Theory behind testing whether $\mu \in \mathbb{Q}$ for $X \sim \mathcal{N}(\mu, \sigma^2)$ Suppose that $X_i \stackrel{\mbox{i.i.d.}}{\sim} \mathcal{N} (\mu, \sigma^2)$, where $\sigma^2$ is known. Using this data, we wish to test whether $\mu \in \mathbb{Q}$, that is, whether the mean $\mu$ is a rational number. 
It seems intuitively clear that we cannot do this, since any noise will be too much noise. I imagine that any test will have type II error rate $\beta = 0$ and type I error rate $\alpha = 1$ or vice versa. But, I don't understand how to make theoretical statements about this hypotheis testing problem. How does this problem fit into a more general framework that illustrates when testing is "difficult"?
 A: Your intuition is correct: When you have a null set for the mean that is dense with respect to the total space, you won't be able differentiate the null and alternative sets with continuous data.  This is because, for any mean value in the alternative hypothesis, we can always get one that is "arbitrarily close" in the null set.  Hence, there should never be any evidence for the alternative hypothesis.
To get a formal demonstration of this result, you need to go through the motions of constructing this as a composite hypothesis test.  This is a bit tricky, since you have to make an argument for some test statistic, and there are some plausible objections here.

Formal construction of classical hypothesis test: For this test the hypotheses are:
$$\begin{equation} \begin{aligned}
H_0 &: \mu \in \mathbb{Q}, \\[4pt]
H_A &: \mu \notin \mathbb{Q}.
\end{aligned} \end{equation}$$
The first problem you run into is constructing a test statistic.  The likelihood ratio (LR) statistic for this problem is always equal to one, since the rationals are dense in the reals.  We have:
$$\sup_{\mu \in \mathbb{Q}} \sup_{\sigma >0} \prod_{i=1}^n \text{N}(x_i | \mu, \sigma^2) = \Big( \frac{n}{2 \pi \sum x_i^2} \Big)^{n/2} \exp \Big( - \frac{n}{2} \Big) = \sup_{\mu \notin \mathbb{Q}} \sup_{\sigma >0} \prod_{i=1}^n \text{N}(x_i | \mu, \sigma^2),$$
so that the ratio of these supremums is unity.  This means that the standard LR statistic fails to serve as a measure of evidence for the hypotheses, and so we need a custom test statistic.
Now, for these hypotheses, the ordinal ranking of evidence falls into only two categories: if the sample mean is rational (which occurs with probability zero), this is greater evidence for the null hypothesis; if the sample mean is irrational (which occurs with probability one), this is greater evidence for the alternative hypothesis.  Hence, the appropriate test statistic for the test is $T \equiv T(X_1, ..., X_n) \equiv \mathbb{I}(\bar{X} \notin \mathbb{Q})$, with higher values of this (indicator) test statistic constituting greater evidence for the alternative.
Since $\bar{X} \sim \text{N}(\mu, \sigma^2 /n)$ is continuous, we have $\mathbb{P}(T = 0 | \mu, \sigma) = \mathbb{P}(\bar{X} \in \mathbb{Q} | \mu, \sigma) = 0$ over all the parameter values (this follows from the fact that the rationals have Lebesgue measure zero).  This means that the test statistic has the same distribution regardless of the parameter values.
If we observe $\bar{x} \notin \mathbb{Q}$ (i.e., the sample mean is irrational) then the p-value for the test is:
$$p \equiv \mathbb{P}(T(\bar{X}) \geqslant t(\bar{x}) | H_0) = \mathbb{P}(T \geqslant 1 | \mu \in \mathbb{Q}) = 1.$$
If we observe $\bar{x} \in \mathbb{Q}$ (i.e., the sample mean is rational) then the p-value for the test is:
$$\begin{equation} \begin{aligned}
p \equiv \mathbb{P}(T(\bar{X}) \geqslant t(\bar{x}) | H_0) &= \mathbb{P}(T \geqslant 0 | \mu \in \mathbb{Q}) = 1.
\end{aligned} \end{equation}$$
So we see that even with a custom test statistic that tries to differentiate the hypotheses, we never get any evidence against the null.  This is intuitively reasonable, since for any mean value in the alternative hypothesis, we can always get one that is "arbitrarily close" in the null set.
