[proofplan] Write $\omega = a(y)\, dy_1 \wedge \cdots \wedge dy_n$ for a compactly supported continuous coefficient $a: V \to \mathbb{R}$. The proof has three main ingredients. First, we compute the pullback of a top-degree form: $f^*\omega = (a \circ f)(x)\,\det(Jf_x)\, dx_1 \wedge \cdots \wedge dx_n$, where the determinant appears because pullback commutes with the wedge product and acts on $1$-forms via the Jacobian. Second, the definition of integration of an $n$-form on an oriented open subset of $\mathbb{R}^n$ converts both sides into ordinary Lebesgue integrals of their coefficient functions. Third, the [Change of Variables formula](/theorems/22) for Lebesgue integrals, which contains $|\det Jf_x|$, is reconciled with the signed Jacobian by tracking the sign $\varepsilon(f)$ — constant on $U$ by the orientation hypothesis. [/proofplan]

[step:Reduce the form to coefficient times the standard volume form]Since $V \subseteq \mathbb{R}^n$ is open and $\omega$ is a continuous compactly supported $n$-form on $V$, there exists a unique continuous, compactly supported function $a: V \to \mathbb{R}$ with $\operatorname{supp}(a) = \operatorname{supp}(\omega)$ such that \begin{align*} \omega = a(y)\, dy_1 \wedge \cdots \wedge dy_n \qquad \text{on } V. \end{align*} This is because $\Lambda^n(\mathbb{R}^n)^*$ is one-dimensional with basis $dy_1 \wedge \cdots \wedge dy_n$, so every $n$-form on an open subset of $\mathbb{R}^n$ is a scalar multiple of the standard volume form by a uniquely determined coefficient function. Extending $a$ by zero outside its support (which is a compact subset of $V$) we may regard $a$ as a continuous function on all of $\mathbb{R}^n$ — but for integration purposes we work on $V$.[/step]

[guided]We want to extract the analytic content of the $n$-form $\omega$. The space of alternating $n$-linear forms on $\mathbb{R}^n$, denoted $\Lambda^n(\mathbb{R}^n)^*$, has dimension $\binom{n}{n} = 1$, with the canonical basis element $dy_1 \wedge \cdots \wedge dy_n$ (the standard volume form). Therefore at every point $y \in V$, the value $\omega_y \in \Lambda^n(T_y^*V) = \Lambda^n(\mathbb{R}^n)^*$ is a unique scalar multiple of $(dy_1 \wedge \cdots \wedge dy_n)_y$: \begin{align*} \omega_y = a(y) \, (dy_1 \wedge \cdots \wedge dy_n)_y. \end{align*} The function $a: V \to \mathbb{R}$ inherits continuity from $\omega$, and its support coincides with $\operatorname{supp}(\omega)$, hence is compact in $V$. This reduces the form $\omega$ to a single scalar coefficient, which is the form in which the classical change of variables theorem will apply.[/guided]

[step:Compute the pullback of the standard volume form via the Jacobian determinant]We claim that for $f: U \to V$ a $C^1$ map with Jacobian matrix $Jf_x \in \mathbb{R}^{n \times n}$ (entries $(Jf_x)_{ij} = \partial_{x_j}f_i(x)$), \begin{align*} f^*(dy_1 \wedge \cdots \wedge dy_n) = \det(Jf_x)\, dx_1 \wedge \cdots \wedge dx_n. \end{align*} [claim:Pullback of the top volume form equals the Jacobian determinant times the standard volume form on $U$] For every $C^1$ map $f: U \to V$, one has \begin{align*} f^*(dy_1 \wedge \cdots \wedge dy_n) = \det(Jf_x)\, dx_1 \wedge \cdots \wedge dx_n. \end{align*} [/claim] [proof] For each $x \in U$ and vectors $v_1, \dots, v_n \in T_xU = \mathbb{R}^n$, \begin{align*} \bigl(f^*(dy_1 \wedge \cdots \wedge dy_n)\bigr)_x(v_1, \dots, v_n) &= (dy_1 \wedge \cdots \wedge dy_n)_{f(x)}\bigl(df_x(v_1), \dots, df_x(v_n)\bigr) \\ &= \det\bigl[(dy_i)_{f(x)}(df_x(v_j))\bigr]_{i,j=1}^n. \end{align*} The second equality uses the standard determinantal formula for the action of a wedge product of $1$-forms on a tuple of vectors: \begin{align*} (\alpha_1 \wedge \cdots \wedge \alpha_n)(w_1, \dots, w_n) = \det\bigl[\alpha_i(w_j)\bigr]_{i,j=1}^n, \end{align*} which is the defining property of the wedge product of $1$-forms — equivalently, it follows from the antisymmetric multilinear extension of the [tensor product](/page/Tensor%20Product) $\alpha_1 \otimes \cdots \otimes \alpha_n$. Now $(dy_i)_{f(x)}(df_x(v_j)) = (df_x(v_j))_i$, the $i$-th component of $df_x(v_j) \in \mathbb{R}^n$. By the definition of the [total derivative](/theorems/320) and the Jacobian matrix (see notation standards §0), $df_x: \mathbb{R}^n \to \mathbb{R}^n$ is the [linear map](/page/Linear%20Map) whose matrix in the standard basis is the Jacobian $Jf_x$, with entries $(Jf_x)_{ij} = \partial_{x_j}f_i(x)$. Hence $df_x(v_j) = Jf_x \, v_j$, so \begin{align*} \bigl(df_x(v_j)\bigr)_i = (Jf_x \, v_j)_i = \sum_{k=1}^n (Jf_x)_{ik}\, (v_j)_k. \end{align*} Therefore the matrix in the determinant equals $Jf_x \cdot V$, where $V$ is the $n \times n$ matrix with columns $v_1, \dots, v_n$. By multiplicativity of the determinant, \begin{align*} \det\bigl[(df_x(v_j))_i\bigr] = \det(Jf_x \, V) = \det(Jf_x) \cdot \det(V). \end{align*} On the other hand, \begin{align*} \bigl(\det(Jf_x)\, dx_1 \wedge \cdots \wedge dx_n\bigr)_x(v_1, \dots, v_n) = \det(Jf_x) \cdot \det(V), \end{align*} by the same determinantal formula $(\alpha_1 \wedge \cdots \wedge \alpha_n)(w_1, \dots, w_n) = \det[\alpha_i(w_j)]$ applied with $\alpha_i = dx_i$ and $w_j = v_j$, so that $dx_i(v_j) = (v_j)_i = V_{ij}$. The two expressions agree pointwise, proving the claim. [/proof] Multiplying by the scalar $a(f(x))$ and using $\mathbb{R}$-linearity of pullback, \begin{align*} f^*\omega = f^*\bigl(a(y)\, dy_1 \wedge \cdots \wedge dy_n\bigr) = (a \circ f)(x) \cdot \det(Jf_x) \, dx_1 \wedge \cdots \wedge dx_n. \end{align*}[/step]

