[proofplan] The proof has two points. First, each product $\rho_i s_i$ extends smoothly by zero because outside $\operatorname{supp}\rho_i$ it is identically zero, while inside $U_i$ it is the product of a smooth function and a smooth local section. Second, the local finiteness of the supports ensures that near every point only finitely many summands occur, so the global sum is locally a finite sum of smooth sections. [/proofplan] [step:Show that each zero extension is a smooth global section] Let $\pi:E\to M$ denote the smooth vector bundle projection. Fix $i\in I$. Define $\widetilde{s}_i:M\to E$ to be the zero extension of $\rho_i s_i$ from the statement. For every $p\in M$, if $p\in U_i$ then $\widetilde{s}_i(p)=\rho_i(p)s_i(p)\in E_p$, while if $p\notin U_i$ then $\widetilde{s}_i(p)=0_p\in E_p$. Hence $\pi(\widetilde{s}_i(p))=p$ for every $p\in M$, so $\widetilde{s}_i$ is a global section as a set-theoretic map. We prove smoothness locally. If $p\in U_i$, then on the open neighbourhood $U_i$ the section $\widetilde{s}_i$ agrees with the product $\rho_i s_i$, where $\rho_i|_{U_i}:U_i\to\mathbb{R}$ is smooth and $s_i:U_i\to E|_{U_i}$ is smooth. Smooth scalar multiplication in the vector bundle therefore implies that $\widetilde{s}_i$ is smooth near $p$. If $p\notin \operatorname{supp}\rho_i$, then $M\setminus \operatorname{supp}\rho_i$ is an open neighbourhood of $p$, and $\rho_i$ vanishes on this neighbourhood. Since $\operatorname{supp}\rho_i\subset U_i$, the definition of $\widetilde{s}_i$ gives $\widetilde{s}_i(q)=0_q$ for every $q\in M\setminus \operatorname{supp}\rho_i$. Thus $\widetilde{s}_i$ agrees there with the smooth zero section of $E$. These two cases cover $M$, because $\operatorname{supp}\rho_i\subset U_i$. Therefore $\widetilde{s}_i\in\Gamma(E)$. [guided] Let $\pi:E\to M$ denote the smooth vector bundle projection. Fix an index $i\in I$. The possible difficulty is that $s_i$ is only defined on $U_i$, so we must check that extending $\rho_i s_i$ by zero does not create a loss of smoothness along the boundary of $U_i$. First, $\widetilde{s}_i$ is a section. For each $p\in M$, the value $\widetilde{s}_i(p)$ is either $\rho_i(p)s_i(p)\in E_p$ when $p\in U_i$, or the zero vector $0_p\in E_p$ when $p\notin U_i$. Thus $\pi(\widetilde{s}_i(p))=p$ for every $p\in M$. Now we prove smoothness pointwise. If $p\in U_i$, then near $p$ the section is simply the product of two smooth objects: the smooth function $\rho_i|_{U_i}:U_i\to\mathbb{R}$ and the smooth local section $s_i:U_i\to E|_{U_i}$. Since scalar multiplication in a smooth vector bundle is a smooth map on fibres, the product $\rho_i s_i$ is a smooth section on $U_i$. The remaining points are those outside $U_i$ or near its boundary. Here the support condition is exactly what prevents a discontinuity. Because $\operatorname{supp}\rho_i\subset U_i$, every point $p\notin U_i$ also satisfies $p\notin \operatorname{supp}\rho_i$. More generally, if $p\notin \operatorname{supp}\rho_i$, then the [open set](/page/Open%20Set) $M\setminus \operatorname{supp}\rho_i$ is a neighbourhood of $p$, and $\rho_i$ vanishes on that neighbourhood. On this neighbourhood, the extended section satisfies $\widetilde{s}_i(q)=0_q$ for every $q\in M\setminus \operatorname{supp}\rho_i$. Thus $\widetilde{s}_i$ agrees locally with the smooth zero section. Every point of $M$ is either in $U_i$ or outside $\operatorname{supp}\rho_i$, because $\operatorname{supp}\rho_i\subset U_i$. Hence $\widetilde{s}_i$ is smooth on a neighbourhood of every point of $M$, so $\widetilde{s}_i\in\Gamma(E)$. [/guided] [/step] [step:Use local finiteness to make the infinite sum locally finite] Let $p\in M$. Since the family $(\operatorname{supp}\rho_i)_{i\in I}$ is locally finite, there exists an open neighbourhood $V\subset M$ of $p$ and a finite subset $J\subset I$ such that \begin{align*} V\cap \operatorname{supp}\rho_i=\varnothing \end{align*} for every $i\in I\setminus J$. For $q\in V$ and $i\in I\setminus J$, we have $q\notin \operatorname{supp}\rho_i$, hence $\widetilde{s}_i(q)=0_q$. Therefore, on $V$, \begin{align*} s(q)=\sum_{i\in I}\widetilde{s}_i(q)=\sum_{i\in J}\widetilde{s}_i(q). \end{align*} The right-hand side is a finite sum of smooth sections on $V$, so it is a smooth section on $V$. [/step] [step:Conclude that the locally finite sum is a smooth global section] For every $p\in M$, the preceding step gives an open neighbourhood $V$ of $p$ on which $s$ is a finite sum of smooth sections. Hence $s$ is smooth in a neighbourhood of every point of $M$. Also, for every $q\in M$, each nonzero term $\widetilde{s}_i(q)$ lies in the fibre $E_q$, so their finite fibrewise sum also lies in $E_q$ and satisfies \begin{align*} \pi(s(q))=q. \end{align*} Thus $s:M\to E$ is a smooth global section, i.e. $s\in\Gamma(E)$. [/step]