[proofplan] We first interpret the right-hand side as a classical Lebesgue integral: the pullback $(\varphi_\alpha^{-1})^*(\rho_\alpha \omega)$ on $\varphi_\alpha(U_\alpha) \subset \mathbb{R}^n$ is $f_\alpha \, dy_1 \wedge \cdots \wedge dy_n$ for some $f_\alpha \in C_c(\varphi_\alpha(U_\alpha))$, and the sum is finite because $\operatorname{supp}(\omega)$ is compact and the partition is locally finite. Linearity in $\omega$ is immediate. For independence, we compare the definitions using atlas $\{(U_\alpha, \varphi_\alpha)\}$ with partition $\{\rho_\alpha\}$ and a second pair $\{(V_j, \psi_j)\}$ with $\{\eta_j\}$; the comparison reduces to proving the identity $\int_{\varphi_\alpha(U_\alpha \cap V_j)} (\varphi_\alpha^{-1})^*(\rho_\alpha \eta_j \omega) = \int_{\psi_j(U_\alpha \cap V_j)} (\psi_j^{-1})^*(\rho_\alpha \eta_j \omega)$, which follows from the classical Euclidean change-of-variables formula applied to the orientation-preserving transition $\psi_j \circ \varphi_\alpha^{-1}$, whose Jacobian has positive determinant and thus no absolute value correction. A double-sum over $\alpha, j$ with $\sum_\alpha \rho_\alpha = 1 = \sum_j \eta_j$ concludes. [/proofplan] [step:Interpret each local integral as a classical Lebesgue integral] Fix an oriented atlas $\{(U_\alpha, \varphi_\alpha)\}$ and a partition of unity $\{\rho_\alpha\}$ subordinate to it, with $\operatorname{supp}(\rho_\alpha) \subset U_\alpha$ and $\sum_\alpha \rho_\alpha \equiv 1$. For $\omega \in \Omega^n_c(M)$, the product $\rho_\alpha \omega$ is an $n$-form on $M$ with $\operatorname{supp}(\rho_\alpha \omega) \subset U_\alpha$, which extends by zero to all of $M$. Its pullback under $\varphi_\alpha^{-1}$ is an $n$-form on $\varphi_\alpha(U_\alpha) \subset \mathbb{R}^n$; since $\Omega^n(\varphi_\alpha(U_\alpha))$ is spanned by $dy_1 \wedge \cdots \wedge dy_n$, we write \begin{align*} (\varphi_\alpha^{-1})^*(\rho_\alpha \omega) = f_\alpha \cdot dy_1 \wedge \cdots \wedge dy_n, \end{align*} where $f_\alpha : \varphi_\alpha(U_\alpha) \to \mathbb{R}$ is smooth and compactly supported (compact support passes through $\varphi_\alpha$, a homeomorphism). We define the local integral as the Lebesgue integral \begin{align*} \int_{\varphi_\alpha(U_\alpha)} (\varphi_\alpha^{-1})^*(\rho_\alpha \omega) := \int_{\varphi_\alpha(U_\alpha)} f_\alpha(y) \, d\mathcal{L}^n(y). \end{align*} The integral is finite because $f_\alpha$ is continuous and compactly supported. [/step] [step:Verify the global sum is finite and linear in $\omega$] Let $K := \operatorname{supp}(\omega) \subset M$. By compactness of $K$ and local finiteness of $\{\operatorname{supp}(\rho_\alpha)\}$, the set $\{\alpha : \operatorname{supp}(\rho_\alpha) \cap K \ne \varnothing\}$ is finite; call it $F$. For $\alpha \notin F$, $\rho_\alpha \omega \equiv 0$ and the corresponding term in $\sum_\alpha \int_{\varphi_\alpha(U_\alpha)} (\varphi_\alpha^{-1})^*(\rho_\alpha \omega)$ is zero. Hence \begin{align*} \int_M \omega = \sum_{\alpha \in F} \int_{\varphi_\alpha(U_\alpha)} (\varphi_\alpha^{-1})^*(\rho_\alpha \omega), \end{align*} a finite sum. Linearity in $\omega$ follows from $\mathbb{R}$-linearity of pullback, linearity of the Lebesgue integral, and finiteness of the sum: if $\omega, \omega' \in \Omega^n_c(M)$ and $c \in \mathbb{R}$, then $\rho_\alpha (\omega + c \omega') = \rho_\alpha \omega + c \rho_\alpha \omega'$, pullback and $\int$ commute with this decomposition, so $\int_M (\omega + c \omega') = \int_M \omega + c \int_M \omega'$. [/step] [step:Set up the comparison between two atlas-plus-partition choices] Let $\{(U_\alpha, \varphi_\alpha)\}$ with partition $\{\rho_\alpha\}$ and $\{(V_j, \psi_j)\}$ with partition $\{\eta_j\}$ be two oriented-atlas-plus-partition-of-unity choices on $M$. The two candidate definitions are \begin{align*} I_1 := \sum_\alpha \int_{\varphi_\alpha(U_\alpha)} (\varphi_\alpha^{-1})^*(\rho_\alpha \omega), \qquad I_2 := \sum_j \int_{\psi_j(V_j)} (\psi_j^{-1})^*(\eta_j \omega). \end{align*} Using $\sum_j \eta_j \equiv 1$ inside $I_1$ and $\sum_\alpha \rho_\alpha \equiv 1$ inside $I_2$, both finite sums because $\operatorname{supp}(\omega)$ is compact: \begin{align*} I_1 &= \sum_\alpha \int_{\varphi_\alpha(U_\alpha)} (\varphi_\alpha^{-1})^*\Big(\rho_\alpha \cdot \sum_j \eta_j \cdot \omega\Big) = \sum_{\alpha, j} \int_{\varphi_\alpha(U_\alpha)} (\varphi_\alpha^{-1})^*(\rho_\alpha \eta_j \omega), \\ I_2 &= \sum_j \int_{\psi_j(V_j)} (\psi_j^{-1})^*\Big(\sum_\alpha \rho_\alpha \cdot \eta_j \cdot \omega\Big) = \sum_{\alpha, j} \int_{\psi_j(V_j)} (\psi_j^{-1})^*(\rho_\alpha \eta_j \omega). \end{align*} Pulling the scalar multiplication out of the pullback and exchanging the finite double sum with the integral justifies both manipulations. [/step] [step:Reduce term-by-term equality to a Euclidean change-of-variables identity] Fix $\alpha, j$ and note $\operatorname{supp}(\rho_\alpha \eta_j \omega) \subset U_\alpha \cap V_j$. Since the integrand is zero outside this overlap, we may restrict both integrals to the image of $U_\alpha \cap V_j$: \begin{align*} \int_{\varphi_\alpha(U_\alpha)} (\varphi_\alpha^{-1})^*(\rho_\alpha \eta_j \omega) &= \int_{\varphi_\alpha(U_\alpha \cap V_j)} (\varphi_\alpha^{-1})^*(\rho_\alpha \eta_j \omega), \\ \int_{\psi_j(V_j)} (\psi_j^{-1})^*(\rho_\alpha \eta_j \omega) &= \int_{\psi_j(U_\alpha \cap V_j)} (\psi_j^{-1})^*(\rho_\alpha \eta_j \omega). \end{align*} Term-by-term equality $I_1 = I_2$ reduces to showing \begin{align*} \int_{\varphi_\alpha(U_\alpha \cap V_j)} (\varphi_\alpha^{-1})^*(\rho_\alpha \eta_j \omega) = \int_{\psi_j(U_\alpha \cap V_j)} (\psi_j^{-1})^*(\rho_\alpha \eta_j \omega). \end{align*} [/step] [step:Apply Euclidean change of variables via the orientation-preserving transition map] Set $T := \psi_j \circ \varphi_\alpha^{-1} : \varphi_\alpha(U_\alpha \cap V_j) \to \psi_j(U_\alpha \cap V_j)$, a smooth diffeomorphism. By orientation-compatibility of the atlases, $\det(JT) > 0$ everywhere on $\varphi_\alpha(U_\alpha \cap V_j)$. Write \begin{align*} (\psi_j^{-1})^*(\rho_\alpha \eta_j \omega) = g \, dy_1 \wedge \cdots \wedge dy_n \quad \text{on } \psi_j(U_\alpha \cap V_j), \end{align*} for some $g \in C_c(\psi_j(U_\alpha \cap V_j))$. Naturality of pullback gives \begin{align*} (\varphi_\alpha^{-1})^*(\rho_\alpha \eta_j \omega) = (T \circ \varphi_\alpha \circ \varphi_\alpha^{-1})^{-1 \ *} (\cdot) = T^* \big( (\psi_j^{-1})^*(\rho_\alpha \eta_j \omega)\big) = T^*(g \, dy_1 \wedge \cdots \wedge dy_n). \end{align*} Computing the pullback of the top form under $T$ in coordinates $x$ on the source and $y$ on the target: \begin{align*} T^*(g \, dy_1 \wedge \cdots \wedge dy_n) = (g \circ T) \cdot \det(JT) \, dx_1 \wedge \cdots \wedge dx_n. \end{align*} Therefore, \begin{align*} \int_{\varphi_\alpha(U_\alpha \cap V_j)} (\varphi_\alpha^{-1})^*(\rho_\alpha \eta_j \omega) = \int_{\varphi_\alpha(U_\alpha \cap V_j)} (g \circ T)(x) \cdot \det(JT)(x) \, d\mathcal{L}^n(x). \end{align*} We apply the [Euclidean change of variables formula](/theorems/???) for the diffeomorphism $T$: this requires $T$ to be a smooth diffeomorphism between open sets of $\mathbb{R}^n$, which holds; and that the integrand $g$ be integrable on the image, which holds because $g$ is continuous and compactly supported. The classical formula gives \begin{align*} \int_{\varphi_\alpha(U_\alpha \cap V_j)} (g \circ T)(x) \cdot |\det(JT)(x)| \, d\mathcal{L}^n(x) = \int_{\psi_j(U_\alpha \cap V_j)} g(y) \, d\mathcal{L}^n(y). \end{align*} Because $\det(JT) > 0$, we have $|\det(JT)| = \det(JT)$, so the two expressions agree without any sign correction: \begin{align*} \int_{\varphi_\alpha(U_\alpha \cap V_j)} (\varphi_\alpha^{-1})^*(\rho_\alpha \eta_j \omega) = \int_{\psi_j(U_\alpha \cap V_j)} g(y) \, d\mathcal{L}^n(y) = \int_{\psi_j(U_\alpha \cap V_j)} (\psi_j^{-1})^*(\rho_\alpha \eta_j \omega). \end{align*} [guided] The orientation hypothesis on the atlas is consumed at exactly one point: the classical Euclidean change-of-variables formula involves the absolute value $|\det(JT)|$, whereas the pullback formula for top forms produces a signed $\det(JT)$. Without orientation-compatibility, these would differ by a sign on regions where $\det(JT) < 0$, and the integration rule would depend on the choice of atlas. The assumption $\det(JT) > 0$ makes $|\det(JT)| = \det(JT)$ identically, so the signed pullback matches the absolute-value change of variables, and the two local integrals coincide. This is the structural reason why integration of differential forms is defined for oriented manifolds — the sign of the determinant carries the geometric information that orientation has to reconcile with the classical unsigned change-of-variables formula. [/guided] [/step] [step:Conclude independence and linearity] Summing the term-by-term equality established in Step 5 over the finite set of pairs $(\alpha, j)$ contributing to the support of $\omega$: \begin{align*} I_1 = \sum_{\alpha, j} \int_{\varphi_\alpha(U_\alpha \cap V_j)} (\varphi_\alpha^{-1})^*(\rho_\alpha \eta_j \omega) = \sum_{\alpha, j} \int_{\psi_j(U_\alpha \cap V_j)} (\psi_j^{-1})^*(\rho_\alpha \eta_j \omega) = I_2. \end{align*} Thus $\int_M \omega$ is independent of the atlas and partition of unity used to define it. Combined with the linearity established in Step 2, this completes the proof that $\int_M : \Omega^n_c(M) \to \mathbb{R}$ is a well-defined $\mathbb{R}$-linear map. [/step]