**Proof plan.** The strategy is to reduce both sides to $\mathcal{L}^m$-[integrals](/page/Integral) over the parameter domain $A$ by applying the [Area Formula (Classical)](/theorems/25) twice — once to $\sigma$ (which parametrizes $M$) and once to $\Psi \circ \sigma$ (which parametrizes $\Psi(M)$). The chain rule relates the two Jacobians, and the tangential Jacobian emerges as their ratio. A separate argument using the QR decomposition shows that this ratio is independent of the parametrization.
**Step 1: Apply the area formula to $\Psi \circ \sigma$.**
The composition $\Psi \circ \sigma: A \to \mathbb{R}^n$ is $C^1$ (as a composition of $C^1$ maps) and injective (since both $\sigma$ and $\Psi|_M$ are injective). By the [Area Formula (Classical)](/theorems/25) applied to the injective $C^1$ map $\Psi \circ \sigma$ on the [open set](/page/Open%20Set) $A \subseteq \mathbb{R}^m$:
\begin{align*}
\int_{\Psi(M)} f(z) \, d\mathcal{H}^m(z) = \int_A f(\Psi(\sigma(y))) \, \sqrt{\det\bigl(J(\Psi \circ \sigma)(y)^T \, J(\Psi \circ \sigma)(y)\bigr)} \, d\mathcal{L}^m(y). \tag{1}
\end{align*}
**Step 2: Factor the integrand using the tangential Jacobian.**
By definition of $J_M \Psi$, for each $y \in A$:
\begin{align*}
\sqrt{\det\bigl(J(\Psi \circ \sigma)(y)^T \, J(\Psi \circ \sigma)(y)\bigr)} = J_M \Psi(\sigma(y)) \cdot \sqrt{\det\bigl(J\sigma(y)^T \, J\sigma(y)\bigr)}.
\end{align*}
Substituting into (1):
\begin{align*}
\int_{\Psi(M)} f(z) \, d\mathcal{H}^m(z) = \int_A f(\Psi(\sigma(y))) \cdot J_M \Psi(\sigma(y)) \cdot \sqrt{\det\bigl(J\sigma(y)^T \, J\sigma(y)\bigr)} \, d\mathcal{L}^m(y). \tag{2}
\end{align*}
**Step 3: Apply the area formula to $\sigma$ in the reverse direction.**
By the [Area Formula (Classical)](/theorems/25) applied to the injective $C^1$ immersion $\sigma: A \to \mathbb{R}^n$, for any Borel measurable function $g: M \to [0, \infty]$:
\begin{align*}
\int_A g(\sigma(y)) \, \sqrt{\det\bigl(J\sigma(y)^T \, J\sigma(y)\bigr)} \, d\mathcal{L}^m(y) = \int_M g(x) \, d\mathcal{H}^m(x). \tag{3}
\end{align*}
The integrand in (2) has the form of the left-hand side of (3) with $g(x) := f(\Psi(x)) \cdot J_M \Psi(x)$. The function $g$ is Borel measurable: $f \circ \Psi$ is Borel measurable as a composition of a Borel measurable function with a [continuous](/page/Continuity) function, and $J_M \Psi$ is continuous on $M$ (as a ratio of continuous, strictly positive [functions](/page/Function)). Applying (3):
\begin{align*}
\int_{\Psi(M)} f(z) \, d\mathcal{H}^m(z) = \int_M f(\Psi(x)) \cdot J_M \Psi(x) \, d\mathcal{H}^m(x).
\end{align*}
**Step 4: Verify parametrization independence of $J_M \Psi$.**
[claim:Parametrisation Independence]
Let $\tilde{\sigma}: \tilde{A} \to \mathbb{R}^n$ be another injective $C^1$ immersion with $\tilde{\sigma}(\tilde{A}) = M$. Then the tangential Jacobian computed via $\tilde{\sigma}$ agrees with the one computed via $\sigma$ at every point of $M$.
[/claim]
[proof]
**Step 4a.** Fix $x \in M$ and let $y = \sigma^{-1}(x) \in A$, $\tilde{y} = \tilde{\sigma}^{-1}(x) \in \tilde{A}$. Since $\sigma$ and $\tilde{\sigma}$ are injective $C^1$ immersions with the same image, the transition map
\begin{align*}
\varphi := \sigma^{-1} \circ \tilde{\sigma}: \tilde{A} \to A
\end{align*}
is a $C^1$ diffeomorphism (by the [implicit function theorem](/page/Implicit%20Function%20Theorem) applied to the immersion $\sigma$), and $\sigma \circ \varphi = \tilde{\sigma}$.
