Applying Bayes' rule in a more technical way when densities don't exist Say $y \mid x \sim \text{Normal}(Ax, B)$ and $x \sim \text{Normal}(c,D)$. Let's assume further that $y \in \mathbb{R}^1$ and $x \in \mathbb{R}^2$.
To find $p(x \mid y)$ we can usually do
\begin{align*}
p(x \mid y) &\propto p(y \mid x) p(x) \\
&\propto \exp\left[ -\frac{1}{2}\left\{ (y - Ax)^\intercal B^{-1}(y-Ax) + (x-c)^\intercal D^{-1}(x-c) \right\} \right] \\
&\propto \exp\left[ -\frac{1}{2}\left\{  x^\intercal A^\intercal B^{-1} A x  - 2 x^\intercal A^\intercal B^{-1} y +  x^\intercal D^{-1}x -2 x^\intercal D^{-1} c  \right\} \right] 
\end{align*}
and end up with a posterior precision of $\left[A^\intercal B^{-1} A + D^{-1} \right] $, and a posterior mean of $\left[A^\intercal B^{-1} A + D^{-1} \right]^{-1}\left[y^\intercal B^{-1}A + c^\intercal D^{-1}  \right]$. 
However, when $B$ is the $0$ matrix, $y$ follows a "singular" or "degenerate" multivariate normal distribution, and none of the above work is valid because $p(y \mid x)$ doesn't exist (it isn't dominated by Lebesgue's measure).
We can apply the more general Bayes' rule
$$
p(x \in S \mid y) = \frac{\int_S f(x)\mathbb{1}(Ax=y)dx}{\int_{\mathbb{R}} f(x)\mathbb{1}(Ax=y)dx}
$$ 
but I'm having some trouble with the integral. Can anyone help? 
Edit:
The last expression isn't valid--taking the expectation of that indicator is $0$. It's kind of like asking for the probability that a continuous random variable is exactly equal to a specific value. 
 A: Define 
\begin{align*}
S &= \{ x : Ax = y, x_2 \le s\} \\
&= \{ (x_1, x_2) : a_1 x_1 + a_2 x_2 = y, x_2 \le s \} .
\end{align*}
We can say that 
\begin{align*}
p(x \in S \mid y) &= \frac{\int_{-\infty}^s \int _{\mathbb{R}} f_{x_1,x_2}(x_1,x_2)\delta_{a_1 x_1 + a_2 x_2}(dy) dx_2 }{\int_{\mathbb{R}} \int _{\mathbb{R}} f_{x_1,x_2}(x_1,x_2)\delta_{a_1 x_1 + a_2 x_2}(dy) dx_2}\\
&=\frac{P(a_1 x_1 + a_2 x_2 = y, x_2 \le s)}{P(a_1 x_1 + a_2 x_2 = y)} \\
\end{align*}
taking care not to write $dx$ or $dx_1dx_2$ anywhere (because that would imply we have a Radon-Nikodym derivative with respect to the product Lebesgue measure).
So we get
$$
p(x \in S \mid y) = \frac{\int_{-\infty}^s f_{x_1,x_2}[(y - a_2x_2)/a_1, x_2] dx_2}{\int_{-\infty}^{\infty} f_{x_1,x_2}[(y - a_2x_2)/a_1, x_2] dx_2} \tag{3}
$$
A: The degenerate normal distribution with zero variance matrix is just a point-mass distribution on its mean (if you take it to be well-defined at all), so you have $\mathbb{P}(Y=Ax|X=x) = 1$.  To facilitate analysis, define the set function:
$$\mathcal{H}(y) \equiv \{ x \in \mathbb{R} | y=Ax \}
\quad \quad \quad \text{for all } y \in \mathbb{R},$$
so we have $f(y|x) = \mathbb{I}(x \in \mathcal{H}(y))$.  The relevant application of Bayes' theorem is:$^\dagger$
$$\begin{aligned}
p(X \in \mathcal{S}|Y=y) 
&= \frac{f(x \in \mathcal{S}, y)}{f(y)} \\[6pt]
&= \frac{\int_\mathcal{S} f(x,y) \ dx}{\int_\mathbb{R} f(x,y) \ dx} \\[6pt]
&= \frac{\int_\mathcal{S} f(y|x) f(x) \ dx}{\int_\mathbb{R} f(y|x) f(x) \ dx} \\[6pt]
&= \frac{\int_\mathcal{S} \mathbb{I}(x \in \mathcal{H}(y)) f(x) \ dx}{\int_\mathbb{R} \mathbb{I}(x \in \mathcal{H}(y)) f(x) \ dx} \\[6pt]
&= \frac{\int_{\mathcal{S} \ \cap \ \mathcal{H}(y)} f(x) \ dx}{\int_{\mathcal{H}(y)} f(x) \ dx}. \\[6pt]
\end{aligned}$$

$^\dagger$ For simplicity, we will ignore the pathological case where $A=\mathbf{0}$ and we condition on $y \neq 0$.  In that pathological case we have $\mathcal{H}(y) = \varnothing$ and so we cannot deploy the equation shown.  To deal with that pathological case, see here.
