If the Derivative of a Curve is Continuous, the Curve Can Be Measured 📂Analysis

If the Derivative of a Curve is Continuous, the Curve Can Be Measured

Theorem¹

If $\gamma ^{\prime}$ is continuous on the interval $[a,b]$, then $\gamma$ forms a rectifiable curve, and the following equation holds:

$$ \Lambda (\gamma) = \int _{a} ^{b} \left| \gamma^{\prime}(t) \right| dt $$

Proof

Part 1.

Let $P=\left\{ a=x_{0},\dots,x_{n}=b \right\}$ be any partition of the interval $[a,b]$. If we state $a\le x_{i-1}<x_{i}\le b$, then the following is true:

$$ \begin{align*} \left| \gamma (x_{i})-\gamma (x_{i-1}) \right| &= \left| \int_{x_{i-1}}^{x_{i}}\gamma^{\prime} (t)dt \right| \\ &\le \int_{x_{i-1}}^{x_{i}} \left| \gamma^{\prime} (t) \right|dt \end{align*} $$

The first line is true due to The Fundamental Theorem of Calculus, Part 2, and the second line is true because the integral of the absolute value is greater than the absolute value of the integral. Therefore, we obtain the following:

$$ \Lambda (P,\gamma) \le \int _{a} ^{b} \left| \gamma^{\prime} (t) \right| dt $$

Since this is true for all subintervals of $[a,b]$, we then obtain:

$$ \Lambda (\gamma) \le \int _{a} ^{b} \left| \gamma^{\prime} (t) \right| dt $$

Part 2.

Let’s consider an arbitrary positive number $\varepsilon >0$. Since a continuous function in a compact metric space is uniformly continuous, $\gamma^{\prime}$ is uniformly continuous. Thus, there exists a positive number $\delta >0$ satisfying the following equation for $\varepsilon$:

$$ \left| s-t \right| < \delta \implies \left| \gamma^{\prime} (s) -\gamma^{\prime} (t) \right| < \varepsilon,\quad s,t\in[a,b] $$

Now, let’s choose a partition $P=\left\{ x_{0},\dots,x_{n} \right\}$ of the interval $[a,b]$ that satisfies $\Delta x_{i}=x_{i}-x_{i-1}<\delta$. Then, for $x_{i-1}\le t \le x_{i}$, the following is true:

$$ \begin{align*} &&\left| \gamma^{\prime} (t) \right| -\left| \gamma^{\prime} (x_{i}) \right| &\le \left| \gamma^{\prime} (t)-\gamma^{\prime}(x_{i}) \right| <\varepsilon \\ \implies && \left| \gamma^{\prime} (t) \right| &\le \left| \gamma^{\prime}(x_{i}) \right|+\varepsilon \end{align*} $$

Hence, integrating both sides gives us the following equation:

$$ \begin{align*} \int_{x_{i-1}}^{x_{i}}\left| \gamma^{\prime} (t) \right|dt &\le \int_{x_{i-1}}^{x_{i}}\left| \gamma^{\prime} (x_{i}) \right|dt +\int_{x_{i-1}}^{x_{i}} \varepsilon dt \\ &= \left| \gamma^{\prime}(x_{i}) \right|\Delta x_{i} + \varepsilon\Delta x_{i} \\ &= \left|\int_{x_{i-1}}^{x_{i}} \gamma^{\prime}(x_{i})dt \right|+\varepsilon\Delta x_{i} \\ &= \left|\int_{x_{i-1}}^{x_{i}}\big( \gamma^{\prime}(t) +\gamma^{\prime} (x_{i})-\gamma^{\prime}(t) \big)dt\right| +\varepsilon\Delta x_{i} \\ &\le \color{green}{\left|\int_{x_{i-1}}^{x_{i}} \gamma^{\prime}(t)dt\right|}+ \color{blue}{\left| \int_{x_{i-1}}^{x_{i}} \gamma^{\prime} (x_{i})-\gamma^{\prime}(t)dt\right|} +\varepsilon\Delta x_{i} \\ &\le \color{green}{\left| \gamma (x_{i})-\gamma (x_{i-1}) \right|} +\color{blue}{\int_{x_{i-1}}^{x_{i}}\varepsilon dt}+ \varepsilon\Delta x_{i} \\ &= \left| \gamma (x_{i})-\gamma (x_{i-1}) \right| + 2\varepsilon\Delta x_{i} \end{align*} $$

The second and third lines are true because $\gamma^{\prime}(x_{i})$ is constant. The fifth line is true due to the triangle inequality. The green part is true because of The Fundamental Theorem of Calculus, Part 2. The blue part is true due to the condition of uniform continuity. Since $\varepsilon$ is any positive number, the following is true:

$$ \int_{x_{i-1}}^{x_{i}}\left| \gamma^{\prime} (t) \right|dt \le \left| \gamma (x_{i})-\gamma (x_{i-1}) \right| $$

Summing the above inequality over all $\Delta x_{i}$ gives us the following equation:

$$ \int _{a} ^{b} \left| \gamma^{\prime} (t) \right| dt \le \sum \limits _{i=1} ^{n} \left|\gamma (x_{i})-\gamma (x_{i-1}) \right|=\Lambda (P,\gamma)\le \Lambda (\gamma) $$

Therefore, based on Part 1. and Part 2., the following holds: