[proofplan] We work directly from the definition of the integral of a top-degree form on an oriented open subset of $\mathbb{R}^n$: choose any positively-oriented diffeomorphism $\psi$ onto $U$, pull $\omega$ back along $\psi$, and integrate the resulting top-form coefficient against Lebesgue measure. On $U$ with the standard orientation, the identity map is positively-oriented and the integral reduces to $\int_U f \, d\mathcal{L}^n$. On $-U$, the identity is *negatively*-oriented; we exhibit a positively-oriented chart by reflecting the first coordinate via $\psi(y) = (-y_1, y_2, \ldots, y_n)$. Pulling $\omega$ back along $\psi$ introduces a sign $-1$ from $d(-y_1) = -dy_1$, and the standard Lebesgue change of variables (with $|\det J\psi| = 1$) cancels the geometric reflection. The resulting sign $-1$ is the assertion. [/proofplan] [step:Reduce $\omega$ to its top-form coefficient and recall the integral on the standard-oriented domain] Since $U \subseteq \mathbb{R}^n$ is open, the global coordinate functions $(x_1, \ldots, x_n)$ give a single chart, and by the [Coordinate Basis for Differential Forms on an Open Subset of $\mathbb{R}^n$](/theorems/3562) every $n$-form on $U$ admits a unique representation \begin{align*} \omega = f \, dx_1 \wedge \cdots \wedge dx_n, \end{align*} where $f: U \to \mathbb{R}$ is measurable. The integrability hypothesis is precisely that $f \in L^1(U, \mathcal{L}^n)$. The standard orientation on $U$ is the one for which the identity map \begin{align*} \mathrm{id}_U : U &\to U \\ x &\mapsto x \end{align*} is positively oriented. By the definition recorded in [Integration of Differential Forms](/theorems/1529), choosing $\mathrm{id}_U$ as the positively-oriented chart, \begin{align*} \int_U \omega = \int_U f(x) \, d\mathcal{L}^n(x). \end{align*} [guided] A differential $n$-form on an open subset $U \subseteq \mathbb{R}^n$ has a global coordinate description because $U$ is already a coordinate domain — the inclusion $U \hookrightarrow \mathbb{R}^n$ supplies the chart. The [Coordinate Basis for Differential Forms on an Open Subset of $\mathbb{R}^n$](/theorems/3562) tells us that there is a unique measurable coefficient $f: U \to \mathbb{R}$ with $\omega = f \, dx_1 \wedge \cdots \wedge dx_n$. Integrability of $\omega$ is, by definition, integrability of this coefficient: $f \in L^1(U, \mathcal{L}^n)$. The definition of "integral of a top form on an oriented manifold" is: choose any positively-oriented diffeomorphism $\psi: V \to U$ from an [open set](/page/Open%20Set) $V \subseteq \mathbb{R}^n$, write $\psi^* \omega = g \, dy_1 \wedge \cdots \wedge dy_n$, and set $\int_U \omega := \int_V g \, d\mathcal{L}^n$. The result is independent of the chart by [Independence of the Local Integral of a Top-Degree Form Under Change of Positive Charts](/theorems/3578). For $U$ with the standard orientation, the simplest positively-oriented diffeomorphism is the identity $\mathrm{id}_U$. Its pullback is the identity on forms, so $\mathrm{id}_U^* \omega = \omega = f \, dx_1 \wedge \cdots \wedge dx_n$, and \begin{align*} \int_U \omega = \int_U f \, d\mathcal{L}^n. \end{align*} This is the baseline against which we must compare $\int_{-U} \omega$. [/guided] [/step] [step:Exhibit a positively-oriented diffeomorphism onto $-U$ via reflection of the first coordinate] The opposite orientation on $-U$ is, by definition, the orientation for which a diffeomorphism $\psi: V \to U$ (with $V \subseteq \mathbb{R}^n$ open) is