I just read the nice exposition Fermionic Path Integral on nLab and began to wonder about some details to which references appear to be lacking. Suppose we live on Euclidean space as in the Osterwalder-Schrader approach to QFT:

- Is there a deeper analogy between fermionic and bosonic integration? For bosons, we should consider e.g the space $\mathcal{S}'$ of Schwartz distributions as "path space". What is the corresponding - presumably non-commutative - fermionic space $\mathcal{F}'$?
- If such an analogy exists, what is the corresponding one between probability measures $\mu$ on $\mathcal{S}'$ and Berezin integrals $\nu$ on $\mathcal{F}'$?
- Is there an analogy to the spaces physicists would
**like**to work on? i.e The space $\mathcal{S}$ of Schwartz functions would be nice and obliterate the need for regularisation/renormalisation, but unfortunately we have to work on $\mathcal{S}'$ instead. If such an analogy exists, what is $\mathcal{F}$? - Finally, what features should an interaction $S^{\mathrm{int}}$ have in order to make \begin{equation} S^{\mathrm{eff}} \left( \phi \right) = - \ln \int_{\mathcal{F}'} \exp \left[ - S^{\mathrm{int}} \left( \phi, \psi \right) \right] \mathrm{d} \nu_{\mathrm{Berezin}} \left( \psi \right) \end{equation} well-defined? i.e How to integrate out a fermion?

**EDIT:**
For the analogies, I was expecting something along the lines:

- For bosons, we encode $\mathbb{R}^4$ by a commutative ring of test functions on $\mathbb{R}^4$, e.g $\mathcal{S}$
- For fermions, I would expect something similar to give a non-commutative ring $\mathcal{F}$
- For bosons, consider the cylindrical measure of a free theory on $\mathcal{S}$ which extends to a measure on $\mathcal{S}'$
- For fermions, consider some weird (in some sense positive) linear functional on $C_b( \mathcal{F} )$ that is somehow compatible with the non-commutativity of $\mathcal{F}$, expose a failure of regularity and show that the failure disappears upon prolonging to $C_b( \mathcal{F}' )$ where $\mathcal{F}'$ is something even more untangible than $\mathcal{F}$ - but has a concrete description