Consider the path integral expression for the (unrenormalised) propagator in $\phi^4$ theory with a real scalar field $\phi$ on 4D Minkowski space and with a complex mass term:
$$ G[f,g] = \int\mathcal{D}\phi\,{\textstyle(\int f\phi)(\int g\phi)\exp(i\int[\frac12(\partial_\mu\phi)(\partial^\mu\phi) - \frac12(m^2 - i\epsilon)\phi^2 - \frac1{4!}\lambda\phi^4]}) $$
where $f$ and $g$ are smooth, square-integrable functions on Minkowski space. In the limit $\epsilon\to0$ this is just the propagator of $\phi^4$ theory which I know is UV divergent. My question is whether this expression is also divergent for finite $\epsilon>0$.
I have one argument which shows that $G[f,g]$ is finite and one which shows that it is infinite and I can't figure out which one is wrong:
Argument 1: I know that $\mathcal{D}\phi$ is just a sloppy physicist notation for an ill-defined functional integration measure, but if my understanding of Gaussian functional integration measures (which is somewhat limited and comes mostly from math-ph/0510087) is correct then
$$ d\mu_\epsilon(\phi)\equiv\mathcal{D}\phi{\textstyle\exp(-\frac12\epsilon\int\phi^2)} $$
can be regarded as a well-defined functional integration measure which satisfies
$$ \int d\mu_\epsilon(\phi) = 1 \quad,\quad \int d\mu_\epsilon(\phi){\textstyle(\int f\phi)(\int g\phi)} = \frac1\epsilon{\textstyle\int fg} $$
If this is true we may write
$$ G[f, g] = \int d\mu_\epsilon(\phi){\textstyle(\int f\phi)(\int g\phi)\exp(i\int[\frac12(\partial_\mu\phi)(\partial^\mu\phi) - \frac12m^2\phi^2 - \frac1{4!}\lambda\phi^4]}) $$
and since the expression in the exponent is now purely imaginary the exponential has magnitude 1 and we have
$$ |G[f,g]| \leq \left|\int d\mu_\epsilon(\phi){\textstyle(\int f\phi)(\int g\phi)}\right| = \frac1\epsilon|{\textstyle\int fg}| < \infty $$
Argument 2: The integrals we encounter when we calculate $G[f, g]$ in perturbation theory have the same UV divergences irrespective of whether $\epsilon$ is zero or not. For example, the one-loop 'tadpole' integral is
$$ \int d^4l \frac1{l^2 - m^2 + i\epsilon} = -i\int_0^\infty dr\frac{2\pi^2r^3}{r^2 + m^2 - i\epsilon} $$
where we obtained the right-hand side by Wick rotating and integrating out the angular variables. The UV divergence is associated with the behaviour of the integrand for $r\to\infty$ and for this the value of $\epsilon$ is irrelevant. Thus, if $G[f, g]$ is UV divergent for $\epsilon\to0$ it should be UV divergent for all $\epsilon>0$.
My instinct is to trust the non-perturbative argument 1 over the perturbative argument 2, but this would mean that a complex mass term can serve as a UV regulator and I've never heard of anyone use or even discuss such a regularisation.