positively oriented iff its Jacobian determinant is strictly negative everywhere. We construct such a $\psi$ explicitly. Let \begin{align*} V := \{(y_1, y_2, \ldots, y_n) \in \mathbb{R}^n : (-y_1, y_2, \ldots, y_n) \in U\}, \end{align*} which is open in $\mathbb{R}^n$ because it is the preimage of $U$ under the continuous map $y \mapsto (-y_1, y_2, \ldots, y_n)$. Define \begin{align*} \psi: V &\to U \\ (y_1, y_2, \ldots, y_n) &\mapsto (-y_1, y_2, \ldots, y_n). \end{align*} Then $\psi$ is a smooth bijection with smooth inverse $y \mapsto (-y_1, y_2, \ldots, y_n)$ (the same formula), so $\psi$ is a diffeomorphism. Its Jacobian matrix is $\operatorname{diag}(-1, 1, \ldots, 1)$, hence $\det J\psi_y = -1 < 0$ for every $y \in V$. Therefore $\psi$ is a positively-oriented chart for $-U$. [guided] We need a concrete positively-oriented chart for $-U$ — without one, the integral $\int_{-U} \omega$ is just a symbol. The opposite orientation on a connected oriented manifold is determined by the rule: $\psi$ is positively oriented for $-U$ iff $\det J\psi < 0$. So any diffeomorphism with negative Jacobian works. The simplest is a reflection in one coordinate; we pick the first coordinate for concreteness. Why is $V$ open? It is $\sigma^{-1}(U)$ where $\sigma(y) = (-y_1, y_2, \ldots, y_n)$ is continuous (in fact linear) and $U$ is open, so $V$ is open. Why is $\psi$ a diffeomorphism? Because $\psi \circ \psi = \mathrm{id}_V$ — the reflection is an involution — and $\psi$ is smooth, so it is its own smooth inverse. Why is $\psi$ positively oriented for $-U$? Because $\det J\psi_y = -1$ at every $y \in V$, and "positively oriented for $-U$" *means* "Jacobian determinant negative" since we are using the convention that the standard orientation on $U$ is "Jacobian determinant positive." [/guided] [/step] [step:Compute the pullback $\psi^* \omega$ and pick up the sign $-1$] We compute the pullback of $\omega = f \, dx_1 \wedge \cdots \wedge dx_n$ under $\psi$. By naturality of the wedge product and [exterior derivative](/theorems/1525) under pullback, \begin{align*} \psi^* \omega = (f \circ \psi) \cdot \psi^*(dx_1) \wedge \psi^*(dx_2) \wedge \cdots \wedge \psi^*(dx_n). \end{align*} The component functions of $\psi$ are $\psi_1(y) = -y_1$ and $\psi_i(y) = y_i$ for $i \geq 2$. Hence \begin{align*} \psi^*(dx_1) &= d(\psi_1) = d(-y_1) = -dy_1, \\ \psi^*(dx_i) &= d(\psi_i) = dy_i \qquad (i \geq 2). \end{align*} Substituting and pulling the scalar $-1$ outside the wedge, \begin{align*} \psi^* \omega = (f \circ \psi) \cdot (-dy_1) \wedge dy_2 \wedge \cdots \wedge dy_n = -\,(f \circ \psi) \, dy_1 \wedge dy_2 \wedge \cdots \wedge dy_n. \end{align*} [guided] The pullback acts on a top form $\omega = f \, dx_1 \wedge \cdots \wedge dx_n$ by pulling back the coefficient (composition with $\psi$) and the basis form (wedge of pullbacks of the $dx_i$). The pullback commutes with $d$, so $\psi^*(dx_i) = d(\psi^* x_i) = d(\psi_i)$, where $\psi_i = x_i \circ \psi$ is the $i$-th component of $\psi$. For our reflection $\psi(y) = (-y_1, y_2, \ldots, y_n)$, only the first component flips sign: $\psi_1(y) = -y_1$, so $d\psi_1 = -dy_1$. The remaining $\psi_i$ are the identity in their $i$-th slot, so $d\psi_i = dy_i$. The single sign on $dy_1$ propagates through the wedge product by multilinearity, producing the overall factor $-1$: \begin{align*} (-dy_1) \wedge dy_2 \wedge \cdots \wedge dy_n = -(dy_1 \wedge dy_2 \wedge \cdots \wedge dy_n). \end{align*} This single