Radon–Nikodym Derivative and Bayes' Theorem 
Theorem 1.3.1. (Bayes' theorem):
  Suppose that $X$ has a parametric family $\mathcal{P}_0$ of distributions with parameter space $\Omega$.
  Suppose that $P_\theta \ll \nu$ for all $\theta \in \Omega$, and let $f_{X\mid\Theta}(x\mid\theta)$ be the conditional density (with respect to $\nu$) of $X$ given $\Theta = \theta$.
  Let $\mu_\Theta$ be the prior distribution of $\Theta$.
  Let $\mu_{\Theta\mid X}(\cdot \mid x)$ denote the conditional distribution of $\Theta$ given $X = x$.
  Then $\mu_{\Theta\mid X} \ll \mu_\Theta$, a.s. with respect to the marginal of $X$, and the Radon–Nikodym derivative is
  $$
\frac{\mathrm d\mu_{\Theta\mid X}}{\mathrm d\mu_\Theta}(\theta \mid x)
= \frac{f_{X\mid \Theta}(x\mid \theta)}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, \mathrm d\mu_\Theta(t)}
$$
  for those $x$ such that the denominator is neither $0$ nor infinite.
  The prior predictive probability of the set of $x$ values such that the denominator is $0$ or infinite is $0$, hence the posterior can be defined arbitrarily for such $x$ values.

I tried to derive the right hand side of the Radon–Nikodym derivative above but I got different result, here is my attempt:
\begin{equation} \label{eq1}
\begin{split}
\frac{\mathrm d\mu_{\Theta\mid X}}{\mathrm d\mu_\Theta}(\theta \mid x) &= f_{\Theta\mid X}(\theta\mid x) \mathrm \space \space \space[1]\\
&=\frac{f_{X\mid \Theta}(x\mid \theta) \cdot f_{\Theta}(\theta)}{f_X(x)}\\
&=\frac{f_{X\mid \Theta}(x\mid \theta) \cdot f_{\Theta}(\theta)}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, \cdot f_{\Theta}(t) \space \mathrm dt}\\
&=\frac{f_{X\mid \Theta}(x\mid \theta) \cdot f_{\Theta}(\theta)}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, \mathrm d\mu_\Theta(t)}
\end{split}
\end{equation}
but now, where does $f_{\Theta}(\theta)$ go? 
for $[1]$ see slide $10$ of the following document: http://mlg.eng.cam.ac.uk/mlss09/mlss_slides/Orbanz_1.pdf 
Thanks in advance.
 A: You wrote:
$$
\frac{\mathrm d\mu_{\Theta\mid X}}{\mathrm d\mu_\Theta}(\theta \mid x)
= \frac{f_{X\mid \Theta}(x\mid \theta)}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, \mathrm d\mu_\Theta(t)}
$$
Let's rearrange it a little bit:
$$
\mathrm d\mu_{\Theta\mid X} (\theta \mid x)
= \frac{f_{X\mid \Theta}(x\mid \theta) \, \mathrm d\mu_\Theta}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, \mathrm d\mu_\Theta(t)}
$$
and then:
$$
\frac{\mathrm d\mu_{\Theta\mid X}}{d\nu} (\theta \mid x)
= \frac{f_{X\mid \Theta}(x\mid \theta) \, (\mathrm d\mu_\Theta/d\nu)(\theta)}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, \mathrm d\mu_\Theta(t)}
$$
$${}$$
$$
\frac{d\mu_{\Theta\,\mid\, X=x}}{d\nu}(\theta) = \frac{ \displaystyle \frac{d\mu_{X\,\mid\,\Theta=t}}{d\lambda}(x) \cdot \frac{d\mu_\Theta}{d\nu}(\theta) }{ \displaystyle \int \frac{d\mu_{X\,\mid\,\Theta=t} (x)}{d\lambda} \cdot d\mu_\Theta(t) }
$$
A: You seem to be confused about how to reconcile the familiar version of Bayes' theorem
$$
\tag{1}\label{1}
p(\theta \mid x) = \frac{p(\theta) p(x \mid \theta)}{p(x)}
$$
with the formal version presented here:
$$
\tag{2}
\label{2}
\frac{d\mu_{\Theta\mid X}}{d\mu_\Theta}(\theta \mid x)
= \frac{f_{X\mid \Theta}(x\mid \theta)}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, d\mu_\Theta(t)}.
$$
(I will be using the same notation as at that link.)
On the one hand, the left-hand-side of \eqref{1} is supposed to represent a density of the conditional distribution of $\Theta$ given $X$ with respect to some unspecified dominating measure on the parameter space.
On the other hand, the left-hand-side of \eqref{2} is the density of the conditional distribution of $\Theta$ given $X$ with respect to the prior distribution.
If, in addition, the prior distribution $\mu_\Theta$ has a density $f_\Theta$ with respect to some (let's say $\sigma$-finite) measure $\lambda$ on the parameter space $\Omega$, then $\mu_{\Theta \mid X}(\cdot\mid x)$ is also absolutely continuous with respect to $\lambda$ for $\mu_X$-a.e. $x \in \mathcal{X}$, and if $f_{\Theta \mid X}$ represents a version of the Radon-Nikodym derivative of $d\mu_{\Theta\mid X}/d\lambda$, then \eqref{2} yields
$$
\begin{aligned}
f_{\Theta \mid X}(\theta \mid x)
&= \frac{d \mu_{\Theta \mid X}}{d\lambda}(\theta \mid x) \\
&= \frac{d \mu_{\Theta \mid X}}{d\mu_\Theta}(\theta \mid x) \frac{d \mu_{\Theta}}{d\lambda}(\theta) \\
&= \frac{d \mu_{\Theta \mid X}}{d\mu_\Theta}(\theta \mid x) f_\Theta(\theta) \\
&= \frac{f_\Theta(\theta) f_{X\mid \Theta}(x\mid \theta)}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, d\mu_\Theta(t)} \\
&= \frac{f_\Theta(\theta) f_{X\mid \Theta}(x\mid \theta)}{\int_\Omega f_\Theta(t) f_{X\mid\Theta}(x\mid t) \, d\lambda(t)}.
\end{aligned}
$$
Hopefully this shows you how to get to the familiar form \eqref{1} from \eqref{2}.
