[proofplan] We compare the supremum definition of path length with the integral of the speed. The upper bound follows by applying the coordinatewise [fundamental theorem of calculus](/theorems/632) on each subinterval of an arbitrary partition and then using the triangle inequality for the Euclidean norm. For the reverse bound, [uniform continuity](/page/Uniform%20Continuity) of $\gamma'$ on the compact interval $[0,1]$ lets us choose a partition fine enough that $\gamma'$ is nearly constant on each subinterval; chord lengths then approximate the speed integral up to an arbitrarily small error. The finiteness of the speed integral gives rectifiability. [/proofplan]

[step:Define the partition sums that determine the path length] For a finite partition $P$ of $[0,1]$, write \begin{align*} P=\{0=t_0<t_1<\cdots<t_m=1\}. \end{align*} Define the polygonal length sum associated to $P$ by \begin{align*} S(P,\gamma)=\sum_{i=1}^m |\gamma(t_i)-\gamma(t_{i-1})|. \end{align*} By definition of path length, \begin{align*} L(\gamma)=\sup_P S(P,\gamma), \end{align*} where the supremum is taken over all finite partitions $P$ of $[0,1]$. Let \begin{align*} v:[0,1]\to[0,\infty), \qquad t \mapsto |\gamma'(t)| \end{align*} be the speed function. Since $\gamma\in C^1([0,1];\mathbb{R}^n)$, the derivative map $\gamma':[0,1]\to\mathbb{R}^n$ is continuous. Hence $v$ is continuous, and therefore $v$ is Lebesgue integrable on $[0,1]$ with respect to $\mathcal{L}^1$. [/step]

[step:Bound every polygonal sum by the speed integral]Fix a partition \begin{align*} P=\{0=t_0<t_1<\cdots<t_m=1\}. \end{align*} For each $i\in\{1,\dots,m\}$, the coordinate functions of $\gamma$ are $C^1$ on $[t_{i-1},t_i]$. By the coordinatewise fundamental theorem of calculus for $C^1$ vector-valued functions, \begin{align*} \gamma(t_i)-\gamma(t_{i-1})=\int_{t_{i-1}}^{t_i}\gamma'(t)\,d\mathcal{L}^1(t). \end{align*} Taking the Euclidean norm and using the triangle inequality for coordinatewise integrals gives \begin{align*} |\gamma(t_i)-\gamma(t_{i-1})|\leq \int_{t_{i-1}}^{t_i}|\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*} Summing this estimate over $i=1,\dots,m$ and using additivity of the [Lebesgue integral](/page/Lebesgue%20Integral) over adjacent intervals, \begin{align*} S(P,\gamma)\leq \int_0^1|\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*} Taking the supremum over all partitions $P$ yields \begin{align*} L(\gamma)\leq \int_0^1|\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*}[/step]

[guided]Fix a partition \begin{align*} P=\{0=t_0<t_1<\cdots<t_m=1\}. \end{align*} The length sum $S(P,\gamma)$ is built from the chord vectors $\gamma(t_i)-\gamma(t_{i-1})$. To compare these chord vectors with the derivative $\gamma'$, we express each chord as an integral of the velocity over the corresponding time interval. For each $i\in\{1,\dots,m\}$, the restriction of $\gamma$ to $[t_{i-1},t_i]$ is still a $C^1$ map into $\mathbb{R}^n$. Applying the one-variable fundamental theorem of calculus to each coordinate function gives the vector identity \begin{align*} \gamma(t_i)-\gamma(t_{i-1})=\int_{t_{i-1}}^{t_i}\gamma'(t)\,d\mathcal{L}^1(t). \end{align*} This integral is understood coordinatewise: if $\gamma=(\gamma_1,\dots,\gamma_n)$, then the $j$-th coordinate of the integral is $\int_{t_{i-1}}^{t_i}\gamma_j'(t)\,d\mathcal{L}^1(t)$. Now take the Euclidean norm. The triangle inequality for integrals of continuous $\mathbb{R}^n$-valued functions gives \begin{align*} \left|\int_{t_{i-1}}^{t_i}\gamma'(t)\,d\mathcal{L}^1(t)\right|\leq \int_{t_{i-1}}^{t_i}|\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*} Combining the last two displayed formulas, \begin{align*} |\gamma(t_i)-\gamma(t_{i-1})|\leq \int_{t_{i-1}}^{t_i}|\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*} Summing over all subintervals in the partition gives \begin{align*} S(P,\gamma)=\sum_{i=1}^m |\gamma(t_i)-\gamma(t_{i-1})|\leq \sum_{i=1}^m\int_{t_{i-1}}^{t_i}|\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*} Since the intervals $[t_{i-1},t_i]$ cover $[0,1]$ up to shared endpoints, and shared endpoints have $\mathcal{L}^1$-measure zero, additivity of the integral gives \begin{align*} \sum_{i=1}^m\int_{t_{i-1}}^{t_i}|\gamma'(t)|\,d\mathcal{L}^1(t)=\int_0^1|\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*} Thus every partition sum satisfies \begin{align*} S(P,\gamma)\leq \int_0^1|\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*} Because $L(\gamma)$ is the supremum of these sums over all partitions $P$, we obtain \begin{align*} L(\gamma)\leq \int_0^1|\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*}[/guided]

