Length formula for Lipschitz curve, i.e. ${\rm Length}\ \gamma = \int_a^b|\gamma '(t)| dt $ I want to prove the following problem. But I think I can only
complete the half : We need to prove the part ${\rm Length}\ \gamma
\geq \int_a^b|\gamma '(t)| dt $
Problem : If $\gamma :[a,b]\rightarrow \mathbb{R}^3$ is
$L$-Lipschitz map with $\gamma(0)=(0,0,0)$, then we define ${\rm Length}\ \gamma =\sup_{P} \bigg\{\sum_{i=1}^N |\gamma(t_i)-\gamma(t_{i-1} )| \bigg\}$ over all partition $P=\{t_i\}_{i=0}^N$. Then prove that $$ {\rm
Length}\ \gamma =\int_a^b|\gamma '(t)| dt
$$
(This is an exercise in the book, What is differential geometry : curves and surfaces - Petrunin and Barrera)
Proof :
Step 1 : If $\gamma$ is $L$-Lipschitz, then each
component functions $x,\ y,\ z$ of $\gamma$ is $L$-Lispchitz :
\begin{align*}L|s-t|&\geq |\gamma (s)-\gamma (t)| \\&= \sqrt{ (x(s)-x(t))^2 +
(y(s)-y(t))^2 + (z(s)-z(t))^2} \\& \geq | x(s)-x(t) |
\end{align*}
Step 2 : By Rademacher Theorem, we have
measurable functions $ x',\ y',\ z'$ with a bound $M$.
By Lusin Theorem on $[a,b]$, there are continuous functions
$X_\epsilon, \ Y_\epsilon,\ Z_\epsilon$ with a bound $M$ s.t. $x',\
X_\epsilon$ coincides outside some set of measure $< T $ and so does
$Y_\epsilon,\
Z_\epsilon$
Step 3 : We define $\alpha = (f,g,h)(t)$ where $$
f(t)=\int_a^t X_\epsilon (s)ds,\ g(t)=\int_a^t Y_\epsilon (s)ds,\
 h(t)=\int_a^t Z_\epsilon (s)ds $$
Hence $\alpha$ is in $C^1$-class i.e., continuous first derivatives, so that we have ${\rm Length}\ \alpha = \int_a^b |\alpha'(t)|\ dt
 $ in the same book (I think I have already proved by using fundamental theorem of calculus, but I think it is still messy)
Furthermore, by Lusin Theorem, $$ \bigg|
 \int_a^b |\alpha'(t)| dt -\int_a^b|\gamma'(t)|dt\bigg|<2\sqrt{3}M T $$
By Rademacher Theorem, $$
|x(t)-f(t)| =\bigg|\int_a^t (x'(s)- X_\epsilon (s)) ds \bigg|<2MT $$
Hence $$|\alpha (t)-\gamma(t)| \leq 2\sqrt{3} MT
$$
$$\bigg||\gamma (t_i)-\gamma (t_{i+1}) | - |\alpha (t_i)-
\alpha (t_{i+1} )| \bigg|  < 4\sqrt{3} MT $$
Assume that given $\epsilon>0$, there is a partition $\{
t_i\}_{i=0}^N$ for $\gamma$ s.t.
$$\bigg|{\rm Length}\ \gamma - \sum_{i=1}^N\ |\gamma
(t_i)-\gamma (t_{i+1}) | \bigg|<\epsilon$$
If $4\sqrt{3}MNT<\epsilon$, then $$ \bigg|{\rm Length}\ \gamma
-\sum_i\ |\alpha (t_i)- \alpha (t_{i+1} )|\bigg|  < 2\epsilon  $$
so that \begin{align*}{\rm Length}\ \gamma &\leq\sum_{i=1}^N \
\bigg|\alpha (t_i)- \alpha (t_{i-1} )\bigg| + 2\epsilon
\\&\leq \int_a^b|\alpha'(t)|dt +2\epsilon \\&\leq
\int_a^b|\gamma'(t)|dt +3\epsilon
 \end{align*}
How could we prove the remaining ?
 A: I think the following will complete the proof. (If you have another one, I'll be welcome)
If $I_i$ is a disjoint open interval in $[a,b]$ s.t.
$\bigcup_i I_i$ with $\sum_i\ {\rm Length}\ I_i <2T<\epsilon$ covers
the set $K$ s.t. $x',\ X_\epsilon$ (respectively $y',\ Y_\epsilon$
and $z',\ Z_\epsilon$) coincides on the compliment $\widetilde{K}$.
Since $x'=X_\epsilon$ on $\widetilde{K}$, if $[a',b']$ is in
$\widetilde{\bigcup_i I_i}$, then $\gamma |[a',b']$ is a translation
of $\alpha |[a',b']$.
Hence since $|\int_a^b|\gamma'(t)|dt-\int_a^b|\alpha'(t)|dt|\leq 2\sqrt{3}MT$, then \begin{align*}\int_a^b|\gamma'(t)|dt -2\sqrt{3}MT
-\sqrt{3}M\epsilon &\leq {\rm Length}\ \alpha -\sqrt{3}M \epsilon
\\&\leq {\rm Length}\ \alpha|\widetilde{\bigcup_i I_i} \\&= {\rm Length}\
\gamma|\widetilde{\bigcup_i I_i} \\
&\leq {\rm Length}\ \gamma\end{align*} so that we have $\int_a^b|\gamma'(t)|dt \leq {\rm Length}\
\gamma$.
