Invariance of integration measure under shifts in field

["Don't panic!"]

I've been trying to teach myself the path integral formulation of quantum field theory and there's a point that's really bugging me: why is the integration measure $\mathcal{D}\phi(x)$ invariant under shifts in the field of the form $$\phi(x)\rightarrow\tilde{\phi}(x)=\phi(x)+\int d^{4}y\Delta(x-y)J(y),\qquad\mathcal{D}\phi(x)\rightarrow\mathcal{D}\tilde{\phi}(x)$$ (where $\Delta(x-y)$ is a Green's function corresponding to the differential operator $\Box + m^{2}$ and $J(x)$ is a source).
Is it simply because the shift term $\int d^{4}y\Delta(x-y)J(y)$ is independent of the field configuration $\phi(x)$ at each spacetime point, or is there more to it than that?

 

[vanhees71]

Yes, that's it! If you are in doubt about some operation in path integration, just think about it in the "lattice version", i.e., consider space and time to be a finite 4D grid of discrete points. Then you have just a usual multidimensional integral. Then it's indeed clear that the integration measure does not change by just shifting all integration variables by arbitrary constants.

 

["Don't panic!"]

Yes, that's it! If you are in doubt about some operation in path integration, just think about it in the "lattice version", i.e., consider space and time to be a finite 4D grid of discrete points. Then you have just a usual multidimensional integral. Then it's indeed clear that the integration measure does not change by just shifting all integration variables by arbitrary constants.

Ah OK ,great, I thought that was probably the case, but just wanted to make sure. Thanks for your help!