Say that you place an asymptotically Euclidean metric on $\mathbb{R}^3,$ e.g. $\mathbb{R}^3$ is endowed Riemannian metric $g$ such that $\text{supp}(g^{ij}-\delta^{ij})\subseteq\{|x|\leq R\}$ for some large $R>0.$ Does it follow that $(\mathbb{R}^3,g)$ is geodesically complete? This seems to be applied in PDE literature quite often when working with the flow generated by the principle symbol of the Laplace-Beltrami operator, but I am unaware of a reference. It seems intuitively true: $\mathbb{R}^3=\{|x|\leq R\}\cup \{|x|>R\}.$ The first set in the union is compact, and the other is where our metric is just the standard metric.


If this is a simple question, please let me know in the comments and I will delete.



Completeness of asymptotically Euclidean manifolds
Yes, and it follows from [Hopf-Rinow](https://en.wikipedia.org/wiki/Hopf%E2%80%93Rinow_theorem) (let's assume $g$ is at least $C^2$ so that there's no ambiguity about the geodesic flow).


Since $g$ and $\delta$ differs only on a compact set, you have the the lengths defined by $g$ and $\delta$ are globally comparable, that is, there exists $\lambda, \Lambda > 0$ such that for every $x\in \mathbb{R}^3$ and $\xi\in T\_x\mathbb{R}^3$ you have


$$ \lambda \delta(\xi,\xi) \leq g(\xi,\xi) \leq \Lambda \delta(\xi,\xi) $$


and so the metric topologies are equivalent. Thus


A set $K$ is closed and bounded w.r.t. $g$


IFF


$K$ is closed and bounded w.r.t $\delta$


IFF


$K$ is compact.


And by Hopf-Rinow you have $g$ is complete.


