$\newcommand\th\theta\newcommand\R{\mathbb R}$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 **whatever** versions of the densities. 

Usually, though, it is assumed that versions $f_\th$ of the densities **can** be chosen so that the likelihood $L_x(\th):=f_\th(x)$ be regular enough in $\th$. In particular, in proofs of the [consistency][1] of the MLE, one of the standard assumptions is that the likelihood $L_x(\th)$ be continuous in $\th$; in proofs of [central limit theorems][2] for the MLE, one of the standard assumptions is that the likelihood $L_x(\th)$ be twice differentiable in $\th$.

For some textbook statistical models (that is, for some families $(\mu_\th)$), it is impossible to make $L_x(\th)$ continuous in $\th$. E.g., for real $\th>0$, let $\mu_\th$ be the joint distribution of $n$ iid random variables each with the uniform distribution over the interval $[0,\th]$. Choosing the versions $f_\th$ of the densities of $\mu_\th$ (with respect to the Lebesgue measure over $\R^n$) so that for $x=(x_1,\dots,x_n)\in\R^n$ we have $f_\th(x)=\th^{-n}\,1(\max_i x_i\le\th)$, and so, the MLE is $\max_i x_i$. However, if we choose the versions $f_\th$ of the densities of $\mu_\th$ so that $f_\th(x)=\th^{-n}\,1(\max_i x_i<\th)$, 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].  


  [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