Why group elements associated with gauge transformations of finite action field configurations in QCD don't depend in $r$?

Question

I am reading the chapter on instantons in Coleman's Aspects of Symmetry. I am puzzled by an argument i don't quite follow. In section 3.2, Coleman considers configurations of the gauge field with finite action. Using some arguments that (I think) are not relevant in what follows he concludes that for the action to be finite $F_{\mu\nu}$ must fall off at infinity like $O(1/r^3)$. He then says that this doesn't imply that $A_{\mu}$ falls off like $O(1/r^2)$ but rather that in general it can be a gauge transform of a configuration that does fall like $O(1/r^2)$. So far so good.

He then explicitly writes the gauge transformed field

$$A_{\mu}\to{}gO(1/r^2)g^{-1}+g\partial_{\mu}g^{-1}$$

$$A_{\mu}\to{}g\partial_{\mu}g^{-1}+O(1/r^2)$$

After writing this equation he says that $g$ denotes an element of the gauge group that does not depend on $r$. I don't see why this is legit. I mean, in principle we should be able to consider a general $g$, shouldn't we?