$\newcommand\th\theta\newcommand\R{\mathbb R}$Indeed, the densities $f_\th:=d\mu_\th/d\nu$ are defined only $\nu$-a.e., and hence it makes no sense to define a maximum likelihood estimate (MLE) as a maximizer of $f_\th(x)$ in $\th$ for **arbitrary** versions of the densities. 

Proofs usually assume that versions $f_\th$ of the densities **can** be chosen to make the likelihood $L_x(\th):=f_\th(x)$ regular enough in $\th$.

- Proofs that MLE is [consistent][1] standardly assume that the likelihood $L_x(\th)$ is continuous in $\th$.

- Proofs of the [central limit theorems][2] for the MLE standardly assume that the likelihood $L_x(\th)$ is twice differentiable in $\th$. 

- Proofs may only discuss a family $(f_\th)$ (regular in $\th$ to some extent), ignoring the family $(d\mu_\th)=(f_\th\,d\nu)$, and avoiding the ambiguities of measures entirely. 

There are some examples where $L_x(\th)$ can not be made continuous in $\th$. E.g., for real $\th>0$, let $\mu_\th$ be the joint distribution of $n$ iid random variables each uniformly distributed over $[0,\th]$, whose derivatives with respect to Lebesgue measure over $\R^n$ give the densities $f_\th$.

- If we define $f_\th(x)=\th^{-n}\,1(\max_i x_i\le\th)$, and so the MLE is $\max_i x_i$.
- if we define $f_\th(x)=\th^{-n}\,1(\max_i x_i<\th)$, then $L_x(\th)=f_\th(x)$ does not attain a maximum in $\th$ for any $x$, so that now an MLE will not exist, strictly speaking. 

A generalized definition of an MLE is given in Section 2 of [this paper][3]. With that definition, even for $f_\th(x)=\th^{-n}\,1(\max_i x_i<\th)$, $\max_i x_i$ will be the generalized MLE.  


  [1]: https://en.wikipedia.org/wiki/Maximum_likelihood_estimation#Consistency
  [2]: https://en.wikipedia.org/wiki/Maximum_likelihood_estimation#Efficiency
  [3]: https://projecteuclid.org/journals/electronic-journal-of-statistics/volume-11/issue-1/Optimal-order-uniform-and-nonuniform-bounds-on-the-rate-of/10.1214/17-EJS1264.full