A question about a proof of Neyman's factorization theorem This question comes from the proof of Neyman's factorization theorem in Robert V. Hogg, Joseph W. McKean, Allen T. Craig, "Introduction to Mathematical Statistics", 6th edition, pp 376-377.

In the proof, a one-to-one transformation is used which is indicated by the red line. But I could not understand why such a one-to-one transformation surely exists. Can you tell me?
Thank you for any help!
 A: I wrote this stuff as my personal notes in some class as an adaptation of the proof in Theory of Statistics by Mark Schervish.


*

*R.N. stands for Radon Nikodym

*T.L. stands for the Tower Property of conditional expectation
We use this definition of Sufficiency:
$\textbf{Definition 1.1}:$ A statistic $T$ is $\textit{sufficient}$ for $\Theta$ if $\forall A \in \sigma(X)$, $\exists$ a version of $P_{\theta}(A|T)$ functionally independent of $\theta$. In this case, we call $P_{\theta}(A|T)$ by $P(A|T)$.
This is the more abstract version of $P_{\theta}(X=x|T=t)$ not depending on $\theta$.
Next, we need these Lemmas (I can write the proof if it helps):
$\textbf{Lemma 1.1}$: If $\forall \theta \in \Omega$, $P_{\theta} << \nu$ for a $\sigma$-finite $\nu$, then $\exists (c_{i})_{i=1}^{\infty}$ in $[0,1]$ s.t. $\sum_{i=1}^{\infty}{c_{i}} = 1$ and $(\theta_{i})_{i=1}^{\infty}$ in $\Omega$ s.t. $\nu^{*} := \sum_{i=1}^{\infty}{c_{i}P_{\theta_{i}}}$ is a probability measure and $\forall \theta \in \Omega$, $P_{\theta} << \nu^{*} << \nu$.
$\textbf{Lemma 1.2}$: Suppose $\exists$ $\sigma$-finite $\nu$ on $(\chi,\beta_{X})$ s.t. $\forall \theta \in \Omega, P_{\theta} << \nu$ and that $T$ is sufficient for $\Theta$. Take $\nu^{*}$ as in Lemma $1.1$. $\forall \theta \in \Omega$, $\nu^{*}(A|T)$ is a $\nu^{*}$-version of $P_{\theta}(A|T)$.
$\textbf{Lemma 1.3}$: Suppose $\exists$ $\sigma$-finite $\nu$ on $(\chi,\beta_{X})$ s.t. $\forall \theta \in \Omega, P_{\theta} << \nu$ and take $\nu^{*}$ as in Lemma $1.1$. There $\exists$ functions $m_{1} \& m_{2,\theta}$ s.t. $\forall \theta \in \Omega$, a $\nu$-version of $\frac{dP_{\theta}}{d\nu}$ has $\forall x \in \chi$, $\frac{dP_{\theta}}{d\nu} = m_{1}m_{2,\theta}$, $m_{1}$ functionally independent of $\theta$ and $m_{2}$ $\sigma(T)$-measurable iif $\forall \theta \in \Omega$, there exists a $\sigma(T)$-measurable $\nu^{*}$-version of $\frac{dP_{\theta}}{d\nu^{*}}$. Whenever there is no confusion, we will also call this version by $\frac{dP_{\theta}}{d\nu^{*}}$.
$\textbf{Proof}$: $\textbf{Stage}$ 1: ``if''
$ \ $
First, observe that:
$$\frac{d\nu^{*}}{d\nu} = \sum_{j}{c_{j}\frac{dP_{\theta_{j}}}{d\nu}} =^{\nu} m_{1}\sum_{j}{c_{j}m_{2,\theta_{j}}}$$
Next, observe that, since $\nu^{*} << \nu$:
$$m_{1}m_{2,\theta} =^{\nu^{*}} \frac{dP_{\theta}}{d\nu^{*}} = \frac{dP_{\theta}}{d\nu} \frac{d\nu^{*}}{d\nu} =^{\nu^{*}} \frac{dP_{\theta}}{d\nu} m_{1}\sum_{j}{c_{j}m_{2,\theta_{j}}}$$
Observe that $A = \{x \in \chi: m_{1}\sum_{j}{c_{j}m_{2,\theta_{j}}} = 0\}$ is such that $\nu^{*}(A) = 0$, see\footnote{$\nu^{*}(A) = \int_{A}{d\nu^{*}} = \int_{A}{\frac{d\nu^{*}}{d\nu}d\nu} = \int_{A}{m_{1}\sum_{j}{c_{j}m_{2,\theta_{j}}}d\nu} = 0$}. Hence,
$$\frac{dP_{\theta}}{d\nu^{*}} =^{\nu^{*}} \frac{m_{1}m_{2,\theta}}{m_{1}\sum_{j}{c_{j}m_{2,\theta_{j}}}} = \frac{m_{2,\theta}}{\sum_{j}{c_{j}m_{2,\theta_{j}}}}$$
