$\newcommand\th\theta\newcommand\R{\mathbb R}$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 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 for the MLE, one of the standard assumptions is that the likelihood $L_x(\th)$ be twice differentiable in $\th$.
In fact, usually it is a family $(f_\th)$ (regular in $\th$ to some extent) that is assumed, and the family $(d\mu_\th)=(f_\th\,d\nu)$ may come next, if ever. So, the difficulty that concerns you hardly ever arises.
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. 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.