Radon-Nikodym Derivative and Bayes' Theorem

[Jatex]

Theorem 1.31 (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
$$
\frac{d\mu_{\Theta\mid X}}{d\mu_\Theta}(\theta \mid x)
= \frac{f_{X\mid \Theta}(x\mid \theta)}{\int_\Omega f_{X\mid\Theta}(x\mid t) \, d\mu_\Theta(t)}
$$
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[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.

 

[Stephen Tashi]

but now, where does $f_{\Theta}(\theta)$ go?

The $f_{\Theta}(\theta)$ is present in the expression for $f_{\Theta|X}(t|x)$ according to http://web.stanford.edu/class/stats200/Lecture20.pdf eq. 20.1.

Is the error in saying $\frac{ d \mu_{\Theta|X}}{d \mu_\Theta} (t,x)= f_{\Theta|X}(t,x)$? What is he definition of $\frac{ d \mu_{\Theta|X}}{d \mu_\Theta}$?

Or is there a typo in the notes you quoted?