**Step 4b.** By the chain rule, $J\tilde{\sigma}(\tilde{y}) = J\sigma(y) \cdot J\varphi(\tilde{y})$, where $J\varphi(\tilde{y}) \in \mathbb{R}^{m \times m}$ is invertible. Set $P := J\varphi(\tilde{y})$. Then:
\begin{align*}
\det\bigl(J\tilde{\sigma}^T J\tilde{\sigma}\bigr) = \det\bigl(P^T J\sigma^T J\sigma \, P\bigr) = (\det P)^2 \cdot \det\bigl(J\sigma^T J\sigma\bigr).
\end{align*}
Similarly, $J(\Psi \circ \tilde{\sigma})(\tilde{y}) = J(\Psi \circ \sigma)(y) \cdot P$, so:
\begin{align*}
\det\bigl(J(\Psi \circ \tilde{\sigma})^T J(\Psi \circ \tilde{\sigma})\bigr) = (\det P)^2 \cdot \det\bigl(J(\Psi \circ \sigma)^T J(\Psi \circ \sigma)\bigr).
\end{align*}
Taking square roots and forming the ratio, the factors $|\det P|$ cancel:
\begin{align*}
\frac{\sqrt{\det\bigl(J(\Psi \circ \tilde{\sigma})^T J(\Psi \circ \tilde{\sigma})\bigr)}}{\sqrt{\det\bigl(J\tilde{\sigma}^T J\tilde{\sigma}\bigr)}} = \frac{\sqrt{\det\bigl(J(\Psi \circ \sigma)^T J(\Psi \circ \sigma)\bigr)}}{\sqrt{\det\bigl(J\sigma^T J\sigma\bigr)}}.
\end{align*}
[/proof]
[remark:Application To Submanifolds Without A Global Chart]
If $M$ is a compact $C^1$ submanifold of $\mathbb{R}^n$ that cannot be covered by a single chart (e.g. $M = \partial B(0,1) \subseteq \mathbb{R}^n$), one partitions $M$ into finitely many $\mathcal{H}^m$-measurable pieces $M_1, \ldots, M_k$, each contained in the image of a chart $\sigma_\alpha: A_\alpha \to \mathbb{R}^n$. The formula is applied to each piece, and the results are summed by $\sigma$-additivity of $\mathcal{H}^m$. The conclusion $\int_{\Psi(M)} f \, d\mathcal{H}^m = \int_M (f \circ \Psi) \cdot J_M \Psi \, d\mathcal{H}^m$ follows, with $J_M \Psi$ well-defined by parametrisation independence.
[/remark]
[remark:Affine Maps And Spheres]
When $\Psi: \mathbb{R}^n \to \mathbb{R}^n$ is the affine map $\Psi(x) = z + rx$ for fixed $z \in \mathbb{R}^n$ and $r > 0$, the [derivative](/page/Derivative) is $D\Psi(x) = rI_n$ for all $x$. For any $C^1$ parametrisation $\sigma: A \to \mathbb{R}^n$ of an $m$-dimensional surface $M$:
\begin{align*}
J(\Psi \circ \sigma)(y) = r \cdot J\sigma(y),
\end{align*}
so $\det(J(\Psi \circ \sigma)^T J(\Psi \circ \sigma)) = r^{2m} \det(J\sigma^T J\sigma)$, giving $J_M \Psi = r^m$. The formula becomes
\begin{align*}
\int_{\Psi(M)} f \, d\mathcal{H}^m = r^m \int_M f(z + rx) \, d\mathcal{H}^m(x).
\end{align*}
In particular, for $M = \partial B(0,1)$, $\Psi(M) = \partial B(z, r)$, and $m = n - 1$:
\begin{align*}
\int_{\partial B(z, r)} f(y) \, d\mathcal{H}^{n-1}(y) = r^{n-1} \int_{\partial B(0, 1)} f(z + r\omega) \, d\mathcal{H}^{n-1}(\omega),
\end{align*}
which is the change-of-variables identity for spherical integrals.
[/remark]