In the Ashok Das QFT book] pg. 212-213 (pdf), the section on Noether's theorem, the author considered general infinitesimal transformations of the form $$x^\mu\rightarrow x'^\mu,$$ $$\phi(x)\rightarrow\phi'(x'),$$ $$\partial_\mu\phi(x)\rightarrow\partial'_\mu\phi'(x')\tag{6.3}$$ where $x^\mu$ are spacetime coordinates and $\phi(x)$ is a field.
For such transformations, the invariance of the action $S$ would imply $$\delta S=\int d^4x' \mathcal L(\phi'(x'),\partial'_\mu\phi'(x')) - \int d^4x\mathcal L(\phi(x),\partial_\mu\phi(x))=0,$$ or, $$\delta S= \int d^4x \mathcal L(\phi'(x),\partial'_\mu\phi'(x)) - \int d^4x\mathcal L(\phi(x),\partial_\mu\phi(x))=0,$$ $$\delta S= \int d^4x [\mathcal L(\phi'(x),\partial'_\mu\phi'(x)) - \mathcal L(\phi(x),\partial_\mu\phi(x))]=0.\tag{6.7}$$
The author then said that the last equality will hold if $$\mathcal L(\phi'(x),\partial'_\mu\phi'(x)) - \mathcal L(\phi(x),\partial_\mu\phi(x))=\partial_\mu K^\mu,\tag{6.8}$$ where $\partial_\mu K^\mu$ is a total divergence.
Why is that so? According to the Gauss theorem, for the integral $\int\partial_\mu K^\mu d^4x$ to vanish, $K^\mu$ needs to vanish at the surface boundary, which I assume is at infinity. How can we be sure that $K^\mu$ will vanish at infinity when we don't know what $K^\mu$ is?
