I would like to find the probability of an event given a signal (update using Bayes' rule).
The events are $x\in \mathbb{R}$ with pdf $g:\mathbb{R}\rightarrow \mathbb{R}_+$, with $g(x)>0\;\forall x$. The signals are $s\in \mathbb{R}$. Event $x$ leads to signal $s=x$ with probability $1-\epsilon$, and with probability $\epsilon$, $s$ is drawn from some pdf $f:\mathbb{R}\rightarrow \mathbb{R}_+$, with $f(s)>0\;\forall s$ and $f$ independent of $x$.
Writing down the standard Bayes' rule, I see at least 2 possibilities: $\Pr(x|s=x)=\frac{(1-\epsilon)g(x)}{(1-\epsilon)g(x)+1\cdot\epsilon f(x)}$ or $\frac{(1-\epsilon)g(x)+\epsilon f(x)g(x)}{(1-\epsilon)g(x)+\epsilon f(x)g(x)+1\cdot\epsilon f(x)}$. The problem is that a particular event $x^*$ has probability zero (being drawn from a continuous pdf), but leads to the signal $s^*=x^*$ with positive probability, while the complementary event $\mathbb{R}\setminus x^*$ has probability one, but leads to $s^*$ with probability zero (since $s^*$ is drawn from a continuous pdf). For $\Pr(x|s\neq x)$, there is no problem, it is just $\frac{g(x)f(s)}{\int g(x)f(s)dx}=\frac{g(x)f(s)}{f(s)}=g(x)$.
What is the correct Bayesian updating formula in such cases of mixed discrete-continuous distributions? Is there a way to write Bayes' rule with cdf-s that avoids the ambiguity?
The conditional pdf of $s$ given $x$ is $$p(s|x)=(1-\epsilon)\delta(s-x)+\epsilon f(s).$$
The a-posteriori pdf of $x$ given observation $s$ is \begin{align} p(x|s)&=\frac{p(s|x)g(x)}{\int_{\mathbb{R}} p(s|x)g(x)dx} \\ &=\frac{(1-\epsilon)\delta(s-x) g(x) +\epsilon f(s)g(x)}{(1-\epsilon)g(s)+\epsilon f(s)}. \end{align}
Asking for $\Pr(x|s=x)$ makes no sense since, given that observation $s$ equals $x$, $x$ is no longer a random variable! A meaningful quantity to ask for would be $\Pr(x=s|s)$, i.e., what is the probability of $x$ being equal to $s$ given that $s$ is observed. However, it is clear from the system model that $\Pr(x=s|s)=\Pr(x=s)=1-\epsilon, \forall s$.