Why is it that in gauge theories the assumption "all fields decay sufficiently rapidly at infinity" not justified anymore? I read that in gauge theories the assumption that "all fields decay sufficiently rapidly at infinity" is not justified anymore and therefore, one needs to consider boundary terms that ultimately lead to asymptotic symmetries.
But why is the assumption not possible in gauge theories?
 A: One idea would be that even at infinity your gauge symmetry still has to hold and so for example even if you set out with a gauge field $A_\mu \rightarrow 0$ at infinity through a gauge transformation the fields $A_\mu \rightarrow i\Omega \partial_\mu \Omega^{-1}$ has to be included in your path integral as well which might is non vanishing on the boundary.
So normally you can just set the boundary conditions on your path integral such that no such fields are included but in gauge theories, this is no longer possible as for a given equivalence classes of fields some representatives might not necessarily fulfill this.
Edit: Note why we normally impose this condition is such that we have configurations of finite action. However if we have a purely gauge configuration this will not contribute to the action of the gauge field.
A: Usually, one uses some argument related to finiteness energy/angular momentum/charges, etc. to show that $A \to 0$ at infinity. However, all of these arguments remain unchanged if $A \to pure~gauge$ so we could in principle allow such configurations.
Now, these configurations do actually have an effect on the path integral due to boundary effects. For instance, the gauge field couples to a conserved charge current through the term
$$
S \ni \int A_\mu J^\mu
$$
If $A$ is pure gauge
$$
S \ni \int \partial_\mu \epsilon J^\mu = \int \partial_\mu ( \epsilon J^\mu )
$$
Now, usually we drop such terms because we assume things fall-off sufficiently fast near the boundary. However, we have to be careful here. The current falls of as $r^{-2}$, but then the integration measure grows as $r^2$. Thus, if $\epsilon$ is finite at infinity (i.e. if the gauge field does NOT vanish at infinity), then such a boundary term is non-zero and leads to observable differences in the theory. It follows that including these modes in your phase space leads to interesting new physics.
One question you might ask whether including these modes is necessary, i.e. can we not consistently discuss all of physics without ever introducing these large gauge modes? The answer to this is -- as we have learnt through Strominger's work -- NO! These large gauge modes are canonically conjugate to soft photons. Consequently, if you have soft-photons in your phase space (which we always do in traditional QFT!), then we must also keep the large gauge modes. In other words, soft photons source the large gauge dof so we cannot consistently set one to zero without also doing the same to the other.
