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$\newcommand\th\theta\newcommand\R{\mathbb R}$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.

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 $(f_\th)$ of densities (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.

$\newcommand\th\theta\newcommand\R{\mathbb R}$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.

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 $(f_\th)$ of densities (regular in $\th$ to some extent) that is assumed, whereas the family $(\mu_\th)$ with $d\mu_\th=f_\th\,d\nu$ may come next, if ever. Also, the choice of the reference measure $\nu$ usually matters little -- because, for any two mutually absolutely continuous reference measures, the likelihood profiles $\th\mapsto L_x(\th)$ will be proportional to each other (with the proportionality coefficient depending only on $x$), which usually will not affect statistical inference based on the likelihood. 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.

$\newcommand\th\theta\newcommand\R{\mathbb R}$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.

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.

$\newcommand\th\theta\newcommand\R{\mathbb R}$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.

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 $(f_\th)$ of densities (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.

$\newcommand\th\theta\newcommand\R{\mathbb R}$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 standardly assume that the likelihood $L_x(\th)$ is continuous in $\th$.

• Proofs of the central limit theorems 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. 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.

$\newcommand\th\theta\newcommand\R{\mathbb R}$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.

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.