Arc-length. Let \(f: \mathbb{R} ightarrow \mathbb{R}\) be a twice continuously differentiable function and denote by \(\Gamma_{f}:=\{(t, f(t)): t \in \mathbb{R}\}\) its graph. Define a function \(\Phi: \mathbb{R} ightarrow \mathbb{R}^{2}\) by \(\Phi(x):=(x, f(x))\). Then

(i) \(\Phi: \mathbb{R} ightarrow \Gamma_{f}\) is a \(C^{1}\)-diffeomorphism and \(|D \Phi(x)|=\sqrt{1+\left(f^{\prime}(x)ight)^{2}}\);

(ii) \(\sigma:=\Phi\left(|D \Phi| \lambda^{1}ight)\) is a measure on \(\Gamma_{f}\);

(iii) \(\int_{\Gamma_{f}} u(x, y) d \sigma(x, y)=\int_{\mathbb{R}} u(t, f(t)) \sqrt{1+\left(f^{\prime}(t)ight)^{2}} d \lambda^{1}(t)\) with the understanding that, whenever one side of the equality makes sense (measurability!) and is finite, so does the other.

The measure \(\sigma\) is called canonical surface measure on \(\Gamma_{f}\). This name is justified by the following compatibility property w.r.t. \(\lambda^{2}\). Let \(n(x)\) be a unit normal vector to \(\Gamma_{f}\) at point \((x, f(x))\) and define a map \(\widetilde{\Phi}: \mathbb{R} \times \mathbb{R} ightarrow \mathbb{R}^{2}\) by \(\widetilde{\Phi}(x, r):=\Phi(x)+r n(x)\).

(iv) Then \(n(x)=\left(-f^{\prime}(x), 1ight) / \sqrt{1+\left(f^{\prime}(x)ight)^{2}}\) and

\[\operatorname{det} D \widetilde{\Phi}(x, r)=\sqrt{1+\left(f^{\prime}(x)ight)^{2}}-\frac{r f^{\prime \prime}(x)}{1+\left(f^{\prime}(x)ight)^{2}}\]

Conclude that for every compact interval \([c, d]\) there exists some \(\epsilon>0\) such that \(\left.\widetilde{\Phi}ight|_{(c, d) \times(-\epsilon, \epsilon)}\) is a \(C^{1}\)-diffeomorphism.

(v) Let \(C \subset \Gamma_{f \mid(c, d)}\) and \(r<\epsilon\) with \(\epsilon\) as in (iv). Make a sketch of the set \(C(r):=\) \(\widetilde{\Phi}\left(\Phi^{-1}(C) \times(-r, r)ight)\) and show that it is Borel measurable.

(vi) Show that for every \(x \in(c, d)\)

\[
\lim _{r \downarrow 0} \frac{1}{2 r} \int_{(-r, r)}|\operatorname{det} D \widetilde{\Phi}(x, s)| \lambda^{1}(d s)=|\operatorname{det} D \widetilde{\Phi}(x, 0)|
\]

(vii) Use the general transformation theorem, Tonelli's theorem, (vi) and dominated convergence to show that
\[
\lim _{r \downarrow 0} \frac{1}{2 r} \lambda^{2}(C(r))=\int_{\Phi^{-1}(C)}|\operatorname{det} D \widetilde{\Phi}(x, 0)| \lambda^{1}(d x)
\]

(viii) Conclude that \(\int \sqrt{1+\left(f^{\prime}(t)ight)^{2}} d t\) is the arc-length of the graph of \(\Gamma_{f}\).