[guided]Pullback of forms is defined so that for any $k$-form $\eta$ on $V$ and any $k$ tangent vectors $v_1, \dots, v_k$ at $x \in U$, \begin{align*} (f^*\eta)_x(v_1, \dots, v_k) = \eta_{f(x)}\bigl(df_x(v_1), \dots, df_x(v_k)\bigr). \end{align*} We apply this to the top-degree form $dy_1 \wedge \cdots \wedge dy_n$. The action of a wedge product of $1$-forms on a tuple of vectors is the determinant of the matrix of pairings: \begin{align*} (\alpha_1 \wedge \cdots \wedge \alpha_n)(w_1, \dots, w_n) = \det\bigl[\alpha_i(w_j)\bigr]. \end{align*} Here $\alpha_i = dy_i$ and $w_j = df_x(v_j) = Jf_x \, v_j$, so $\alpha_i(w_j)$ is the $i$-th component of $Jf_x \, v_j$. The matrix of these pairings is therefore exactly $Jf_x \cdot V$ where $V$ has columns $v_j$. Multiplicativity of $\det$ factors out $\det(Jf_x)$, and the remaining $\det(V)$ is precisely what the standard volume form $dx_1 \wedge \cdots \wedge dx_n$ extracts from $(v_1, \dots, v_n)$. The conclusion is that pullback of the top volume form just multiplies by the Jacobian determinant — a signed scalar. This signed-ness, as opposed to the absolute value that appears in measure-theoretic change of variables, is the source of the orientation sign in the theorem. Pullback is also $\mathbb{R}$-linear in the form (and respects multiplication by scalar functions on $V$), so multiplying the form by $a(y)$ multiplies its pullback by $a(f(x))$, giving the displayed formula.[/guided]

[step:Express both integrals as Lebesgue integrals of their coefficients]By the [definition of the integral of a top-degree form](/theorems/1529) on an oriented open subset of $\mathbb{R}^n$ (with the standard orientation), the integral of a compactly supported $n$-form $\alpha = b\, dz_1 \wedge \cdots \wedge dz_n$ on an [open set](/page/Open%20Set) $W \subseteq \mathbb{R}^n$ is \begin{align*} \int_W \alpha := \int_W b(z) \, d\mathcal{L}^n(z). \end{align*} Applying this with $W = V$, $\alpha = \omega$, $b = a$: \begin{align*} \int_V \omega = \int_V a(y) \, d\mathcal{L}^n(y). \end{align*} Applying it with $W = U$, $\alpha = f^*\omega$, $b(x) = (a \circ f)(x)\, \det(Jf_x)$ (which is continuous and compactly supported in $U$ since $a$ is continuous and compactly supported in $V$, $f$ is a homeomorphism, and $\det Jf_x$ is continuous): \begin{align*} \int_U f^*\omega = \int_U (a \circ f)(x) \, \det(Jf_x) \, d\mathcal{L}^n(x). \end{align*}[/step]

