Let $C$ be a compact subset of a 2 dimensional manifold $M$. $g_1$ and $g_2$ are two Riemannian metrices on $M$ such that for all $p\in M\setminus C$ and for all $v\in T_pM$, we have $g_1(v,v)\leq g_2(v,v)$. Moreover $(M,g_1)$ is a complete Riemannian manifold.
Does it imply $(M,g_2)$ complete?
If it was given that $g_1\le g_2$ on $M$ then closed and bounded subset with respect to $g_2$, will be closed and bounded with respect to $g_1$, and therefore by Hopf-Rinow, it is compact. So we are done. But it is not given, we have $g_1\le g_2$ only on $M\setminus C$. I think I am missing a very trivial thing..
By the Hopf-Rinow theorem it suffices to show that every $g_2$-bounded set is $g_1$-bounded.
Take a set $A$ having diameter $D < \infty$ with respect to $g_2$ and let $x,y \in A$. Take a path $\gamma$ joining $x$ to $y$ and having $g_2$ length almost minimal, so less than $D+\epsilon$. If this path doesn't enter $C$ then its $g_1$-length is also less than $D+\epsilon$ and we are done. If it does enter $C$, then let $z_1$ be the first point where it enters, and $z_2$ the last point where it exits. By the previous argument, $d_1(x, z_1) < D+\epsilon$ and $d_1(z_2, y) < D+\epsilon$, and $d_1(z_1, z_2) < \operatorname{diam}_{g_1}(C)$ which is finite because $C$ is compact. Thus by the triangle inequality, the $g_1$-diameter of $A$ is at most $2(D+\epsilon)+\operatorname{diam}_{g_1}(C)$ which is finite.