Can the Fisher factorization theorem be understood as a product of densities? Let $T$ be some random variable on a probability space $\Omega$. Then we have, for $x\in\Omega$:
$$P(x) = P(x|T=T(x))P(T = T(x))$$
This equation is nonsense in an arbitrary probability space but informally it makes some kind of sense, and in a discrete space it's actually correct. Writing $g(t)=P(T=t)$ and $h(x)=P(x|T=T(x))$, we have
$$P(x)=g(T(x))h(x)$$
Which looks suspiciously like the Fisher-Neyman factorization theorem. If $T$ is sufficient for a family of distributions $P_\theta$, then we should instead write $g_\theta$ for $g$, but $h_\theta(x)=P_\theta(x|T=T(x))$ is indeed independent of $\theta$ by sufficiency, and the resemblance is even closer.
Can this interpretation be made rigorous, perhaps by thinking $g$ and $h$ as probability density functions?

Let $T$ be a random variable on some space $(\Omega, P)$ where $P$ is dominated by a measure $\mu$. Can the identity
  $$\frac{dP}{d\mu}(x) = P(x|T=T(x))P(T = T(x))$$
  be made rigorous by replacing the meaningless probabilities on the right with appropriate pdfs?

 A: Pollard argues that disintegrations, which give conditional probability distributions under fairly general conditions, are the right way to think about the factorisation theorem and sufficiency in the continuous case.
A: You could say that a statistic $T(x)$ is sufficient if you could model the sampling like first drawing the sufficient statistic from a distribution that depends on the parameters
$$T \sim f(t\vert \theta)$$
and then draw the observations $X$ from a distribution that depends only on $T$ and is independent from $\theta$
$$X \vert T \sim g(x \vert T)$$
In that case you have that the data $X$ does not tell anything more about $\theta$ than $T$.
And indeed for the distribution of X you get your expression where you multiply them. But, you would have to integrate it. To get a marginal distribution.
$$X\vert \theta \sim \int_{t \in \Omega} g(x\vert t) f(t \vert \theta) dt$$
Your expression is the joint distribution
$$h(x,t) = g(x\vert t) f(t \vert \theta)$$
For instance. To take a sample of size $n$ from a uniform distribution $U(0,a)$, you could first sample the maximum $t$ from a beta distribution and multiply with $a$, and then sample $n-1$ values from a uniform distribution $U(0,t)$.
The distribution of those $n-1$ variables tell nothing more about the value of $a$ than $t$.
