A riemannian distance with no cross component finite value

Given: The two disjoint Euclidean lines with their disjoint-union smooth structure.

Extended riemannian distance on a disconnected manifold: The extended Riemannian distance on arbitrary $M$ is the componentwise Riemannian distance when two points are in the same component, and $+\mathrm{\infty }$ otherwise. Within each component use thm-riemannian-distance-is-a-metric. Components are open, since small coordinate balls are connected. A continuous curve cannot meet two components because its connected interval image is connected, so the cross-component curve family is empty, with $\inf \mathrm{\varnothing }=+\mathrm{\infty }$. This is an extended metric: if two endpoints are in different components, any third point is in a different component from at least one of them, so the triangle inequality has infinite right side. It is a finite metric precisely when there are no distinct components. Empty and singleton manifolds retain their unique distances.

Newton–Leibniz needs only continuity on $[a,b]$, differentiability on $(a,b)$, and a Riemann-integrable extension of the interior derivative: Let $a<b$. Suppose $G:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f:[a,b]\to \mathbb{R}$ is Riemann integrable and $f(x)=G^{\mathrm{\prime }}(x)(a<x<b),$ then ${\int }_a^{b}f=G(b)-G(a)\mathrm{.}$ No derivative of $G$ at either endpoint is assumed, and the two endpoint values assigned to the integrable extension $f$ do not enter the conclusion.