[proofplan] We show the two candidate values agree by inserting the identity $\sum_j \sigma_j = 1$ (valid on a neighbourhood of $\operatorname{supp}\omega$) into the first sum and the dual identity $\sum_i \rho_i = 1$ into the second. The cross terms $\rho_i \sigma_j \omega$ are compactly supported in $U_{\alpha(i)} \cap V_{\beta(j)}$, hence integrable in either chart, and the two local-integral values agree by the orientation-preserving change of variables formula. Compactness of $\operatorname{supp}\omega$ reduces every sum to a finite sum, so the rearrangement is purely algebraic and no convergence issues arise. [/proofplan] [step:Reduce all sums to finite sums via compactness of $\operatorname{supp}\omega$] Let $K := \operatorname{supp}\omega \subseteq M$; by hypothesis $K$ is compact. Since $\{\rho_i\}_{i \in I}$ is a [partition of unity](/page/Partition%20of%20Unity), the family of supports $\{\operatorname{supp}\rho_i\}_{i \in I}$ is locally finite, so the open cover $\{M \setminus \operatorname{supp}\rho_i : \rho_i|_K \equiv 0\} \cup \{U_{\alpha(i)} : \operatorname{supp}\rho_i \cap K \neq \varnothing\}$ admits a finite subcover of $K$. In particular the set \begin{align*} I_K := \{i \in I : \operatorname{supp}\rho_i \cap K \neq \varnothing\} \end{align*} is finite. Analogously $J_K := \{j \in J : \operatorname{supp}\sigma_j \cap K \neq \varnothing\}$ is finite. For $i \notin I_K$ the product $\rho_i \omega$ vanishes identically on $M$, hence $\int_{U_{\alpha(i)}} \rho_i \omega = 0$, and similarly for $j \notin J_K$. Therefore \begin{align*} \sum_{i \in I} \int_{U_{\alpha(i)}} \rho_i \, \omega = \sum_{i \in I_K} \int_{U_{\alpha(i)}} \rho_i \, \omega, \qquad \sum_{j \in J} \int_{V_{\beta(j)}} \sigma_j \, \omega = \sum_{j \in J_K} \int_{V_{\beta(j)}} \sigma_j \, \omega, \end{align*} and both sides are finite sums. [guided] Before any algebraic manipulation we must verify that the symbols on both sides of the desired equality make sense as ordinary finite sums; only then is the rearrangement we plan to do legitimate. The partitions $\{\rho_i\}_{i \in I}$ and $\{\sigma_j\}_{j \in J}$ are a priori indexed by arbitrary (possibly uncountable) sets, but local finiteness combined with the compactness of $K = \operatorname{supp}\omega$ collapses the relevant index sets to finite ones. Concretely, local finiteness of $\{\operatorname{supp}\rho_i\}_{i \in I}$ means that each $p \in M$ has an open neighbourhood $W_p$ meeting only finitely many of the sets $\operatorname{supp}\rho_i$. Covering the compact set $K$ by finitely many such neighbourhoods $W_{p_1}, \dots, W_{p_N}$, the index set \begin{align*} I_K := \{i \in I : \operatorname{supp}\rho_i \cap K \neq \varnothing\} \end{align*} is contained in the union $\bigcup_{k=1}^N \{i : \operatorname{supp}\rho_i \cap W_{p_k} \neq \varnothing\}$ of finitely many finite sets, hence is finite. The same argument applied to $\{\sigma_j\}_{j \in J}$ shows $J_K := \{j \in J : \operatorname{supp}\sigma_j \cap K \neq \varnothing\}$ is finite. Now for any index $i \notin I_K$ we have $\operatorname{supp}\rho_i \cap \operatorname{supp}\omega = \varnothing$, so $\rho_i \omega \equiv 0$ on $M$; hence the local integral $\int_{U_{\alpha(i)}} \rho_i \omega$ is zero. The sum $\sum_{i \in I} \int_{U_{\alpha(i)}} \rho_i \omega$ therefore reduces to the finite sum $\sum_{i \in I_K} \int_{U_{\alpha(i)}} \rho_i \omega$, and similarly on the $\sigma_j$ side. Every "sum" appearing below is therefore a finite sum, and the rearrangements we perform require no convergence argument. [/guided] [/step] [step:Establish the coordinate-independence of the local integral on a chart overlap] [claim:Local integrals on overlaps are chart-independent] Let $(U, \varphi)$ and $(V, \psi)$ be two positively oriented charts on $M$ and let $\eta \in \Omega^n_c(U \cap V)$. Then \begin{align*} \int_U \eta \;=\; \int_V \eta. \end{align*} [/claim] [proof] Let $W := U \cap V$ and let $\tau := \psi \circ \varphi^{-1}: \varphi(W) \to \psi(W)$, which is an orientation-preserving smooth diffeomorphism between open subsets of $\mathbb{R}^n$ (composition of orientation-preserving diffeomorphisms). Write \begin{align*} (\varphi^{-1})^*\eta &= f(x)\, dx_1 \wedge \cdots \wedge dx_n, & (\psi^{-1})^*\eta &= g(y)\, dy_1 \wedge \cdots \wedge dy_n, \end{align*} with $f \in C^\infty_c(\varphi(W))$ and $g \in C^\infty_c(\psi(W))$ (extended by zero to $\varphi(U)$ and $\psi(V)$ respectively). Pulling back the second display by $\tau$ and using the standard formula \begin{align*} \tau^*(dy_1 \wedge \cdots \wedge dy_n) = \det(J\tau_x) \, dx_1 \wedge \cdots \wedge dx_n, \end{align*} where $J\tau_x \in \mathbb{R}^{n \times n}$ is the Jacobian matrix of $\tau$ at $x$, we obtain \begin{align*} (\varphi^{-1})^*\eta = (\tau^* \circ (\psi^{-1})^*)\eta = (g \circ \tau)(x) \cdot \det(J\tau_x) \, dx_1 \wedge \cdots \wedge dx_n. \end{align*} Uniqueness of the coefficient of $dx_1 \wedge \cdots \wedge dx_n$ gives $f(x) = (g \circ \tau)(x) \cdot \det(J\tau_x)$ on $\varphi(W)$. Since $\tau$ is orientation-preserving, $\det(J\tau_x) > 0$ everywhere on $\varphi(W)$, so $|\det(J\tau_x)| = \det(J\tau_x)$. Applying the [Change of Variables (general)](/theorems/22) formula to the diffeomorphism $\tau: \varphi(W) \to \psi(W)$ with integrand $g$, \begin{align*} \int_{\psi(W)} g(y) \, d\mathcal{L}^n(y) = \int_{\varphi(W)} (g \circ \tau)(x) \, |\det(J\tau_x)| \, d\mathcal{L}^n(x) = \int_{\varphi(W)} f(x) \, d\mathcal{L}^n(x). \end{align*} Because $f$ vanishes outside $\varphi(W)$ and $g$ vanishes outside $\psi(W)$, these integrals coincide with $\int_{\varphi(U)} f \, d\mathcal{L}^n$ and $\int_{\psi(V)} g \, d\mathcal{L}^n$ respectively. By the definition of the local integral, $\int_U \eta = \int_V \eta$. [/proof] [guided] The claim says that for any compactly supported $n$-form whose support lies in two positively oriented charts simultaneously, the local integral is the same regardless of which chart we use to compute it. This is the heart of why integration of $n$-forms on oriented manifolds is well-defined; without it, the global construction would be ambiguous on each overlap and the partition-of-unity argument would have nothing to glue together. Let $W = U \cap V$ and consider the transition diffeomorphism \begin{align*} \tau := \psi \circ \varphi^{-1} : \varphi(W) \to \psi(W). \end{align*} Because both $\varphi$ and $\psi$ are orientation-preserving (this is the role of the hypothesis "positively oriented atlas"), so is $\tau$, meaning $\det(J\tau_x) > 0$ for every $x \in \varphi(W)$. This positivity is what allows us to drop the absolute value bars in the change-of-variables formula — a step that is silently used in textbook accounts but is the precise place where orientability enters. Now expand $\eta$ in