[proofplan] We construct a weak-$*$ convergent subsequence by a diagonal argument. First, we select a countable dense subset $\{f_\ell\}_{\ell \geq 1}$ of $C_c(\mathbb{R}^n)$ (which exists because $C_c(\mathbb{R}^n)$ is separable in the supremum norm). The local boundedness hypothesis ensures that the numerical sequences $(\int f_\ell \, d\mu_k)_k$ are bounded for each $\ell$. Successive applications of Bolzano--Weierstrass and a diagonal extraction produce a single subsequence $(\mu_{k_j})$ such that $\int f_\ell \, d\mu_{k_j}$ converges for every $\ell$. We extend this convergence to all of $C_c(\mathbb{R}^n)$ by density, verify that the resulting limit functional is positive and linear, and then invoke the [Riesz Representation Theorem for Positive Functionals](/theorems/3036) to obtain the limiting Radon measure $\mu$. [/proofplan] [step:Establish that the numerical sequence $\left(\int f \, d\mu_k\right)_k$ is bounded for each $f \in C_c(\mathbb{R}^n)$] Fix $f \in C_c(\mathbb{R}^n)$. Let $K = \operatorname{supp}(f)$, which is compact. Since $(\mu_k)$ is locally bounded, the constant $M_K := \sup_{k \geq 1} \mu_k(K) < \infty$ is finite. For every $k$, \begin{align*} \left|\int_{\mathbb{R}^n} f \, d\mu_k\right| \leq \int_{\mathbb{R}^n} |f| \, d\mu_k = \int_K |f| \, d\mu_k \leq \|f\|_\infty \cdot \mu_k(K) \leq \|f\|_\infty \cdot M_K. \end{align*} Hence the sequence of real numbers $\left(\int f \, d\mu_k\right)_{k \geq 1}$ is bounded by $\|f\|_\infty \cdot M_K$. [/step] [step:Select a countable dense subset of $C_c(\mathbb{R}^n)$ in the supremum norm] The space $C_c(\mathbb{R}^n)$ is separable in the topology induced by the supremum norm $\|\cdot\|_\infty$. To see this, write $\mathbb{R}^n = \bigcup_{j=1}^\infty \overline{B}(0, j)$ and note that for each $j$, the space $C(\overline{B}(0,j))$ is separable by the Stone--Weierstrass theorem (polynomials with rational coefficients are dense). Functions in $C_c(\mathbb{R}^n)$ supported in $\overline{B}(0,j)$ are dense in $\{f \in C_c(\mathbb{R}^n) : \operatorname{supp}(f) \subset \overline{B}(0,j)\}$, and the union over $j$ is dense in all of $C_c(\mathbb{R}^n)$. A countable union of countable dense sets is countable, so we obtain a countable dense subset $\{f_\ell\}_{\ell \geq 1} \subset C_c(\mathbb{R}^n)$. [/step] [step:Extract a diagonal subsequence along which $\int f_\ell \, d\mu_{k_j}$ converges for every $\ell$] We apply the Bolzano--Weierstrass theorem iteratively. For $\ell = 1$: the sequence $\left(\int f_1 \, d\mu_k\right)_{k}$ is bounded (by the first step), so there exists a subsequence $(\mu_{k}^{(1)})$ of $(\mu_k)$ such that $\int f_1 \, d\mu_k^{(1)}$ converges to some $L_1 \in \mathbb{R}$. For $\ell = 2$: the sequence $\left(\int f_2 \, d\mu_k^{(1)}\right)_{k}$ is bounded, so there exists a further subsequence $(\mu_k^{(2)})$ of $(\mu_k^{(1)})$ such that $\int f_2 \, d\mu_k^{(2)}$ converges to some $L_2 \in \mathbb{R}$. Continuing inductively, at stage $\ell$ we extract $(\mu_k^{(\ell)})$ as a subsequence of $(\mu_k^{(\ell-1)})$ such that $\int f_\ell \, d\mu_k^{(\ell)} \to L_\ell$. Define the diagonal subsequence $\mu_{k_j} = \mu_j^{(j)}$. For each fixed $\ell$, the sequence $(\mu_{k_j})_{j \geq \ell}$ is a subsequence of $(\mu_k^{(\ell)})$, so \begin{align*} \lim_{j \to \infty} \int_{\mathbb{R}^n} f_\ell \, d\mu_{k_j} = L_\ell \end{align*} for every $\ell \geq 1$. [/step] [step:Extend convergence from the dense set $\{f_\ell\}$ to all of $C_c(\mathbb{R}^n)$] Fix an arbitrary $f \in C_c(\mathbb{R}^n)$ with $K = \operatorname{supp}(f)$. Let $\varepsilon > 0$. By density of $\{f_\ell\}$ in $C_c(\mathbb{R}^n)$, choose $f_\ell$ with $\|f - f_\ell\|_\infty < \varepsilon$ and $\operatorname{supp}(f_\ell) \subset K'$ for some compact $K'$ containing $K$. Set $M_{K'} = \sup_{k} \mu_k(K')$. For all $j, j'$ sufficiently large, \begin{align*} \left|\int f \, d\mu_{k_j} - \int f \, d\mu_{k_{j'}}\right| &\leq \left|\int (f - f_\ell) \, d\mu_{k_j}\right| + \left|\int f_\ell \, d\mu_{k_j} - \int f_\ell \, d\mu_{k_{j'}}\right| + \left|\int (f_\ell - f) \, d\mu_{k_{j'}}\right| \\ &\leq \varepsilon \cdot M_{K'} + \left|\int f_\ell \, d\mu_{k_j} - \int f_\ell \, d\mu_{k_{j'}}\right| + \varepsilon \cdot M_{K'}. \end{align*} Since $\int f_\ell \, d\mu_{k_j}$ converges, the middle term is less than $\varepsilon$ for $j, j'$ large enough. The total is at most $(2M_{K'} + 1)\varepsilon$. As $\varepsilon > 0$ was arbitrary, $\left(\int f \, d\mu_{k_j}\right)_j$ is a Cauchy sequence in $\mathbb{R}$ and therefore converges to a limit $\Lambda(f) \in \mathbb{R}$. [/step] [step:Verify that $\Lambda : C_c(\mathbb{R}^n) \to \mathbb{R}$ is a positive linear functional] **Linearity**: For $f, g \in C_c(\mathbb{R}^n)$ and $\alpha, \beta \in \mathbb{R}$, \begin{align*} \Lambda(\alpha f + \beta g) = \lim_{j \to \infty} \int (\alpha f + \beta g) \, d\mu_{k_j} = \alpha \lim_{j \to \infty} \int f \, d\mu_{k_j} + \beta \lim_{j \to \infty} \int g \, d\mu_{k_j} = \alpha \Lambda(f) + \beta \Lambda(g). \end{align*} **Positivity**: If $f \geq 0$, then $\int f \, d\mu_{k_j} \geq 0$ for every $j$ (since each $\mu_{k_j}$ is a positive Radon measure), so $\Lambda(f) = \lim_j \int f \, d\mu_{k_j} \geq 0$. [/step] [step:Apply the Riesz Representation Theorem to obtain the limiting Radon measure $\mu$] Since $\Lambda : C_c(\mathbb{R}^n) \to \mathbb{R}$ is a positive linear functional, the [Riesz Representation Theorem for Positive Functionals](/theorems/3036) guarantees the existence of a unique Radon measure $\mu$ on $\mathbb{R}^n$ such that \begin{align*} \Lambda(f) = \int_{\mathbb{R}^n} f \, d\mu \end{align*} for all $f \in C_c(\mathbb{R}^n)$. The hypotheses of the Riesz theorem require $\Lambda$ to be a positive linear functional on $C_c(\mathbb{R}^n)$, which we verified in the previous step. Therefore \begin{align*} \lim_{j \to \infty} \int_{\mathbb{R}^n} f \, d\mu_{k_j} = \int_{\mathbb{R}^n} f \, d\mu \end{align*} for all $f \in C_c(\mathbb{R}^n)$, which is the definition of $\mu_{k_j} \overset{*}{\rightharpoonup} \mu$. [/step]