Practical example of a non-measurable maximum likelihood estimator This post gives an example of a situation where the MLE is not measurable. However, this doesn't seem to be a situation that you would ever encounter in practice.
Is there a more practical example of a situation where the MLE is not measurable? To be more specific, is there an example that does not rely on the axiom of choice? If such an example exists, and we happen to be put in a situation where we want to estimate a parameter, but the MLE is not measurable, what should we do?
 A: So I think whether or not an example is practical is a little ambiguous. However, with regard to the axiom of choice (AC), I think the answer is no. That is, if we want to construct an example in which $\hat\theta$ is non-measurable this will require AC. The reason for this is simply that in order to construct a non-measurable set of real numbers we need to employ AC. And, of course, we will need a non-measurable set inside our measure space to construct a non-measurable function.
Interestingly, my understanding is that while we cannot construct an example without AC it is still possible that a non-measurable $\hat\theta$ could exist! The reason for this is that the existence of a non-Lebesgue measurable is consistent with Zermelo-Fraenkel (ZF) axioms. That is, a non-measurable set does not imply AC. I don't think this really answers your question about practicality but it is an interesting point (assuming my reading is correct).
So what do we do about this? Well, I am going to give a bit of a dodging :) answer here. One of the more classical assumptions in MLE is that $\Theta$ is compact and the criterion is continuous in $\theta$. This will be enough for the measurability of $\hat\theta$. Another possibility is to work with outer probability measures when generalizing to the case of extremeum estimators (there is an example of this generalization in Chapter 5 of Van der Vaart (1998)). Specifically, we will relax uniform convergence to,
$$P^*[\sup_{\theta\in\Theta} |\hat Q(\theta)-Q(\theta)|>\epsilon]\to 0$$
a.s. in outer probability. where each of these events $\{\sup_{\theta\in\Theta} |\hat Q(\theta)-Q(\theta)|>\epsilon\}$ need not be measurable. But this only lets us bypass measurability issues as we approach the limit. It seems that as long as $(\Theta, \mathcal{A})$ is a separable measurable space then that is enough to guarantee measurability.