[step:Use uniform continuity of the velocity to build accurate polygonal approximations]Let $\varepsilon>0$. Since $\gamma':[0,1]\to\mathbb{R}^n$ is continuous and $[0,1]$ is compact, $\gamma'$ is uniformly continuous. Hence there exists $\delta>0$ such that for all $s,t\in[0,1]$, \begin{align*} |s-t|<\delta\implies |\gamma'(s)-\gamma'(t)|<\varepsilon. \end{align*} Choose a partition \begin{align*} P=\{0=t_0<t_1<\cdots<t_m=1\} \end{align*} such that \begin{align*} t_i-t_{i-1}<\delta \end{align*} for every $i\in\{1,\dots,m\}$. For each such $i$, set $a_i=t_{i-1}$ and $b_i=t_i$. Then for every $t\in[a_i,b_i]$, \begin{align*} |\gamma'(t)-\gamma'(a_i)|<\varepsilon. \end{align*} Using the [reverse triangle inequality](/theorems/2300) in $\mathbb{R}^n$, \begin{align*} |\gamma'(t)|\leq |\gamma'(a_i)|+\varepsilon \end{align*} for all $t\in[a_i,b_i]$. Therefore \begin{align*} \int_{a_i}^{b_i}|\gamma'(t)|\,d\mathcal{L}^1(t)\leq |\gamma'(a_i)|(b_i-a_i)+\varepsilon(b_i-a_i). \end{align*} Again by the coordinatewise fundamental theorem of calculus, \begin{align*} \gamma(b_i)-\gamma(a_i)=\int_{a_i}^{b_i}\gamma'(t)\,d\mathcal{L}^1(t). \end{align*} Subtracting the integral of the constant vector $\gamma'(a_i)$ gives \begin{align*} \gamma'(a_i)(b_i-a_i)-(\gamma(b_i)-\gamma(a_i))=\int_{a_i}^{b_i}(\gamma'(a_i)-\gamma'(t))\,d\mathcal{L}^1(t). \end{align*} Taking norms and using the triangle inequality for integrals, \begin{align*} \left|\gamma'(a_i)(b_i-a_i)-(\gamma(b_i)-\gamma(a_i))\right|\leq \varepsilon(b_i-a_i). \end{align*} Thus \begin{align*} |\gamma'(a_i)|(b_i-a_i)\leq |\gamma(b_i)-\gamma(a_i)|+\varepsilon(b_i-a_i). \end{align*} Combining the two estimates on $[a_i,b_i]$, \begin{align*} \int_{a_i}^{b_i}|\gamma'(t)|\,d\mathcal{L}^1(t)\leq |\gamma(b_i)-\gamma(a_i)|+2\varepsilon(b_i-a_i). \end{align*} Summing over $i=1,\dots,m$ gives \begin{align*} \int_0^1|\gamma'(t)|\,d\mathcal{L}^1(t)\leq S(P,\gamma)+2\varepsilon. \end{align*} Since $S(P,\gamma)\leq L(\gamma)$, \begin{align*} \int_0^1|\gamma'(t)|\,d\mathcal{L}^1(t)\leq L(\gamma)+2\varepsilon. \end{align*} Letting $\varepsilon\downarrow0$ yields \begin{align*} \int_0^1|\gamma'(t)|\,d\mathcal{L}^1(t)\leq L(\gamma). \end{align*}[/step]

