[proofplan] We convert the metric problem into a one-dimensional variation problem. The total variation function $m$ of the curve is finite and absolutely continuous because the curve has an $L^1$ control. The one-dimensional metric differentiation theorem for curves then identifies $m'(t)$ with the metric derivative for almost every $t$, which gives the length bound by integrating $m'$. Finally, any other interval control $g$ bounds the metric difference quotients, and the Lebesgue differentiation theorem turns those averaged bounds into the pointwise minimality inequality $|\gamma'|\le g$. [/proofplan] [step:Build the total variation function controlled by an admissible integrable speed] Let $a\in L^1((0,1),\mathcal B((0,1)),\mathcal L^1)$ be a nonnegative admissible control for $\gamma$, so that \begin{align*} d(\gamma(s),\gamma(t))\le \int_s^t a(r)\,d\mathcal L^1(r) \end{align*} for every $0\le s\le t\le 1$. Define the total variation function $m:[0,1]\to[0,\infty)$ by \begin{align*} m(t):=\sup\left\{\sum_{i=1}^N d(\gamma(t_{i-1}),\gamma(t_i)):\ 0=t_0<t_1<\cdots<t_N=t,\ N\in\mathbb N\right\}. \end{align*} For every partition $0=t_0<t_1<\cdots<t_N=t$, summing the admissible control inequality gives \begin{align*} \sum_{i=1}^N d(\gamma(t_{i-1}),\gamma(t_i))\le \sum_{i=1}^N\int_{t_{i-1}}^{t_i}a(r)\,d\mathcal L^1(r)=\int_0^t a(r)\,d\mathcal L^1(r). \end{align*} Taking the supremum over all partitions gives \begin{align*} m(t)\le \int_0^t a(r)\,d\mathcal L^1(r), \end{align*} so $m(t)<\infty$ for every $t\in[0,1]$. Fix $0\le s\le t\le1$. To compare $m(t)$ and $m(s)$, let $P$ be any partition $0=t_0<t_1<\cdots<t_N=t$ of $[0,t]$. Refine $P$ by inserting $s$ if necessary. The part of the refined partition lying in $[0,s]$ contributes at most $m(s)$ by the definition of $m(s)$, while summing the admissible control inequality over the part lying in $[s,t]$ contributes at most $\int_s^t a(r)\,d\mathcal L^1(r)$. Hence every such partition satisfies \begin{align*} \sum_{i=1}^N d(\gamma(t_{i-1}),\gamma(t_i))\le m(s)+\int_s^t a(r)\,d\mathcal L^1(r). \end{align*} Taking the supremum over partitions of $[0,t]$ gives \begin{align*} m(t)-m(s)\le \int_s^t a(r)\,d\mathcal L^1(r). \end{align*} Hence $m$ is absolutely continuous on $[0,1]$. By the real-variable theorem that absolutely continuous functions are differentiable almost everywhere and recover their increments by integrating their derivative, $m'(t)$ exists for $\mathcal L^1$-a.e. $t\in(0,1)$, $m'\in L^1((0,1),\mathcal B((0,1)),\mathcal L^1)$, and \begin{align*} m(t)-m(s)=\int_s^t m'(r)\,d\mathcal L^1(r) \end{align*} for every $0\le s\le t\le1$. This uses the classical real-variable theorem that absolutely continuous real-valued functions are differentiable almost everywhere and equal the integral of their derivative. [/step] [step:Compare every metric increment with the variation increment] Fix $0\le s\le t\le1$. The partition $s<t$ of the interval $[s,t]$ contributes the single increment $d(\gamma(s),\gamma(t))$ to the variation on $[s,t]$. Therefore \begin{align*} d(\gamma(s),\gamma(t))\le m(t)-m(s). \end{align*} Using the absolute-continuity identity for $m$, we obtain \begin{align*} d(\gamma(s),\gamma(t))\le \int_s^t m'(r)\,d\mathcal L^1(r). \end{align*} Thus $m'$ is an $L^1$ upper control for the curve. [/step] [step:Identify the metric derivative with the derivative of the variation function] We use the one-dimensional metric differentiation theorem for absolutely continuous curves in metric spaces: if a metric-space-valued curve has a finite absolutely continuous variation function $m$, then the metric derivative \begin{align*} \lim_{h\to0,\ t+h\in[0,1]}\frac{d(\gamma(t+h),\gamma(t))}{|h|} \end{align*} exists for $\mathcal L^1$-a.e. $t\in(0,1)$ and equals $m'(t)$ at those points. The hypotheses are satisfied here because the previous step proved that $m(t)<\infty$ for every $t\in[0,1]$ and that $m$ is absolutely continuous on $[0,1]$. This theorem applies to curves in arbitrary metric spaces, so no completeness hypothesis is needed. Applying this theorem to the curve $\gamma$ and the absolutely continuous variation function $m$ constructed above gives \begin{align*} |\gamma'|(t)=m'(t) \end{align*} for $\mathcal L^1$-a.e. $t\in(0,1)$. Since $m'\in L^1((0,1),\mathcal B((0,1)),\mathcal L^1)$, it follows that $|\gamma'|\in L^1((0,1),\mathcal B((0,1)),\mathcal L^1)$. Combining this identity with the estimate from the previous step yields \begin{align*} d(\gamma(s),\gamma(t))\le \int_s^t |\gamma'|(r)\,d\mathcal L^1(r) \end{align*} for every $0\le s\le t\le1$. [/step] [step:Differentiate any admissible interval control to prove minimality] Let $g\in L^1((0,1),\mathcal B((0,1)),\mathcal L^1)$ be a representative satisfying \begin{align*} d(\gamma(s),\gamma(t))\le \int_s^t g(r)\,d\mathcal L^1(r) \end{align*} for every $0\le s\le t\le1$. Let $E\subset(0,1)$ be the set of points $t$ at which the metric derivative $|\gamma'|(t)$ exists and $t$ is a Lebesgue point of $g$. By the previous step and the Lebesgue differentiation theorem for the integrable function $g$, $\mathcal L^1((0,1)\setminus E)=0$. At every such Lebesgue point, the one-sided averages over intervals $[t,t+h]$ also converge to $g(t)$ as $h\downarrow0$. Fix $t\in E$. For $h>0$ with $t+h\le1$, the admissible-control inequality gives \begin{align*} \frac{d(\gamma(t+h),\gamma(t))}{h}\le \frac{1}{h}\int_t^{t+h}g(r)\,d\mathcal L^1(r). \end{align*} Taking $\limsup$ as $h\downarrow0$ and using that $t$ is a Lebesgue point of $g$, we get \begin{align*} |\gamma'|(t)\le g(t). \end{align*} This proves the desired inequality for every $t\in E$, hence for $\mathcal L^1$-a.e. $t\in(0,1)$. [guided] We now prove that no admissible control can be smaller than the metric derivative on a set of positive measure. Let \begin{align*} g\in L^1((0,1),\mathcal B((0,1)),\mathcal L^1) \end{align*} be a representative satisfying \begin{align*} d(\gamma(s),\gamma(t))\le \int_s^t g(r)\,d\mathcal L^1(r) \end{align*} for every $0\le s\le t\le1$. The pointwise argument can only be made at points where two independent differentiability facts hold. First, the metric derivative $|\gamma'|(t)$ must exist. This holds for $\mathcal L^1$-a.e. $t$ by the metric differentiation step above. Second, the averages of $g$ over shrinking intervals must converge to $g(t)$. This holds for $\mathcal L^1$-a.e. $t$ by the Lebesgue differentiation theorem, applied to the integrable function $g$. Therefore the set \begin{align*} E:=\{t\in(0,1): |\gamma'|(t)\text{ exists and }t\text{ is a Lebesgue point of }g\} \end{align*} has full $\mathcal L^1$-measure in $(0,1)$. Fix $t\in E$. We use the interval inequality with $s=t$ and $t$ replaced by $t+h$, where $h>0$ and $t+h\le1$. This gives \begin{align*} d(\gamma(t),\gamma(t+h))\le \int_t^{t+h}g(r)\,d\mathcal L^1(r). \end{align*} Dividing by $h$ gives \begin{align*} \frac{d(\gamma(t+h),\gamma(t))}{h}\le \frac{1}{h}\int_t^{t+h}g(r)\,d\mathcal L^1(r). \end{align*} The left-hand side converges to $|\gamma'|(t)$ as $h\downarrow0$ because the metric derivative exists at $t$. The right-hand side converges to $g(t)$ because the Lebesgue differentiation theorem gives convergence of these one-sided averages at the Lebesgue point $t$. Passing to the limit gives \begin{align*} |\gamma'|(t)\le g(t). \end{align*} Since $E$ has full measure, the inequality holds for $\mathcal L^1$-a.e. $t\in(0,1)$. Notice that this also forces $g(t)\ge0$ for almost every such point, because $|\gamma'|(t)\ge0$ by definition. [/guided] [/step] [step:Conclude the length estimate and the minimality property] The preceding steps prove that the metric derivative exists for $\mathcal L^1$-a.e. $t\in(0,1)$, belongs to $L^1((0,1),\mathcal B((0,1)),\mathcal L^1)$, and satisfies \begin{align*} d(\gamma(s),\gamma(t))\le \int_s^t |\gamma'|(r)\,d\mathcal L^1(r) \end{align*} for every $0\le s\le t\le1$. They also prove that every integrable representative $g$ satisfying the same interval-control inequality obeys \begin{align*} |\gamma'|(t)\le g(t) \end{align*} for $\mathcal L^1$-a.e. $t\in(0,1)$. This is exactly the asserted length control and minimality of the metric derivative. [/step]