This question was on a HW in my Statistical Theory class and I find the professor's answer and explanation to be unsatisfactory. Please give me some guidance as to why
$\bar{x}$ is the MLE if this is the case, or
Let me know if I am correct to think that both $\bar{x}$ and $1-\bar{x}$ maximize the likelihood function and therefore the MLE is not unique, or
If the problem is even well defined as it is posed. I feel like this could be the case also.
I understand the definition of the MLE in more regular circumstances, as detailed in the Principles section of http://en.wikipedia.org/wiki/Maximum_likelihood, but the weird form of the PDF brings up issues with supremums that I am not used to dealing with.
Question: Let $X_1,...,X_n$ be an i.i.d. sequence of 0-1 valued RV's with the probabilities
$$ P(X_1=1)=\begin{cases} \theta, & \theta\in\mathbb{Q}\\1-\theta, & \theta\notin\mathbb{Q} \end{cases} $$
where $\theta\in(0,1)$. Does the MLE of $\theta$ exist?
Outline of Professor's solution: This is the main idea of my professors solution. The likelihood function is
$$ L(\theta|x_1,...,x_n)=\{\theta^{\sum{x_j}}(1-\theta)^{n-\sum{x_j}}\chi_{\theta\in\mathbb{Q}} +\theta^{n-\sum{x_j}}(1-\theta)^{\sum{x_j}}\chi_{\theta\notin\mathbb{Q}}\} $$
where $\chi_A$ is the indicator function of the set $A$. We have
$$ \begin{eqnarray} \underset{\theta\in[0,1]}{\sup} L(\theta|x_1,...,x_n)&=&\underset{\theta\in[0,1]}{\sup} {\{\theta^{\sum{x_j}}(1-\theta)^{n-\sum{x_j}}\chi_{\theta\in\mathbb{Q}} +\theta^{n-\sum{x_j}}(1-\theta)^{\sum{x_j}}\chi_{\theta\notin\mathbb{Q}}\}} \\&=& \max\{\underset{\theta\in\mathbb{Q}}{\sup} \{\theta^{\sum{x_j}}(1-\theta)^{n-\sum{x_j}}\},\underset{\theta\notin\mathbb{Q}}{\sup} \{\theta^{n-\sum{x_j}}(1-\theta)^{\sum{x_j}}\}\} \\&=& \max\{\bar{x}^{\sum{x_j}}(1-\bar{x})^{n-\sum{x_j}},(1-\bar{x})^{n-\sum{x_j}}\bar{x}^{\sum{x_j}}\} \\&=& \bar{x}^{\sum{x_j}}(1-\bar{x})^{n-\sum{x_j}} \end{eqnarray} $$
Professor: At this point, the professor argues that the supremum in the second term,$\underset{\theta\notin\mathbb{Q}}{\sup} \{\theta^{n-\sum{x_j}}(1-\theta)^{\sum{x_j}}\}\}$ is not attained since $\bar{x}$ is a rational number. Since the data consists of rational numbers, the supremum of the first term , $\underset{\theta\in\mathbb{Q}}{\sup} \{\theta^{\sum{x_j}}(1-\theta)^{n-\sum{x_j}}\}$ is attained at $\hat{\theta}_1=\bar{x}$ and that this is the MLE of $\theta$.
Me: It seems like we should consider the supremum over the closure of the sets $\mathbb{Q}$ and $\mathbb{R}\backslash\mathbb{Q}$, which would be $[0,1]$, in which case $\sup L(\theta|x_1,...,x_n)$ is achieved at both $\hat{\theta}_1=\bar{x}$ and $\hat{\theta}_2=1-\bar{x}$. Otherwise, we are essentially assuming that $\theta$ is rational and ignoring irrational $\theta$. Is this the case? If so, is this an undesirable property of the Likelihood Principle in weird cases like this? Is there any plausible situation where issues like this occur? Should I stop worrying about weird problems like this?
As an aside, considering $[0,1]\backslash\mathbb{Q}$ has Lebesgue measure 1 and $[0,1]\cap\mathbb{Q}$ has Lebesgue measure 0, it seems like $\bar{x}$ is a bad estimator, since it is an estimate of $\theta$ if it is in a very small set. Also, if $\theta\in\mathbb{R}\backslash\mathbb{Q}$, $\hat{\theta}_2=1-\bar{x}$ is consistent, so I can't think of a good reason why $\bar{x}$ is better.
Edit As @cardinal pointed out, the $x_i$ are obviously rational, so this is not an issue. This addressed my first (silly) misunderstanding, which involved assuming the estimator $\bar{x}$ could be irrational or rational.
