The question in the title seems a natural one to ask, and I suspect that it is already considered, even completely solved, somewhere in the literature. Although I prefer some explanation of the idea(s), relevant references are also appreciated.



Is every Riemannian metric conformally equivalent to one that is geodesically complete?
This is not hard to prove. The idea of a proof is as follows. Take any metric $g$ on a manifold $M$. Define $d:M\to \mathbb{R}\_+$ by saying that $d(x)$ is the infimum of all lengths of curves $\gamma:[0,b)\to M$ with $\gamma(0)=x$, such that $\gamma$ is proper (In particular: For every compact subset K\subset M, there is a $t\in [0,b)$ with $\gamma(t)\not\in K$). Choose a smooth function $f:M\to (0,1]$ with $0<f<d$. Then $\bar g:= f^{-2}g$ is complete.


There is much more possible, namely bounded geometry. This was published in  

Müller, Olaf; Nardmann, Marc;
Every conformal class contains a metric of bounded geometry.
Math. Ann. 363 (2015), no. 1-2, 143–174.


