quantum field theory - masslessness of Goldstone boson, Effective action, and functional-integral measure

I have difficulty in understanding the path-integral formalism of SSB, and that of Effective Action.

Let's say a complex scalar field theory has the global $U(1)$ SSB, $$L(\phi)=(\partial^\mu \phi)^2-m^2\phi^\dagger\phi-\frac{1}{4}(\phi^\dagger\phi)^2$$

The proof of the masslessness of Goldstone bosons associated with this theory usually uses the $U(1)$ symmetry of the Effective action $\Gamma[\phi]$, which usually is proven by assuming that in the generating functional $$\tag{1} Z[J]=\int [D\phi D\phi^\dagger]e^{i\int d^4x[L(\phi)+J^\dagger\phi+\phi^\dagger J]}$$ the measure $[D\phi D\phi^\dagger]$ is also invariant under $\phi \rightarrow e^{i\theta}\phi$, and thereby proving the symmetry of $Z[J]$ under $J \rightarrow e^{i\theta}J$.

But the problem is, if we assume $[D\phi D\phi^\dagger]$ is invariant, we could easily get $$\langle\Omega|\phi|\Omega\rangle=e^{i\theta}\langle\Omega|\phi|\Omega\rangle$$, where $\Omega$ is the vacuum, and thus $$\tag{2} \langle\Omega|\phi|\Omega\rangle=0$$,which obviously contradicts SSB.

Where did I get wrong, or how to correctly prove that the effective action $\Gamma[\phi]$ indeed has the $U(1)$ symmetry?

PS:
let me make my question clearer.
The fact is that we don't really need the generating functional $Z[J]$ to calculate the VEV, we can simply go back to the functional-integral version of the VEV:
$$\tag{3} \text{VEV}=\int [D\phi D\phi^\dagger]\space \phi\space e^{iS[\phi]}$$ without doting about $J$.
Now the measure $[D\phi D\phi^\dagger]$ (and its region/asymptotic behavior) appearing in $(3)$ is the same as that in $(1)$, so if we assume its invariance to prove the symmetry of $Z[J]$, we can as well use it to prove $(2)$. The proof (if I'm right) goes like this:

$(3)$ just calculates VEV as a expect value of $\phi$, since everything is in the integral, the variables $\phi$ are dummy, we can just rename $\phi$ as $\phi^\prime$ without affecting anything: $$\tag{4} \text{VEV}=\int [D\phi^\prime D\phi^{\prime\dagger}]\space \phi^\prime\space e^{iS[\phi^\prime]}$$ next is the real stuff:
we make a variable transformation
$$\tag{5} \phi^\prime=e^{i\theta}\phi$$ and get $$\tag{6} \text{VEV}=\int [D(e^{i\theta}\phi) D(e^{-i\theta}\phi^\dagger)]\space e^{i\theta}\phi\space e^{iS[e^{i\theta}\phi]}$$ now according to our assumption $$\tag{7}\int[D(e^{i\theta}\phi) D(e^{-i\theta}\phi^\dagger)]=\int[D\phi D\phi^\dagger]$$ $$\text{and}\space\space\space\space\space\space S[e^{i\theta}\phi]=S[\phi]$$ we have, from $(6)$, $$\tag{8} \text{VEV}=\int [D\phi D\phi^\dagger]\space e^{i\theta}\phi\space e^{iS[\phi]}=e^{i\theta}\int [D\phi D\phi^\dagger]\space \phi\space e^{iS[\phi]}=e^{i\theta}\text{VEV}$$ thus we see that $$\text{VEV}=0$$ as in $(2)$.
Looking back, $(7)$ holds only if the functional integral is over an symmetrical region/class of field configurations $\phi(x)$, for example, if the integral is over all field configurations that vanish at infinity, $\phi(\infty)=0$.

So my conclusion is that, since $(8)$ contradicts SSB, the functional integration region shouldn't be invariant under $\phi \rightarrow e^{i\theta}\phi$, and thus picks up a particular/favored "orientation" in field's configurations, and thereby yields an nonzero VEV.
But if this is the case, I don't understand how the effective action $\Gamma[\phi]$ has the delightful property that it has the same symmetry as the classical action $S[\phi]$.