$ are mutually singular, which can be arranged by passing to the Jordan decomposition), so $|\mu|(\mathbb{R}^n) \leq \mu^+(\mathbb{R}^n) + \mu^-(\mathbb{R}^n) < \infty$. For every $f \in C_0(\mathbb{R}^n)$, \begin{align*} \int_{\mathbb{R}^n} f \, d\mu = \int_{\mathbb{R}^n} f \, d\mu^+ - \int_{\mathbb{R}^n} f \, d\mu^- = \Lambda^+(f) - \Lambda^-(f) = \Lambda(f). \end{align*} This establishes the representation. [/step] [step:Prove the norm identity $\|\Lambda\| = |\mu|(\mathbb{R}^n)$] **Upper bound:** For any $f \in C_0(\mathbb{R}^n)$ with $\|f\|_\infty \leq 1$, \begin{align*} |\Lambda(f)| = \left|\int_{\mathbb{R}^n} f \, d\mu\right| \leq \int_{\mathbb{R}^n} |f| \, d|\mu|(x) \leq \|f\|_\infty \cdot |\mu|(\mathbb{R}^n) \leq |\mu|(\mathbb{R}^n). \end{align*} Taking the supremum over such $f$ gives $\|\Lambda\| \leq |\mu|(\mathbb{R}^n)$. **Lower bound:** Let $\mu = \mu^+ - \mu^-$ be the Jordan decomposition with $\mu^+ \perp \mu^-$. There exist disjoint Borel sets $P, N$ with $P \cup N = \mathbb{R}^n$, $\mu^+(N) = 0$, and $\mu^-(P) = 0$. Define $\sigma = \mathbb{1}_P - \mathbb{1}_N$, the sign function of $\mu$. For each $\varepsilon > 0$, use inner regularity of $\mu^+$ and $\mu^-$ to find compact sets $K_P \subset P$ and $K_N \subset N$ with $\mu^+(P \setminus K_P) < \varepsilon$ and $\mu^-(N \setminus K_N) < \varepsilon$. Since $K_P$ and $K_N$ are disjoint compact sets, [Urysohn's Lemma](/theorems/???) provides $f_\varepsilon \in C_0(\mathbb{R}^n)$ with $\|f_\varepsilon\|_\infty \leq 1$, $f_\varepsilon = 1$ on $K_P$, and $f_\varepsilon = -1$ on $K_N$. Then \begin{align*} \Lambda(f_\varepsilon) &= \int_{\mathbb{R}^n} f_\varepsilon \, d\mu^+ - \int_{\mathbb{R}^n} f_\varepsilon \, d\mu^- \\ &\geq \int_{K_P} 1 \, d\mu^+ - \int_{P \setminus K_P} 1 \, d\mu^+ - \int_{K_N} 1 \, d\mu^- + \int_{K_N} 1 \, d\mu^- \\ &= \mu^+(K_P) + \mu^-(K_N) - \mu^+(P \setminus K_P) \\ &\geq \mu^+(P) - \varepsilon + \mu^-(N) - \varepsilon - \varepsilon \\ &= |\mu|(\mathbb{R}^n) - 3\varepsilon. \end{align*} Since $\|f_\varepsilon\|_\infty \leq 1$, this gives $\|\Lambda\| \geq |\mu|(\mathbb{R}^n) - 3\varepsilon$. Since $\varepsilon > 0$ is arbitrary, $\|\Lambda\| \geq |\mu|(\mathbb{R}^n)$. [/step] [step:Establish uniqueness of $\mu$] Suppose $\nu$ is another signed Radon measure with finite total variation satisfying $\int f \, d\nu = \Lambda(f)$ for all $f \in C_0(\mathbb{R}^n)$. Then $\int f \, d(\mu - \nu) = 0$ for all $f \in C_0(\mathbb{R}^n)$. Since $C_c(\mathbb{R}^n) \subset C_0(\mathbb{R}^n)$, this holds in particular for all $f \in C_c(\mathbb{R}^n)$. The signed measure $\mu - \nu$ has finite total variation. The positive and negative parts $(\mu - \nu)^+$ and $(\mu - \nu)^-$ are Radon measures satisfying $\int f \, d(\mu - \nu)^+ = \int f \, d(\mu - \nu)^-$ for all $f \in C_c(\mathbb{R}^n)$. By uniqueness in the [Riesz Representation Theorem for Positive Functionals](/theorems/3036), $(\mu - \nu)^+ = (\mu - \nu)^-$, which forces $\mu - \nu = 0$, i.e., $\mu = \nu$. [/step]