A: Let me refine your argument and also show the opposite direction.
Since $\gamma$ is Lipschitz continuous, it is absolutely continuous and hence the fundamental theorem of calculus holds for $\gamma$, that is, $\gamma'$ is defined a.e. on $[a, b]$, integrable, and satisfies
$$ \forall [c, d] \subseteq [a, b] \ : \quad \int_{c}^{d} \gamma'(t) \, \mathrm{d}t = \gamma(d) - \gamma(c). $$
(You may already know that this type of statement holds in 1D. The above 3D version is an easy corollary of the 1D counterpart.) So, for any partition $\Pi=\{a=t_0<t_1<\ldots<t_n=b\}$ of $[a, b]$,
$$ \sum_{i=1}^{n} \|\gamma(t_i) - \gamma(t_{i-1})\|
= \sum_{i=1}^{n} \left\| \int_{t_{i-1}}^{t_i} \gamma'(t) \, \mathrm{d}t \right\|
\leq \int_{a}^{b} \|\gamma'(t)\| \, \mathrm{d}t.
$$
Taking supremum over all possible choices of $\Pi$, we obtain
$$ \operatorname{Length}(\gamma) \leq \int_{a}^{b} \|\gamma'(t)\| \, \mathrm{d}t. \tag{1} $$
To show the opposite inequality, let $\alpha$ and $\beta$ be any two absolutely continuous $\mathbb{R}^3$-valued functions on $[a, b]$. Then by arguing similarly as above, we find that (see the proof below)
$$ \left| \operatorname{Length}(\alpha) - \operatorname{Length}(\beta) \right| \leq \int_{a}^{b} \| \alpha'(t) - \beta'(t) \| \, \mathrm{d}t. \tag{2} $$
Moreover, if $\alpha$ is $C^1$, then we have
$$ \operatorname{Length}(\alpha) = \int_{a}^{b} \| \alpha'(t) \| \, \mathrm{d}t. \tag{3} $$
Finally, choose a sequence of $C^1$ curves $\gamma_n$ such that $\int_{a}^{b} \| \gamma_n'(t) - \gamma'(t) \| \, \mathrm{d}t \to 0$. Then by the triangle inequality,
\begin{align*}
&\left| \operatorname{Length}(\gamma) - \int_{a}^{b} \|\gamma'(t)\| \, \mathrm{d}t \right| \\
&\qquad\leq \left| \operatorname{Length}(\gamma) - \operatorname{Length}(\gamma_n) \right|
+ \smash[t]{\overbrace{\left| \operatorname{Length}(\gamma_n) - \int_{a}^{b} \|\gamma_n'(t)\| \, \mathrm{d}t \right|}^{=0 \text{ by (3)}}} \\
&\qquad\qquad + \left| \int_{a}^{b} (\|\gamma'(t)\| - \|\gamma_n'(t)\|) \, \mathrm{d}t \right|\\
&\qquad\leq 2\int_{a}^{b} \| \gamma_n'(t) - \gamma'(t) \| \, \mathrm{d}t
\to 0,
\end{align*}
where the last inequality follows from $\text{(2)}$. Therefore the desired claim follows.
(In summary, the two functionals $\alpha' \mapsto \operatorname{Length}(\alpha)$ and $\alpha' \mapsto \int_{a}^{b} \|\alpha'(t)\| \, \mathrm{d}t$ are continuous with respect to the $L^1$-norm of $\alpha'$ and coincide on the dense subset of their domains. So they must agree.)

1) To show the inequality $(2)$, note that for any curves $\alpha$ and $\beta$ defined on the interval $[a, b]$ and for any partition $\Pi = \{t_i\}_{i=1}^{n}$ of $[a, b]$,
\begin{align*}
&\sum_{i=1}^{n} \| (\alpha(t_i) + \beta(t_i)) - (\alpha(t_{i-1}) + \beta(t_{i-1}) \| \\
&\quad \leq \sum_{i=1}^{n} \| \alpha(t_i) - \alpha(t_{i-1} \| + \sum_{i=1}^{n} \| \beta(t_i) - \beta(t_{i-1} \| \\
&\quad \leq \operatorname{Length}(\alpha) + \operatorname{Length}(\beta),
\end{align*}
and taking supremum over all possible $\Pi$, we get
$$ \operatorname{Length}(\alpha + \beta) \leq \operatorname{Length}(\alpha) + \operatorname{Length}(\beta) $$
Now assume that $\alpha$ and $\beta$ are absolutely continuous. Then the above inequality gives
\begin{align*}
\operatorname{Length}(\alpha) &\leq \operatorname{Length}(\alpha - \beta) + \operatorname{Length}(\beta), \\
\operatorname{Length}(\beta) &\leq \operatorname{Length}(\beta - \alpha) + \operatorname{Length}(\alpha).
\end{align*}
Since all the indicated lengths are finite, we can rearrange the above inequality to obtain the reverse triangle inequality:
$$ \left| \operatorname{Length}(\alpha) - \operatorname{Length}(\beta) \right|
\leq \operatorname{Length}(\alpha - \beta)
\leq \int_{a}^{b} \| \alpha'(t) - \beta'(t) \| \, \mathrm{d}t, $$
where the last inequality has already been proved in the above answer.
