[proofplan] We reduce the equality of the two local integrals to the Euclidean [Change of Variables (general)](/theorems/22) applied to the transition diffeomorphism $\tau = \psi \circ \varphi^{-1}$. The key ingredient is the transformation law for top-degree forms under a smooth map: the pullback by $\tau$ acts on the volume form $dy_1 \wedge \cdots \wedge dy_n$ by multiplication by the Jacobian determinant $\det J\tau$. Combining the identity $\varphi^{-1} = \psi^{-1} \circ \tau$ with this pullback formula expresses $f$ in terms of $g$ via $f = (g \circ \tau)\,\det J\tau$. The positivity hypothesis on the two charts forces $\det J\tau > 0$, so $|\det J\tau| = \det J\tau$, and the Euclidean change-of-variables theorem closes the argument. [/proofplan] [step:Set up the transition diffeomorphism and the relation between the two pullbacks] Let $W := U \cap V$, an open subset of $M$. Since $\varphi : U \to \varphi(U)$ and $\psi : V \to \psi(V)$ are diffeomorphisms onto open subsets of $\mathbb{R}^n$, both $\varphi(W)$ and $\psi(W)$ are open in $\mathbb{R}^n$. Define the transition map \begin{align*} \tau : \varphi(W) &\to \psi(W) \\ x &\mapsto (\psi \circ \varphi^{-1})(x). \end{align*} Then $\tau$ is a smooth diffeomorphism with smooth inverse $\tau^{-1} = \varphi \circ \psi^{-1}$, and we have the identity of maps $\varphi^{-1} = \psi^{-1} \circ \tau$ on $\varphi(W)$. Taking pullbacks and using functoriality of the pullback, \begin{align*} (\varphi^{-1})^* \omega \;=\; (\psi^{-1} \circ \tau)^* \omega \;=\; \tau^*\bigl((\psi^{-1})^* \omega\bigr) \quad \text{on } \varphi(W). \end{align*} [guided] We will reduce the equality of local integrals to the Euclidean change-of-variables theorem. The Euclidean theorem requires a diffeomorphism between two open subsets of $\mathbb{R}^n$, so our first task is to manufacture such a diffeomorphism out of the two charts. The natural candidate is the **transition map** \begin{align*} \tau : \varphi(W) &\to \psi(W) \\ x &\mapsto (\psi \circ \varphi^{-1})(x), \end{align*} defined on $W := U \cap V$. Because $\varphi$ and $\psi$ are restrictions of smooth charts to the open subset $W$, each restriction is itself a diffeomorphism onto the open subset $\varphi(W) \subseteq \mathbb{R}^n$ (resp. $\psi(W) \subseteq \mathbb{R}^n$). Their composition $\tau$ is therefore a smooth diffeomorphism with smooth inverse $\tau^{-1} = \varphi \circ \psi^{-1}$. To connect $\tau$ to the two pullbacks $(\varphi^{-1})^*\omega$ and $(\psi^{-1})^*\omega$ that appear in the statement, we rewrite $\varphi^{-1}$ as the composition $\psi^{-1} \circ \tau$ — this is just the identity $\varphi^{-1} = \psi^{-1} \circ (\psi \circ \varphi^{-1})$. By contravariant functoriality of the pullback, $(F \circ G)^* = G^* \circ F^*$, so \begin{align*} (\varphi^{-1})^* \omega \;=\; (\psi^{-1} \circ \tau)^* \omega \;=\; \tau^*\bigl((\psi^{-1})^* \omega\bigr) \quad \text{on } \varphi(W). \end{align*} This is the key relation: the $\varphi$-coordinate representative of $\omega$ is the $\tau$-pullback of the $\psi$-coordinate representative. [/guided] [/step] [step:Apply the transformation law for top-degree forms to relate $f$ and $g$] We compute the pullback $\tau^*\bigl((\psi^{-1})^* \omega\bigr) = \tau^*(g \, dy_1 \wedge \cdots \wedge dy_n)$. The pullback distributes over wedge products: $\tau^*(g \, dy_1 \wedge \cdots \wedge dy_n) = (g \circ \tau) \cdot \tau^*(dy_1) \wedge \cdots \wedge \tau^*(dy_n)$. Writing $\tau = (\tau_1, \dots, \tau_n)$ in components, $\tau^*(dy_i) = d\tau_i = \sum_{j=1}^n \partial_{x_j} \tau_i \, dx_j$, so \begin{align*} \tau^*(dy_1) \wedge \cdots \wedge \tau^*(dy_n) \;=\; \det(J\tau) \, dx_1 \wedge \cdots \wedge dx_n, \end{align*} where $J\tau \in \mathbb{R}^{n \times n}$ is the Jacobian matrix with entries $(J\tau)_{ij} = \partial_{x_j} \tau_i$. Combining with the identity from the previous step, \begin{align*} f \, dx_1 \wedge \cdots \wedge dx_n \;=\; (\varphi^{-1})^*\omega \;=\; \tau^*\bigl((\psi^{-1})^*\omega\bigr) \;=\; (g \circ \tau)\,\det(J\tau)\, dx_1 \wedge \cdots \wedge dx_n, \end{align*} on $\varphi(W)$. Comparing coefficients of the basis $n$-form $dx_1 \wedge \cdots \wedge dx_n$, we conclude that \begin{align*} f(x) \;=\; g(\tau(x))\,\det(J\tau)(x) \qquad \text{for all } x \in \varphi(W). \end{align*} [guided] We now compute the right-hand side $\tau^*\bigl((\psi^{-1})^*\omega\bigr) = \tau^*(g \, dy_1 \wedge \cdots \wedge dy_n)$ in terms of the coordinates $x = (x_1,\dots,x_n)$ on $\varphi(W)$. The pullback is an algebra homomorphism for the wedge product, and it acts on functions by composition: $\tau^*(g) = g \circ \tau$. Hence \begin{align*} \tau^*(g \, dy_1 \wedge \cdots \wedge dy_n) \;=\; (g \circ \tau) \cdot \tau^*(dy_1) \wedge \cdots \wedge \tau^*(dy_n). \end{align*} For each coordinate function $y_i$, we have $\tau^*(dy_i) = d(\tau^* y_i) = d(y_i \circ \tau) = d\tau_i$, where $\tau_i := y_i \circ \tau$ is the $i$-th component of $\tau$. Expanding the differential in the $x$-basis: \begin{align*} \tau^*(dy_i) \;=\; d\tau_i \;=\; \sum_{j=1}^n \frac{\partial \tau_i}{\partial x_j}\, dx_j. \end{align*} Now we wedge these together. The wedge product of $n$ such 1-forms in $\mathbb{R}^n$ collapses, by multilinearity and antisymmetry, into a single multiple of $dx_1 \wedge \cdots \wedge dx_n$, and the multiplier is precisely the determinant of the matrix of coefficients: \begin{align*} \tau^*(dy_1) \wedge \cdots \wedge \tau^*(dy_n) \;=\; \det\!\left(\frac{\partial \tau_i}{\partial x_j}\right)_{i,j=1}^n dx_1 \wedge \cdots \wedge dx_n \;=\; \det(J\tau) \, dx_1 \wedge \cdots \wedge dx_n, \end{align*} where $J\tau \in \mathbb{R}^{n\times n}$ is the Jacobian matrix of $\tau$ with entries $(J\tau)_{ij} = \partial_{x_j} \tau_i$ (see §0 of the notation standards). This is the standard transformation law for top-degree forms: the Jacobian determinant emerges from the algebra of wedge products without any analytic ingredient. Combining everything with the relation from the previous step, \begin{align*} f \, dx_1 \wedge \cdots \wedge dx_n \;=\; (\varphi^{-1})^*\omega \;=\; \tau^*\bigl((\psi^{-1})^*\omega\bigr) \;=\; (g \circ \tau)\,\det(J\tau)\, dx_1 \wedge \cdots \wedge dx_n. \end{align*} The space $\Omega^n(\varphi(W))$ is a free rank-$1$ module over $C^\infty(\varphi(W))$ generated by $dx_1 \wedge \cdots \wedge dx_n$, so two top-forms are equal if and only if their coefficients agree. Reading off the coefficient, \begin{align*} f(x) \;=\; g(\tau(x))\,\det(J\tau)(x) \qquad \text{for all } x \in \varphi(W). \tag{$\ast$} \end{align*} [/guided] [/step] [step:Use positivity of both charts to remove the absolute value on the Jacobian] By hypothesis, $(U,\varphi)$ and $(V,\psi)$ are positive with respect to the orientation of $M$. By [Orientability via Transition Maps](/theorems/1527), the transition diffeomorphism between two positive charts satisfies $\det(J\tau)(x) > 0$ for every $x \in \varphi(W)$. In particular, \begin{align*} |\det(J\tau)(x)| \;=\; \det(J\tau)(x) \qquad \text{for all } x \in \varphi(W), \end{align*} so the identity $(\ast)$ rewrites as $f(x) = g(\tau(x))\,|\det(J\tau)(x)|$. [guided] The Euclidean change-of-variables theorem will produce an absolute value $|\det(J\tau)|$ in the integrand, whereas our identity $(\ast)$ has the *signed* determinant $\det(J\tau)$. The orientation hypothesis is precisely what reconciles these two. By definition, a chart $(U,\varphi)$ is **positive** for the orientation of $M$ if, at each point $p \in U$, the coordinate frame $(\partial_{x_1}|_p, \dots, \partial_{x_n}|_p)$ represents the chosen orientation of $T_pM$. The criterion [Orientability via Transition Maps](/theorems/1527) then says that for any two positive charts $(U,\varphi)$ and $(V,\psi)$, the Jacobian determinant of the transition map $\tau = \psi \circ \varphi^{-1}$ is strictly positive on all of $\varphi(U \cap V)$. We have verified this hypothesis: both charts are positive by assumption, so \begin{align*} \det(J\tau)(x) \;>\; 0 \qquad \text{for all } x \in \varphi(W). \end{align*} Therefore $|\det(J\tau)(x)| = \det(J\tau)(x)$ on $\varphi(W)$, and the identity $(\ast)$ from the previous step can equivalently be written \begin{align*} f(x) \;=\; g(\tau(x))\,|\det(J\tau)(x)| \qquad \text{for all } x \in \varphi(W). \end{align*} This is the form in which the Euclidean change of variables will accept it. [/guided] [/step] [step:Apply the Euclidean change of variables to identify the two integrals] Both integrals in the statement are ordinary Lebesgue integrals of compactly supported smooth functions on open subsets of $\mathbb{R}^n$, so they are well-defined and finite. The map $\tau : \varphi(W) \to \psi(W)$ is a smooth diffeomorphism between open sets of $\mathbb{R}^n$, and $g \in C_c^\infty(\psi(W))$ is in particular Lebesgue-integrable on $\psi(W)$. Applying the [Change of Variables (general)](/theorems/22) to $g$ with the diffeomorphism $\tau$, \begin{align*} \int_{\psi(W)} g(y) \, d\mathcal{L}^n(y) \;=\; \int_{\varphi(W)} g(\tau(x))\,|\det(J\tau)(x)| \, d\mathcal{L}^n(x). \end{align*} By the previous step, $g(\tau(x))\,|\det(J\tau)(x)| = f(x)$ for all $x \in \varphi(W)$, so the right-hand side equals \begin{align*} \int_{\varphi(W)} f(x) \, d\mathcal{L}^n(x). \end{align*} Combining the two equalities yields \begin{align*} \int_{\varphi(W)} f(x) \, d\mathcal{L}^n(x) \;=\; \int_{\psi(W)} g(y) \, d\mathcal{L}^n(y), \end{align*} which is the desired identity. [guided] We are now ready to close the argument. The Euclidean change-of-variables theorem ([Change of Variables (general)](/theorems/22)) states: if $\tau : A \to B$ is a smooth diffeomorphism between open subsets of $\mathbb{R}^n$ and $h : B \to \mathbb{R}$ is Lebesgue-integrable on $B$, then \begin{align*} \int_B h(y) \, d\mathcal{L}^n(y) \;=\; \int_A h(\tau(x))\,|\det(J\tau)(x)| \, d\mathcal{L}^n(x). \end{align*} We apply this with $A = \varphi(W)$, $B = \psi(W)$, and $h = g$. The hypotheses are met: - $A$ and $B$ are open subsets of $\mathbb{R}^n$, as observed in Step 1 (images of opens under the chart homeomorphisms). - $\tau$ is a smooth diffeomorphism between them, by Step 1. - $g \in C_c^\infty(\psi(W))$, so $g$ is continuous with compact support in $\psi(W)$ — in particular $g$ is bounded and supported in a compact subset of $\psi(W)$ of finite Lebesgue measure, hence Lebesgue-integrable on $\psi(W)$. The conclusion gives \begin{align*} \int_{\psi(W)} g(y) \, d\mathcal{L}^n(y) \;=\; \int_{\varphi(W)} g(\tau(x))\,|\det(J\tau)(x)| \, d\mathcal{L}^n(x). \end{align*} By Step 3 we have $g(\tau(x))\,|\det(J\tau)(x)| = f(x)$ pointwise on $\varphi(W)$, so the right-hand integrand may be replaced by $f(x)$, giving \begin{align*} \int_{\psi(W)} g(y) \, d\mathcal{L}^n(y) \;=\; \int_{\varphi(W)} f(x) \, d\mathcal{L}^n(x), \end{align*} which is the desired equality of the two local integral computations. This shows that the local integral of a compactly supported top-degree form, computed in any positive chart whose domain contains the support of the form, is independent of the choice of chart — exactly the well-definedness statement needed to extend the definition of integration over a chart to the global definition of [Integration of Differential Forms](/theorems/1529) on an oriented manifold. [/guided] [/step]