[guided]The bridge from differential geometry to measure theory happens here. The integral $\int_V \omega$ of an $n$-form has been defined as the [Lebesgue integral](/page/Lebesgue%20Integral) of its coefficient with respect to the standard volume form — this is precisely the construction that makes the integral coordinate-invariant up to sign, and the theorem we are proving is the statement that it actually transforms correctly under diffeomorphisms. We must verify that the coefficient $(a \circ f) \det(Jf_x)$ on the $U$ side is genuinely compactly supported and integrable. Since $\operatorname{supp}(a)$ is compact in $V$ and $f$ is a homeomorphism, $f^{-1}(\operatorname{supp}(a))$ is compact in $U$. Outside this preimage $a \circ f$ vanishes, so the product $(a \circ f) \det(Jf_x)$ also vanishes; thus its support lies in the compact set $f^{-1}(\operatorname{supp}(a)) \subset U$. Continuity of $a \circ f$ (composition of continuous maps) and of $\det(Jf_x)$ (a $C^0$ function of the entries of $Jf_x$, themselves continuous since $f$ is $C^1$) gives that the integrand is continuous and compactly supported, hence Lebesgue-integrable on $U$.[/guided]

[step:Apply the classical change of variables formula and resolve the sign]The hypothesis fixes the sign of $\det Jf_x$: writing $\varepsilon(f) \in \{+1, -1\}$ for the orientation sign, \begin{align*} \det(Jf_x) = \varepsilon(f) \cdot |\det(Jf_x)| \qquad \text{for all } x \in U. \end{align*} Indeed, if $f$ is orientation-preserving, $\det Jf_x > 0$ for all $x$, so $\det Jf_x = +|\det Jf_x|$ and $\varepsilon(f) = +1$; if orientation-reversing, $\det Jf_x < 0$ for all $x$, so $\det Jf_x = -|\det Jf_x|$ and $\varepsilon(f) = -1$. Substituting into the result of the previous step, \begin{align*} \int_U f^*\omega = \varepsilon(f) \int_U (a \circ f)(x) \, |\det(Jf_x)| \, d\mathcal{L}^n(x). \end{align*} We now apply the [Change of Variables Formula for Lebesgue Integrals](/theorems/22). This theorem requires: (i) $U, V \subseteq \mathbb{R}^n$ open; (ii) $f: U \to V$ a $C^1$ diffeomorphism; (iii) the integrand $a: V \to \mathbb{R}$ Lebesgue measurable (in fact continuous) and either non-negative or integrable. All three hold by hypothesis: (i) and (ii) are stated; (iii) holds because $a$ is continuous with compact support in $V$, hence bounded and integrable. The conclusion is \begin{align*} \int_U (a \circ f)(x) \, |\det(Jf_x)| \, d\mathcal{L}^n(x) = \int_V a(y) \, d\mathcal{L}^n(y) = \int_V \omega, \end{align*} where the last equality is from the previous step. Combining, \begin{align*} \int_U f^*\omega = \varepsilon(f) \int_V \omega. \end{align*} This is the asserted formula: $+\int_V \omega$ when $f$ is orientation-preserving and $-\int_V \omega$ when $f$ is orientation-reversing. $\blacksquare$[/step]

[guided]The orientation hypothesis is exactly what makes $\det Jf_x$ have a definite sign throughout $U$. Without this hypothesis (for instance, if $f$ were a diffeomorphism with $\det Jf_x$ positive on one connected component of $U$ and negative on another), the statement would have to be local and the relation between $\int_U f^*\omega$ and $\int_V \omega$ would depend on the support of $\omega$. The two cases — strictly positive Jacobian everywhere, strictly negative Jacobian everywhere — are the only ones permitted by the definition of "orientation-preserving" and "orientation-reversing" diffeomorphism between subsets of $\mathbb{R}^n$ with the standard orientation; a diffeomorphism between connected open sets automatically falls into one case or the other, since $x \mapsto \det Jf_x$ is continuous and non-zero. With the sign extracted, we are reduced to integrating $(a \circ f) |\det Jf_x|$, which is exactly the integrand on the left-hand side of the classical change of variables theorem. We verify the three hypotheses of the cited theorem before invoking it: 1. **Openness of $U, V$:** stated. 2. **$C^1$ diffeomorphism:** stated. 3. **Integrability of $a$:** $a$ is continuous on $V$ with compact support, so $|a|$ is bounded by some $M < \infty$ on the compact set $\operatorname{supp}(a)$ and zero elsewhere, giving $\int_V |a(y)|\, d\mathcal{L}^n(y) \le M \cdot \mathcal{L}^n(\operatorname{supp}(a)) < \infty$. The conclusion of the classical change of variables theorem, applied to the integrand $a$ pushed back through $f$, is the explicit identity \begin{align*} \int_U (a \circ f)(x) \, |\det(Jf_x)| \, d\mathcal{L}^n(x) = \int_V a(y) \, d\mathcal{L}^n(y). \end{align*} The right-hand side equals $\int_V \omega$ by the previous step. Substituting into the displayed equation \begin{align*} \int_U f^*\omega = \varepsilon(f) \int_U (a \circ f)(x) \, |\det(Jf_x)| \, d\mathcal{L}^n(x), \end{align*} we obtain $\int_U f^*\omega = \varepsilon(f) \int_V \omega$, which is the stated identity.[/guided]