How to Bayesian update on two events which occur with measure zero? To illustrate what I mean please consider the following hypothetical scenario:
A person's favorite number $x\in[-1,1]$ is randomly distributed with atomless density function $f(x)$. 
Furthermore, suppose that this person (after realizing what their favorite number $x$ is) calls out the absolute value of this favorite number i.e. $|x|$. 
As an observer you know the structure, i.e., distribution of $x$ and the behavior of the person. Thus, after observing say $|x|=0.5$ you know that the person's favorite number is either 0.5 or -0.5. 
But as a Bayesian updater what should you belief be? Does it make sense to say that you believe the persons favorite number is 0.5 with probability
$$\mathbb{P}[x=0.5 \, |\, |x|=0.5]=\frac{\mathbb{P}[|x|=0.5 \,|\, x=0.5] \, f(0.5)}{f(0.5)+f(-0.5)}=\frac{ f(0.5)}{f(0.5)+f(-0.5)} ?$$
I suspect not, since any distribution is equivalent (in various senses) to changes on events of measure zero. But what should be done in such a scenario? 
I would have thought such a problem would arise in economic theory (signaling games)but I have yet to find an reference dealing with this issue (any suggestions here would also be much appreciated).
 A: The paradox is one of measure theory and conditioning rather than one of Bayesian inference (and thus you should modify the title of the question). To quote Andrei Kolmogorov,

"The concept of a conditional probability with regard to an isolated
  hypothesis whose probability equals 0 is inadmissible."

When one defines the density $f$ of the random variable $X$, it can indeed be anything including the null function on any set $A\subset(-1,1)$ of measure zero. However, it seems to me that the easiest explanation is that one cannot choose the set $A$ a posteriori, that is once $X$ or $|X|$ is observed to be $x$, so that $x\in A$. Meaning that the actual observation $x$ (or more precisely the actual realisation $x$ of the random variable $X$) has probability zero to belong to $A$.
When setting
$$\mathbb{P}[X=0.5 \, |\, |X|=0.5]=\frac{\mathbb{P}[|X|=0.5 \,|\, X=0.5] \, f(0.5)}{f(0.5)+f(-0.5)}=\frac{ f(0.5)}{f(0.5)+f(-0.5)}$$
(a) the first equality is an incorrect application of Bayes' formula for sets, since the sets are of measure zero and (b) conditioning on the set of measure zero $\{\omega;|X(\omega)|=0.5\}$ is to be understood, rather than as a conditional probability, as setting the value of the function$$\mathbb{E}[\mathbb{I}_{X=|X|}||X|=x]$$ at $x=0.5$, which is not uniquely defined since the only constraint is the definition of conditional expectations as
$$\mathbb{P}[X=|X|]=\mathbb{E}\{\mathbb{E}[\mathbb{I}_{X=|X|}||X|]\}$$
