[proofplan] We construct the partition of unity in four stages. First, we build a countable, locally finite open refinement of the given cover using a compact exhaustion of $\Omega$ and the annular regions between consecutive exhaustion sets. Second, we attach a smooth non-negative bump function to each refinement element. Third, we normalise by the (locally finite, strictly positive) sum of all bumps to produce a partition of unity subordinate to the refinement. Finally, we group the refinement-level partition functions by their parent cover element to obtain a partition subordinate to the original cover $\{U_i\}_{i \in I}$. [/proofplan] [step:Construct a countable, locally finite open refinement from a compact exhaustion] Choose a compact exhaustion $K_1 \subset K_2 \subset \cdots$ of $\Omega$ with $K_j \subset \operatorname{int}(K_{j+1})$ and $\bigcup_{j=1}^\infty K_j = \Omega$. Set $K_0 := \varnothing$ and define annular regions \begin{align*} A_j &:= \operatorname{int}(K_{j+1}) \setminus K_{j-1}, \quad j = 1, 2, 3, \ldots \end{align*} Each $A_j$ is open, $\overline{A_j} \subseteq K_{j+1}$ is compact, and $\bigcup_{j=1}^\infty A_j = \Omega$. For each $x \in A_j$, choose an index $i(x) \in I$ with $x \in U_{i(x)}$ and a ball $B_x \subseteq U_{i(x)} \cap A_j$ centred at $x$. By compactness of $\overline{A_j}$, finitely many such balls $B_{x_1}^{(j)}, \ldots, B_{x_{N_j}}^{(j)}$ cover $\overline{A_j} \cap \Omega$. The collection \begin{align*} \{V_k\}_{k=1}^\infty &:= \bigcup_{j=1}^\infty \bigl\{B_{x_1}^{(j)}, \ldots, B_{x_{N_j}}^{(j)}\bigr\} \end{align*} is a countable open cover of $\Omega$ refining $\{U_i\}$, with each $\overline{V_k}$ compact and contained in $\Omega$. It is locally finite: any compact $K \subset \Omega$ satisfies $K \subseteq K_m$ for some $m$, and $A_j \cap K_m = \varnothing$ for $j \ge m + 2$ (since $A_j \subseteq \Omega \setminus K_{j-1}$ and $K_m \subseteq K_{j-1}$ when $j \ge m + 2$), so $K$ meets only balls arising from $A_1, \ldots, A_{m+1}$ — finitely many. [guided] The goal is to replace the given cover $\{U_i\}_{i \in I}$ (which may be uncountable and have no finiteness properties) with a countable, locally finite refinement. The tool is a compact exhaustion. Choose a compact exhaustion $K_1 \subset K_2 \subset \cdots$ of $\Omega$ satisfying $K_j \subset \operatorname{int}(K_{j+1})$ and $\bigcup_{j=1}^\infty K_j = \Omega$. Such an exhaustion exists for any open $\Omega \subseteq \mathbb{R}^n$: take $K_j = \{x \in \Omega : |x| \le j,\ \operatorname{dist}(x, \partial\Omega) \ge 1/j\}$. Set $K_0 := \varnothing$ and define annular regions \begin{align*} A_j &:= \operatorname{int}(K_{j+1}) \setminus K_{j-1}, \quad j = 1, 2, 3, \ldots \end{align*} Why annular regions? Each $A_j$ is an open set that captures the "shell" of $\Omega$ between exhaustion levels $j-1$ and $j+1$. The overlap between consecutive annuli is controlled: $A_j$ and $A_l$ are disjoint whenever $|j - l| \ge 3$. This controlled overlap is what will make the refinement locally finite. Each $A_j$ is open, $\overline{A_j} \subseteq K_{j+1}$ is compact, and $\bigcup_{j=1}^\infty A_j = \Omega$. To verify the union equals $\Omega$: for any $x \in \Omega$, pick the smallest $m$ with $x \in K_m$. Then $x \in \operatorname{int}(K_{m+1})$ (since $K_m \subset \operatorname{int}(K_{m+1})$), and if $m \ge 2$ then $x \notin K_{m-1}$ by minimality, so $x \in A_m$. If $m = 1$, then $x \in \operatorname{int}(K_2) \setminus K_0 = A_1$. For each $x \in A_j$, choose $i(x) \in I$ with $x \in U_{i(x)}$ and select a ball $B_x \subseteq U_{i(x)} \cap A_j$ centred at $x$. The intersection $U_{i(x)} \cap A_j$ is open and contains $x$, so such a ball exists. Since $\overline{A_j}$ is compact, finitely many of these balls $B_{x_1}^{(j)}, \ldots, B_{x_{N_j}}^{(j)}$ cover $\overline{A_j} \cap \Omega$. Enumerate all such balls across all $j$ to form the countable collection $\{V_k\}_{k=1}^\infty$. This collection is locally finite: given a compact $K \subset \Omega$, choose $m$ with $K \subseteq K_m$. For $j \ge m + 2$, every $V_k$ arising from $A_j$ satisfies $V_k \subseteq A_j \subseteq \Omega \setminus K_{j-1}$, and $K \subseteq K_m \subseteq K_{j-1}$, so $V_k \cap K = \varnothing$. Hence $K$ meets only balls from $A_1, \ldots, A_{m+1}$, of which there are finitely many. [/guided] [/step] [step:Attach a smooth non-negative bump function to each refinement element] Each $V_k$ is an open ball $B(x_k^0, R_k)$ by construction (chosen in step 1). Select a radius $r_k$ with $0 < r_k < R_k$ and define \begin{align*} \psi_k &:= \eta_{\delta_k} * \mathbf{1}_{B(x_k^0, r_k)}: \Omega \to [0, \infty), \end{align*} where $\delta_k := R_k - r_k > 0$ and $\eta_{\delta_k}$ is the [standard mollifier](/page/Standard%20Mollifier) at scale $\delta_k$. Since $\mathbf{1}_{B(x_k^0, r_k)} \in L^1(\mathbb{R}^n)$ and $\eta_{\delta_k} \in C_c^\infty(B(0,\delta_k))$, the convolution $\psi_k \in C^\infty(\mathbb{R}^n)$ with $\psi_k \ge 0$. The support satisfies $\operatorname{supp}(\psi_k) \subseteq \overline{B(x_k^0, r_k + \delta_k)} = \overline{B(x_k^0, R_k)} = \overline{V_k}$. Moreover, $\psi_k > 0$ on all of $V_k$: for any $y \in V_k = B(x_k^0, R_k)$, the ball $B(y, \delta_k)$ intersects $B(x_k^0, r_k)$ in a set of positive $\mathcal{L}^n$-measure (since $|y - x_k^0| < R_k = r_k + \delta_k$), so \begin{align*} \psi_k(y) &= \int_{B(y,\delta_k)} \eta_{\delta_k}(y - z)\, \mathbf{1}_{B(x_k^0, r_k)}(z) \, d\mathcal{L}^n(z) > 0. \end{align*} Since $\{V_k\}$ covers $\Omega$, every point of $\Omega$ lies in some $V_k$ where the corresponding $\psi_k$ is strictly positive. [/step] [step:Normalise the bump sum to produce a partition of unity for the refinement] Define \begin{align*} \Psi: \Omega &\to \mathbb{R}, \\ x &\mapsto \sum_{k=1}^\infty \psi_k(x). \end{align*} By the locally finite property of $\{V_k\}$, every compact $K \subset \Omega$ meets only finitely many $\operatorname{supp}(\psi_k)$, so $\Psi|_K$ is a finite sum of smooth functions and hence $\Psi \in C^\infty(\Omega)$. For any $x \in \Omega$, some $V_{k_0}$ contains $x$, and $\psi_{k_0}(x) > 0$ by construction, so $\Psi(x) > 0$ on all of $\Omega$. Define $\chi_k: \Omega \to \mathbb{R}$ by $\chi_k := \psi_k / \Psi$. Then $\chi_k \in C^\infty(\Omega)$ (ratio of smooth functions with strictly positive denominator), $\chi_k \ge 0$, $\operatorname{supp}(\chi_k) = \operatorname{supp}(\psi_k) \subseteq V_k$, and \begin{align*} \sum_{k=1}^\infty \chi_k(x) &= \Psi(x)^{-1} \sum_{k=1}^\infty \psi_k(x) = 1 \quad \text{for all } x \in \Omega. \end{align*} Thus $\{\chi_k\}$ is a smooth [partition of unity](/page/Partition%20of%20Unity) subordinate to $\{V_k\}$. [/step] [step:Group the refinement partition by cover index to obtain a partition subordinate to $\{U_i\}$] Each $V_k$ was constructed inside some $U_{i(k)}$. For each $i \in I$, define \begin{align*} \eta_i: \Omega &\to \mathbb{R}, \\ x &\mapsto \sum_{\{k : i(k) = i\}} \chi_k(x). \end{align*} This sum is locally finite (since $\{\operatorname{supp}(\chi_k)\}$ is locally finite), so $\eta_i \in C^\infty(\Omega)$. We verify the required properties: 1. $\eta_i \ge 0$ since each $\chi_k \ge 0$. 2. $\operatorname{supp}(\eta_i) \subseteq \overline{\bigcup_{\{k: i(k)=i\}} \operatorname{supp}(\chi_k)} \subseteq \overline{U_i}$. Note that the support need not be compact when $U_i$ is unbounded, but each $\operatorname{supp}(\chi_k)$ is compact and contained in $U_i$, so $\operatorname{supp}(\eta_i) \subseteq U_i$ (since $U_i$ is open and every $\chi_k$ with $i(k) = i$ vanishes outside $V_k \subseteq U_i$, the closure of the union within $\Omega$ remains in $\overline{U_i}$; but in fact $\eta_i$ vanishes outside $\bigcup_{\{k:i(k)=i\}} V_k \subseteq U_i$, so $\operatorname{supp}(\eta_i) \subseteq \overline{\bigcup_{\{k:i(k)=i\}} V_k}$). In particular, $\operatorname{supp}(\eta_i) \cap K$ is compact for every compact $K \subset \Omega$, since $K$ meets only finitely many $\operatorname{supp}(\chi_k)$. 3. Local finiteness of $\{\operatorname{supp}(\eta_i)\}_{i \in I}$ follows from local finiteness of $\{\operatorname{supp}(\chi_k)\}$: any compact $K \subset \Omega$ meets only finitely many $\operatorname{supp}(\chi_k)$, hence only finitely many indices $i$ satisfy $\operatorname{supp}(\eta_i) \cap K \ne \varnothing$. 4. The partition property: \begin{align*} \sum_{i \in I} \eta_i(x) &= \sum_{i \in I} \sum_{\{k: i(k)=i\}} \chi_k(x) = \sum_{k=1}^\infty \chi_k(x) = 1. \end{align*} [/step]