sign is the entire source of the orientation-reversal identity — the rest of the proof is bookkeeping that propagates it correctly. [/guided] [/step] [step:Apply the definition of $\int_{-U}$ and Lebesgue change of variables to conclude] Since $\psi: V \to U$ is a positively-oriented chart for $-U$, the definition of integration on an oriented manifold (cf. [Integration of Differential Forms](/theorems/1529)) gives \begin{align*} \int_{-U} \omega = \int_V g \, d\mathcal{L}^n, \qquad \text{where } \psi^* \omega = g \, dy_1 \wedge \cdots \wedge dy_n. \end{align*} By the previous step, $g(y) = -f(\psi(y))$. Therefore \begin{align*} \int_{-U} \omega = -\int_V f(\psi(y)) \, d\mathcal{L}^n(y). \end{align*} We now apply the [Change of Variables (general)](/theorems/22) formula for the [Lebesgue integral](/page/Lebesgue%20Integral) under the diffeomorphism $\psi: V \to U$. We verify its hypotheses: $V \subseteq \mathbb{R}^n$ is open, $U \subseteq \mathbb{R}^n$ is open, $\psi$ is a $C^1$ (indeed smooth) bijection with smooth inverse, and $f \in L^1(U, \mathcal{L}^n)$ by hypothesis. The Jacobian satisfies $|\det J\psi_y| = |-1| = 1$ for every $y \in V$. The change of variables formula gives \begin{align*} \int_V f(\psi(y)) \, |\det J\psi_y| \, d\mathcal{L}^n(y) = \int_U f(x) \, d\mathcal{L}^n(x), \end{align*} which, using $|\det J\psi_y| = 1$, reduces to \begin{align*} \int_V f(\psi(y)) \, d\mathcal{L}^n(y) = \int_U f(x) \, d\mathcal{L}^n(x) = \int_U \omega, \end{align*} where the final equality is from Step 1. In particular, $f \circ \psi \in L^1(V, \mathcal{L}^n)$, so $\omega$ is integrable on $-U$ as claimed. Combining, \begin{align*} \int_{-U} \omega = -\int_V f(\psi(y)) \, d\mathcal{L}^n(y) = -\int_U \omega. \end{align*} This is the desired identity, completing the proof. [guided] We have two integrals living on different ambient sets: the integral $\int_{-U} \omega$ unfolds — by its definition as an integral on an oriented manifold — to a [Lebesgue integral](/page/Lebesgue%20Integral) over the *reflected* set $V$, with integrand the *pulled-back* coefficient $g = -f \circ \psi$. We need to relate this back to a [Lebesgue integral](/page/Lebesgue%20Integral) over $U$, the original set. This is exactly what the Lebesgue change of variables theorem does. We verify the hypotheses of [Change of Variables (general)](/theorems/22): (i) $V, U \subseteq \mathbb{R}^n$ are open (Step 2); (ii) $\psi: V \to U$ is a smooth diffeomorphism (Step 2); (iii) $f \in L^1(U, \mathcal{L}^n)$ (theorem hypothesis). All three are met. The Jacobian determinant of $\psi$ is $-1$, but the change of variables formula uses its *absolute value* $|\det J\psi| = 1$. So no extra factor appears. The formula yields \begin{align*} \int_V (f \circ \psi) \, d\mathcal{L}^n = \int_U f \, d\mathcal{L}^n. \end{align*} Why does the sign emerge if the absolute-value Jacobian is $+1$? The sign comes from the *form-pullback* (Step 3), not from the measure-theoretic Jacobian. The form pullback senses the *signed* Jacobian (via $d(-y_1) = -dy_1$), whereas Lebesgue measure senses only the unsigned $|\det J\psi|$. The discrepancy between signed and unsigned Jacobians is precisely the orientation data, and it produces the sign $-1$ in the final identity. Putting the pieces together: \begin{align*} \int_{-U} \omega \;=\; \int_V g \, d\mathcal{L}^n \;=\; -\int_V (f \circ \psi) \, d\mathcal{L}^n \;=\; -\int_U f \, d\mathcal{L}^n \;=\; -\int_U \omega. \end{align*} [/guided] [/step]