both coordinate systems. Writing \begin{align*} (\varphi^{-1})^*\eta = f(x)\, dx_1 \wedge \cdots \wedge dx_n, \qquad (\psi^{-1})^*\eta = g(y)\, dy_1 \wedge \cdots \wedge dy_n, \end{align*} we use the functoriality of pullback, $(\varphi^{-1})^* = \tau^* \circ (\psi^{-1})^*$ (verified by $\varphi^{-1} = \psi^{-1} \circ \tau$), and the standard pullback identity \begin{align*} \tau^*(dy_1 \wedge \cdots \wedge dy_n) = \det(J\tau_x) \, dx_1 \wedge \cdots \wedge dx_n \end{align*} to obtain the pointwise relation $f(x) = (g \circ \tau)(x) \cdot \det(J\tau_x)$. The [Change of Variables (general)](/theorems/22) formula requires (i) a $C^1$-diffeomorphism between open sets of $\mathbb{R}^n$ — satisfied by the smooth $\tau$ — and (ii) Lebesgue-integrability of the transformed integrand — satisfied because $g \in C^\infty_c$ and $|\det(J\tau_x)|$ is continuous. Both hypotheses hold, so \begin{align*} \int_{\psi(W)} g \, d\mathcal{L}^n = \int_{\varphi(W)} (g \circ \tau) \, |\det(J\tau_\cdot)| \, d\mathcal{L}^n, \end{align*} and positivity of $\det(J\tau_\cdot)$ turns the right-hand side into $\int_{\varphi(W)} f \, d\mathcal{L}^n$. Both sides are unchanged if we enlarge the domain to $\psi(V)$ and $\varphi(U)$ respectively, since $f$ and $g$ vanish outside $\varphi(W)$ and $\psi(W)$. By definition of the local integral this yields $\int_U \eta = \int_V \eta$. [/guided] [/step] [step:Expand each summand using the complementary partition] For each $i \in I_K$, the function $\sum_{j \in J_K} \sigma_j$ equals $\sum_{j \in J} \sigma_j$ on a neighbourhood of $\operatorname{supp}\rho_i \cap K$ (by the finite support argument of the first step) and therefore equals $1$ there. Multiplying by $\rho_i \omega$, which is supported in $\operatorname{supp}\rho_i \cap K$, \begin{align*} \rho_i \omega = \rho_i \omega \cdot \sum_{j \in J_K} \sigma_j = \sum_{j \in J_K} \rho_i \sigma_j \, \omega \qquad \text{on } M. \end{align*} By symmetry, for each $j \in J_K$, \begin{align*} \sigma_j \omega = \sum_{i \in I_K} \rho_i \sigma_j \, \omega \qquad \text{on } M. \end{align*} Each cross term $\rho_i \sigma_j \omega$ lies in $\Omega^n_c(M)$ with $\operatorname{supp}(\rho_i \sigma_j \omega) \subseteq \operatorname{supp}\rho_i \cap \operatorname{supp}\sigma_j \subseteq U_{\alpha(i)} \cap V_{\beta(j)}$. [guided] We need to write each $\rho_i \omega$ as a sum of pieces that are simultaneously cut down by some $\sigma_j$, because the chart-independence claim from the previous step applies precisely to forms supported in a chart overlap. The trick is to insert the constant function $1$ in the form $1 = \sum_{j \in J} \sigma_j$. The subtlety is that $\sum_{j \in J} \sigma_j(p) = 1$ holds for every $p \in M$ — this is the defining property of a [partition of unity](/page/Partition%20of%20Unity) — but the relevant index set on which this constant function is nonzero only $J_K$ is genuinely needed when restricted to $\operatorname{supp}\omega$. For $p \in \operatorname{supp}\rho_i \cap K$ we have \begin{align*} \sum_{j \in J_K} \sigma_j(p) = \sum_{j \in J} \sigma_j(p) = 1, \end{align*} the first equality because $\sigma_j(p) = 0$ for $j \notin J_K$ (since $p \in K$ and $\operatorname{supp}\sigma_j \cap K = \varnothing$). Multiplying both sides by $\rho_i(p)\omega(p)$ and noting that $\rho_i\omega$ vanishes outside $\operatorname{supp}\rho_i \cap K$, \begin{align*} \rho_i \omega = \rho_i \omega \cdot \sum_{j \in J_K} \sigma_j = \sum_{j \in J_K} \rho_i \sigma_j \omega \qquad \text{on all of } M. \end{align*} The role of the cut-down to $J_K$ is purely cosmetic — it ensures the sum is literally finite at every point, with no convergence question — but mathematically we could have written $\sum_{j \in J}$ throughout. The dual decomposition $\sigma_j \omega = \sum_{i \in I_K} \rho_i \sigma_j \omega$ follows by exchanging the roles of the two partitions. Finally, each cross term $\rho_i \sigma_j \omega$ has support inside $\operatorname{supp}\rho_i \cap \operatorname{supp}\sigma_j \subseteq U_{\alpha(i)} \cap V_{\beta(j)}$, which is exactly the chart-overlap setting required to apply the claim from the previous step. [/guided] [/step] [step:Sum the cross terms and apply chart-independence on each overlap] Using the decomposition of Step 3 in the first sum, \begin{align*} \sum_{i \in I_K} \int_{U_{\alpha(i)}} \rho_i \, \omega = \sum_{i \in I_K} \int_{U_{\alpha(i)}} \sum_{j \in J_K} \rho_i \sigma_j \, \omega. \end{align*} Because each chart-local integral $\int_{U_{\alpha(i)}}$ is, by its definition, the [Lebesgue integral](/page/Lebesgue%20Integral) of a single smooth coefficient function pulled back through $\varphi_{\alpha(i)}$, and finite sums commute with pulling back and with Lebesgue integration, we may bring the finite sum over $J_K$ outside the chart-local integral: \begin{align*} \sum_{i \in I_K} \int_{U_{\alpha(i)}} \sum_{j \in J_K} \rho_i \sigma_j \, \omega = \sum_{i \in I_K} \sum_{j \in J_K} \int_{U_{\alpha(i)}} \rho_i \sigma_j \, \omega = \sum_{(i,j) \in I_K \times J_K} \int_{U_{\alpha(i)}} \rho_i \sigma_j \, \omega, \end{align*} where the second equality is the rearrangement of a finite double sum. By Step 3, each integrand $\rho_i \sigma_j \omega$ is supported in $U_{\alpha(i)} \cap V_{\beta(j)}$, so the claim from Step 2 applies to the two positively oriented charts $(U_{\alpha(i)}, \varphi_{\alpha(i)})$ and $(V_{\beta(j)}, \psi_{\beta(j)})$ and gives \begin{align*} \int_{U_{\alpha(i)}} \rho_i \sigma_j \, \omega = \int_{V_{\beta(j)}} \rho_i \sigma_j \, \omega \qquad \text{for every } (i,j) \in I_K \times J_K. \end{align*} Substituting and reversing the algebraic manipulation on the $\sigma_j$ side, \begin{align*} \sum_{(i,j) \in I_K \times J_K} \int_{V_{\beta(j)}} \rho_i \sigma_j \, \omega = \sum_{j \in J_K} \int_{V_{\beta(j)}} \sum_{i \in I_K} \rho_i \sigma_j \, \omega = \sum_{j \in J_K} \int_{V_{\beta(j)}} \sigma_j \, \omega. \end{align*} Chaining the equalities, \begin{align*} \sum_{i \in I} \int_{U_{\alpha(i)}} \rho_i \, \omega = \sum_{i \in I_K} \int_{U_{\alpha(i)}} \rho_i \, \omega = \sum_{j \in J_K} \int_{V_{\beta(j)}} \sigma_j \, \omega = \sum_{j \in J} \int_{V_{\beta(j)}} \sigma_j \, \omega, \end{align*} where the outer equalities use the finiteness reduction of Step 1. [guided] We now combine the three preceding ingredients — finite-sum reduction, chart-overlap independence, and the cross-term decomposition — into the equality we want. The computation is a routine but careful interchange of finite sums. Start with the left-hand side, which by Step 1 equals $\sum_{i \in I_K} \int_{U_{\alpha(i)}} \rho_i \omega$. Using the decomposition $\rho_i \omega = \sum_{j \in J_K} \rho_i \sigma_j \omega$ from Step 3, the chart-local integral acquires a sum over $j$ inside it: \begin{align*} \sum_{i \in