Since conservation of charge seems to be a well known concept, I am hoping that I am missing something and that the conclusion is incorrect. However, I have been unable to disprove this. Let me outline the situation as follows.
Overview
Consider the following Lagrangian for scalar QED and field $\varphi$: \begin{align} \mathcal{L}=\overline{D_\mu\varphi}D^\mu\varphi-U(|\varphi|^2)-\frac{1}{4}F_{\mu\nu}F^{\mu\nu},\tag{1} \end{align} where $$D_\mu=\partial_\mu-iA_\mu\varphi.\tag{2}$$
One set of Euler-Lagrange equations is: \begin{align} \frac{\delta\mathcal{L}}{\delta A_\mu}=i(\overline\varphi\partial_\mu\varphi-\varphi\partial_\mu\overline\varphi)+2A^\mu|\varphi|^2+\partial_\nu F^{\mu\nu}=0.\tag{3} \end{align}
If we consider the global gauge symmetry transformation $\varphi\to\varphi+i\epsilon\varphi$, then Noether's first theorem gives the following conservation law (for charge, see e.g. [1]) on solutions of $\delta\mathcal{S}=0$: \begin{align} \partial_\mu j^\mu:=\partial_\mu\left[i(\overline\varphi\partial_\mu\varphi-\varphi\partial_\mu\overline\varphi)+2A^\mu|\varphi|^2\right]=0.\tag{4} \end{align}
Question: does the integral of the charge density $j^0$ vanish identically? If so, does this mean that nothing is conserved?
My attempt at an answer: Yes. To see this, we integrate the continuity equation over $\mathbb{R}^3$ and apply Gauss's Theorem: \begin{align} \partial_0Q:=\partial_0\int_{\mathbb{R}^3}j^0 d^3 x=-\sum_{\mu=1}^3\int_{R^2} j^\mu \bigr\rvert_{|x_\mu|\to\infty}d^2x.\tag{5} \end{align} Here, we say that $Q=\int j^0$ is the total charge.
We assume that the fields $\varphi$ and $F$ vanish as $\max_{\mu\ge 1}(x_\mu)\to\infty$. This is a natural boundary condition for physical observables. Then the fluxes $j^\mu$ clearly vanish in this limit. Thus, our conservation law becomes: \begin{align} \partial_0\int_{\mathbb{R}^3}j^0 d^3 x=0.\tag{6} \end{align} But $$j^0=\delta\mathcal{L}/\delta A_0-\partial_\nu F^{0\nu}=-\partial_\nu F^{0\nu}\tag{7}$$ on solutions, so we can equivalently rewrite the charge integral using Gauss's Theorem as follows: \begin{align} Q=-\sum_{\nu\ge 1}\int_{\mathbb{R}^2}F^{0\nu}\bigr|_{|x_\nu|\to\infty} d^3 x.\tag{8} \end{align} Since the fields vanish at infinity, we obtain: \begin{align} Q\equiv 0.\tag{9} \end{align}
In other words, if correct, then this shows that the charge for this system is identically zero, and that nothing is conserved (it seems incorrect to call $d0/dt\equiv 0$ a conservation law rather than a tautology/identity).
Related Links
There are a few other questions users asked that have some relation to this. However, I do not believe the actual charge integral's conservation was anywhere addressed. One shows that the charge density can be rewritten as a total divergence, from which the argument follows, of course. This article includes the fact that there is no conservation law for the local gauge symmetry, but does not clarify this for the global symmetry case (which should be a special case of the local results). Another gives an overview of charge conservation.
Edit:
Another question also discusses charge conservation, but it seems to be more concerned with the interpretation of the charge density. I am instead asking if the density is conserved at all.