[guided]We now prove the reverse inequality. The idea is that when the partition is fine, the derivative $\gamma'$ changes very little on each subinterval, so the curve over that subinterval behaves almost like the straight segment with velocity $\gamma'(a_i)$. Let $\varepsilon>0$. Since $\gamma':[0,1]\to\mathbb{R}^n$ is continuous and $[0,1]$ is compact, the standard [uniform continuity theorem](/theorems/183) gives a number $\delta>0$ such that \begin{align*} |s-t|<\delta\implies |\gamma'(s)-\gamma'(t)|<\varepsilon \end{align*} for all $s,t\in[0,1]$. Choose a partition \begin{align*} P=\{0=t_0<t_1<\cdots<t_m=1\} \end{align*} whose mesh is less than $\delta$, meaning that \begin{align*} t_i-t_{i-1}<\delta \end{align*} for every $i\in\{1,\dots,m\}$. Fix one subinterval and write $a_i=t_{i-1}$ and $b_i=t_i$. For every $t\in[a_i,b_i]$, we have $|t-a_i|<\delta$, so uniform continuity gives \begin{align*} |\gamma'(t)-\gamma'(a_i)|<\varepsilon. \end{align*} The reverse triangle inequality then implies \begin{align*} |\gamma'(t)|\leq |\gamma'(a_i)|+\varepsilon. \end{align*} Integrating this pointwise estimate over $[a_i,b_i]$ with respect to $\mathcal{L}^1$ gives \begin{align*} \int_{a_i}^{b_i}|\gamma'(t)|\,d\mathcal{L}^1(t)\leq |\gamma'(a_i)|(b_i-a_i)+\varepsilon(b_i-a_i). \end{align*} It remains to compare the quantity $|\gamma'(a_i)|(b_i-a_i)$ with the actual chord length $|\gamma(b_i)-\gamma(a_i)|$. The fundamental theorem of calculus gives \begin{align*} \gamma(b_i)-\gamma(a_i)=\int_{a_i}^{b_i}\gamma'(t)\,d\mathcal{L}^1(t). \end{align*} The constant velocity approximation over the same interval is the vector $\gamma'(a_i)(b_i-a_i)$, which is the integral of the constant map $t\mapsto\gamma'(a_i)$. Subtracting the actual displacement from this constant-velocity displacement, \begin{align*} \gamma'(a_i)(b_i-a_i)-(\gamma(b_i)-\gamma(a_i))=\int_{a_i}^{b_i}(\gamma'(a_i)-\gamma'(t))\,d\mathcal{L}^1(t). \end{align*} Taking Euclidean norms and applying the triangle inequality for coordinatewise integrals, \begin{align*} \left|\gamma'(a_i)(b_i-a_i)-(\gamma(b_i)-\gamma(a_i))\right|\leq \int_{a_i}^{b_i}|\gamma'(a_i)-\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*} The integrand is bounded by $\varepsilon$ on $[a_i,b_i]$, so \begin{align*} \left|\gamma'(a_i)(b_i-a_i)-(\gamma(b_i)-\gamma(a_i))\right|\leq \varepsilon(b_i-a_i). \end{align*} Using the triangle inequality in the form $|u|\leq |w|+|u-w|$ with $u=\gamma'(a_i)(b_i-a_i)$ and $w=\gamma(b_i)-\gamma(a_i)$, we obtain \begin{align*} |\gamma'(a_i)|(b_i-a_i)\leq |\gamma(b_i)-\gamma(a_i)|+\varepsilon(b_i-a_i). \end{align*} Substituting this into the earlier integral estimate gives \begin{align*} \int_{a_i}^{b_i}|\gamma'(t)|\,d\mathcal{L}^1(t)\leq |\gamma(b_i)-\gamma(a_i)|+2\varepsilon(b_i-a_i). \end{align*} Now sum over all subintervals. The integral over $[0,1]$ is the sum of the integrals over the partition intervals, and the sum of the interval lengths is $1$. Hence \begin{align*} \int_0^1|\gamma'(t)|\,d\mathcal{L}^1(t)\leq \sum_{i=1}^m|\gamma(t_i)-\gamma(t_{i-1})|+2\varepsilon. \end{align*} The sum on the right is exactly $S(P,\gamma)$, so \begin{align*} \int_0^1|\gamma'(t)|\,d\mathcal{L}^1(t)\leq S(P,\gamma)+2\varepsilon. \end{align*} Since $L(\gamma)$ is the supremum of all partition sums, $S(P,\gamma)\leq L(\gamma)$. Therefore \begin{align*} \int_0^1|\gamma'(t)|\,d\mathcal{L}^1(t)\leq L(\gamma)+2\varepsilon. \end{align*} Because $\varepsilon>0$ was arbitrary, letting $\varepsilon\downarrow0$ gives \begin{align*} \int_0^1|\gamma'(t)|\,d\mathcal{L}^1(t)\leq L(\gamma). \end{align*}[/guided]

[step:Conclude the equality and rectifiability] The two inequalities proved above give \begin{align*} L(\gamma)=\int_0^1|\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*} Since $t\mapsto|\gamma'(t)|$ is continuous on the compact interval $[0,1]$, it is bounded. Thus there exists $M\geq0$ such that $|\gamma'(t)|\leq M$ for all $t\in[0,1]$, and \begin{align*} \int_0^1|\gamma'(t)|\,d\mathcal{L}^1(t)\leq M<\infty. \end{align*} Therefore $L(\gamma)<\infty$. By definition, $\gamma$ is rectifiable. [/step]