I_K} \int_{U_{\alpha(i)}} \rho_i \omega = \sum_{i \in I_K} \int_{U_{\alpha(i)}} \sum_{j \in J_K} \rho_i \sigma_j \omega. \end{align*} Why can we pull the $\sum_{j \in J_K}$ outside the chart-local integral? By definition, $\int_{U_{\alpha(i)}} \eta$ is the [Lebesgue integral](/page/Lebesgue%20Integral) over $\varphi_{\alpha(i)}(U_{\alpha(i)})$ of the coefficient of $(\varphi_{\alpha(i)}^{-1})^*\eta$ in the standard $n$-form. Pulling back is $\mathbb{R}$-linear, picking off the coefficient is $\mathbb{R}$-linear, and the [Lebesgue integral](/page/Lebesgue%20Integral) over a fixed set is $\mathbb{R}$-linear on integrable functions. The sum $\sum_{j \in J_K}$ is finite, so linearity applies term by term: \begin{align*} \sum_{i \in I_K} \int_{U_{\alpha(i)}} \sum_{j \in J_K} \rho_i \sigma_j \omega = \sum_{i \in I_K} \sum_{j \in J_K} \int_{U_{\alpha(i)}} \rho_i \sigma_j \omega = \sum_{(i,j) \in I_K \times J_K} \int_{U_{\alpha(i)}} \rho_i \sigma_j \omega. \end{align*} The final rearrangement is the standard Fubini for finite double sums. The crucial substitution is now the chart-overlap claim from Step 2. For each pair $(i,j) \in I_K \times J_K$, the cross term $\rho_i \sigma_j \omega$ is supported in $U_{\alpha(i)} \cap V_{\beta(j)}$ (Step 3), so the claim applies — and applies precisely because the atlas is positively oriented, so both transition maps preserve orientation — yielding \begin{align*} \int_{U_{\alpha(i)}} \rho_i \sigma_j \omega = \int_{V_{\beta(j)}} \rho_i \sigma_j \omega. \end{align*} Substituting, \begin{align*} \sum_{(i,j) \in I_K \times J_K} \int_{U_{\alpha(i)}} \rho_i \sigma_j \omega = \sum_{(i,j) \in I_K \times J_K} \int_{V_{\beta(j)}} \rho_i \sigma_j \omega. \end{align*} Now reverse the algebraic manipulation, this time using the dual decomposition $\sigma_j \omega = \sum_{i \in I_K} \rho_i \sigma_j \omega$: \begin{align*} \sum_{(i,j) \in I_K \times J_K} \int_{V_{\beta(j)}} \rho_i \sigma_j \omega = \sum_{j \in J_K} \int_{V_{\beta(j)}} \sum_{i \in I_K} \rho_i \sigma_j \omega = \sum_{j \in J_K} \int_{V_{\beta(j)}} \sigma_j \omega. \end{align*} Putting the chain together and using Step 1 to identify $\sum_{i \in I_K}$ with $\sum_{i \in I}$ and $\sum_{j \in J_K}$ with $\sum_{j \in J}$, we obtain \begin{align*} \sum_{i \in I} \int_{U_{\alpha(i)}} \rho_i \omega = \sum_{j \in J} \int_{V_{\beta(j)}} \sigma_j \omega, \end{align*} which is the required equality. The argument is entirely symmetric in $(\mathcal{U}, \rho)$ and $(\mathcal{V}, \sigma)$ and uses orientability only through the chart-overlap claim of Step 2, where the positivity of $\det(J\tau_\cdot)$ removes the absolute value in the change-of-variables formula. [/guided] [/step] [step:Conclude well-definedness] The two choices $(\mathcal{U}, \{\rho_i\})$ and $(\mathcal{V}, \{\sigma_j\})$ were arbitrary positively oriented atlases on $M$ together with subordinate smooth partitions of unity (whose existence is guaranteed by the [Existence of Smooth Partitions of Unity](/theorems/57) applied to the open cover $\mathcal{U}$ — second-countability of the underlying smooth structure is the standing convention on a smooth manifold, ensuring the hypothesis of paracompactness). Therefore the value \begin{align*} \int_M \omega := \sum_{i \in I} \int_{U_{\alpha(i)}} \rho_i \, \omega \end{align*} depends only on $\omega$ and the orientation of $M$, completing the proof. $\square$ [/step]