Renormalisations in flat spacetime QFT vs curved spacetime QFT In standard QFT (in flat spacetime) textbooks, we've learnt that an action functional, say, $S$, can be expressed in terms of bare quantities, i.e., $S=S_{B}$, or in terms of renormalised quantities and counterterms, i.e., $S=S_{ren}+S_{ct}$. In this case, we have, e.g., $m_{B}=m_{ren}+\delta_{m}$, where $m_{B,ren}$ is the bare/renormalised mass and $\delta_m$ is the coupling constant of the counterterm which is to be determined by computing some specific scattering amplitude. My first question is: Since the renormalised quantities are finite, and the counterterms are divergent, it seems to me that the action $S$ appears to be divergent? (Although the scattering amplitudes are finite.)  A subsequent question is perhaps better illustrated with the following example: In the case of gravity in the presence of, say, a quantum scalar field, we have the total action as $S^{tot}_{B}[g_{\mu\nu},\phi]=S_{B}[g_{\mu\nu}]+S_{B}^{(m)}[g_{\mu\nu},\phi]$. After performing path integral over the scalar field, one obtains $e^{i\Gamma[g_{\mu\nu}]}=\int \mathcal{D}\phi\cdot e^{iS_{B}^{(m)}[g_{\mu\nu},\phi]}$, and thus the total action for gravity in the presence of the scalar field should be: $S^{grav}[g_{\mu\nu}]=S_{B}[g_{\mu\nu}]+\Gamma[g_{\mu\nu}]$. Since $\Gamma[g_{\mu\nu}]$ includes divergent terms, it is often stated that these divergent terms can be cancelled by the bare coupling constants inside $S_{B}[g_{\mu\nu}]$ such that $S^{grav}[g_{\mu\nu}]$ is finite. This then determines the relation between the bare constants and renormalised constants. For example, $S_{B}[g_{\mu\nu}]+\Gamma[g_{\mu\nu}]$ includes terms like: $\frac{R}{16\pi G_B}+\frac{R}{192\pi^2\epsilon^2}$, where $R$ is the Ricci scalar and $\epsilon\rightarrow 0$. The renormalised gravitational constant is then defined as: $ \frac{R}{16\pi G_{ren}}\equiv \frac{R}{16\pi G_B}+\frac{R}{192\pi^2\epsilon^2}$. Here is my confusion: In the standard QFT case, we have the renormalised term $S_{ren}$ that combines with the divergent term $S_{ct}$, which, as raised in the first question, seems to lead to a divergent total action $S$; however, in the gravity case, it is the bare term $S_{B}[g_{\mu\nu}]$ that combines with a divergent term $\Gamma[g_{\mu\nu}]$, which leads to a finite total action $S^{grav}[g_{\mu\nu}]$. How can I appreciate the differences/resemblances of the treatments in the two cases? In addition, in the gravity case, the relation between the bare and renormalised constants can be determined without computing any specific scattering amplitude, which seems quite odd to me.
 A: 
Since the renormalised quantities are finite, and the counterterms are divergent, it seems to me that the action S appears to be divergent? (Although the scattering amplitudes are finite.)

Yes that's right. Although I prefer to think of it as though there is always really some very large cutoff (or very small $\epsilon$ in dimensional regularization) with finite action and we are taking a limiting process.
For your main question, if $S_B[g_{\mu\nu}]$ is really treated as a bare action you are integrating over in the path integral it also needs to be regulated with a cutoff (or continuation in dimension), and the bare action will also diverge as you take the limit. Ignore any subtleties specific to quantum gravity here (I am not an expert in them and I don't think they are essential to your question), such a situation will happen whenever you have a quantum field theory with two interacting fields and you integrate out one of them.
This is true even if there were no $\phi$ field. What the extra divergences due to the $\phi$ field are telling you is that interaction between the $\phi$ field and the $g_{\mu\nu}$ field affects how the action is renormalized. i.e. you can draw divergent diagrams involving both fields interacting.
