Riemannian length is invariant under orientation preserving piecewise c one reparametrization

For each of the finitely many breakpoints $t_j$ of $\gamma$, its fibre under $\phi$ is a closed interval or singleton by monotonicity and continuity. Refine $[c,d]$ at their endpoints and at the breakpoints of $\phi$. Each remaining piece either maps into one $C^1$ piece of $\gamma$ or is a constant fibre; thus the composition is piecewise $C^1$.

On a nonconstant piece $[u,v]$, the chain rule and $\phi \mathrm{\prime }\ge 0$ give $\mathrm{\mid }\frac{d}{dt}(\gamma \circ \phi ){\mathrm{\mid }}_g=\mathrm{\mid }\dot{\gamma }(\phi (t)){\mathrm{\mid }}_g\phi \mathrm{\prime }(t)$. Speed on the target piece is continuous and the derivative of $\phi$ is continuous up to one-sided endpoints, so the substitution theorem applies and gives length ${\int }_{\phi (u)}^{\phi (v)}\mathrm{\mid }\dot{\gamma }(s){\mathrm{\mid }}_{g}ds$. Constant fibres have zero speed and zero endpoint difference. Summing gives the full target integral because monotone surjectivity sends $c$ to $a$ and $d$ to $b$. Partition independence removes the refinements. Degenerate singleton intervals give zero on both sides.