Since $\frac{m_{2,\theta}}{\sum_{j}{c_{j}m_{2,\theta_{j}}}}$ is $\sigma(T)$-measurable, the proof is complete.
$ \ $
$\textbf{Stage}$ 2: ``only if''
$ \ $
We wish to prove that there exist $m_{1}$ and $m_{2,\theta}$ s.t. $\forall A \in \sigma(X)$,
$$\int_{A}{\frac{dP_{\theta}}{d\nu}d\nu} = \int_{A}{m_{1}m_{2,\theta}d\nu}$$
$$\int_{A}{\frac{dP_{\theta}}{d\nu}d\nu} =^{R.N} \int_{A}{dP_{\theta}} =^{R.N} \int_{A}{\frac{dP_{\theta}}{d\nu^{*}}d\nu^{*}} =^{R.N.} \int_{A}{\frac{dP_{\theta}}{d\nu^{*}}\frac{d\nu^{*}}{d\nu}d\nu}$$
Taking $m_{1} = \frac{d\nu^{*}}{d\nu}$ and $m_{2,\theta} = \frac{dP_{\theta}}{d\nu^{*}}$, the proof is complete.
$ \ $
$ \ $
$\textbf{Theorem 1.1}$ (Fisher-Neyman Factorization): Suppose $\exists$ $\sigma$-finite $\nu$ on $(\chi,\beta_{X})$ s.t. $\forall \theta \in \Omega, P_{\theta} << \nu$. Then $T$ is sufficient for $\Theta$ iif $\exists$ functions $m_{1} \& m_{2,\theta}$ s.t. $\forall \theta \in \Omega$, a $\nu$-version of $\frac{dP_{\theta}}{d\nu}$ has $\forall x \in \chi$, $\frac{dP_{\theta}}{d\nu} = m_{1}m_{2,\theta}$, $m_{1}$ functionally independent of $\theta$ and $m_{2}$ is $\sigma(T)$-measurable.
$ \ $
$\textbf{Proof}:$ $\textbf{Stage}$ 1: ``only if''
$ \ $
We wish to find that $m_{1}$ and $m_{2}$ as in the Theorem such that, $\forall A \in \sigma(X)$, $\forall \theta \in \Omega$:
$$\int_{A}{\frac{dP_{\theta}}{d\nu}d\nu} = \int_{A}{m_{1}m_{2,\theta}d\nu}$$
Take an arbitrary $A \in \sigma(X)$ and consider $\nu^{*}$ as in Lemma $1.1$.
$$\int_{A}{\frac{dP_{\theta}}{d\nu}d\nu} =^{R.N} E_{P_{\theta}}(I_{A}) = E_{P_{\theta}}(P_{\theta}(A|T)) = E_{\nu^{*}}(P_{\theta}(A|T) \frac{dP_{\theta}}{d\nu^{*}}) =^ {L.2} E_{\nu^{*}}(\nu^{*}(A|T) \frac{dP_{\theta}}{d\nu^{*}}) =^{T.L.}$$
$$= E_{\nu^{*}}(\nu^{*}(A|T) E(\frac{dP_{\theta}}{d\nu^{*}}|T)) := E_{\nu^{*}}(\nu^{*}(A|T) m_{2,\theta}) = E_{\nu^{*}}(E_{\nu^{*}}(I_{A}m_{2,\theta}|T)) =^{T.L.} E_{\nu^{*}}(I_{A}m_{2,\theta}) =^{R.N.}$$
$$= E_{\nu}(I_{A}\frac{d\nu^{*}}{d\nu}m_{2,\theta}) = \int_{A}{\frac{d\nu^{*}}{d\nu}m_{2,\theta}d\nu} := \int_{A}{m_{1}m_{2,\theta}d\nu}$$
$\textbf{Stage}$ 2: ``if''
$ \ $
We will show that $\forall \theta \in \Omega$, $\nu_{*}(A|T)$ is a $P_{\theta}$-version of $P_{\theta}(A|T)$. Since $\nu^{*}(A|T)$ is functionally independent of $\theta$, the proof will be complete. Take an arbitrary $\theta$, since $\nu_{*}(A|T)$ and $P_{\theta}(A|T)$ are $\sigma(T)$-measurable, we wish to show that:
$$\forall B \in \sigma(T), \int_{B}{P_{\theta}(A|T)dP_{\theta}} = \int_{B}{\nu^{*}(A|T)dP_{\theta}}$$
Take an arbitrary $B \in \sigma(T)$:
$$\int_{B}{P_{\theta}(A|T)dP_{\theta}} = E_{P_{\theta}}(I_{A}I_{B}) =^{R.N.} E_{\nu^{*}}(I_{A}I_{B}\frac{dP_{\theta}}{d\nu^{*}}) =^{T.L.} E_{\nu^{*}}(E_{\nu^{*}}(I_{A}\frac{dP_{\theta}}{d\nu^{*}}|T)I_{B}) =^{L.3}$$
$$= E_{\nu^{*}}(E_{\nu^{*}}(I_{A}|T)\frac{dP_{\theta}}{d\nu^{*}}I_{B}) = E_{\nu^{*}}(\nu^{*}(A|T)I_{B}\frac{dP_{\theta}}{d\nu^{*}}) =^{R.N.} E_{P_{\theta}}(\nu^{*}(A|T)I_{B}) = \int_{B}{\nu^{*}(A|T)dP_{\theta}}$$
Which completes the proof of Theorem $1.1$.
A: The proof is essentially the same as the one in the discrete case. Let $X = X_{1},\ldots,X_{n}$ be random variables with a discrete distribution.
$T$ is sufficient for $\theta$ if $P_{\theta}(X=x|T=t)$ does not depend on $\theta$.
If $T$ is sufficient,
$$P_{\theta}(X=x) = P_{\theta}(X=x,T=t) = P_{\theta}(T=t)P_{\theta}(X=x|T=t) = g_{\theta}(t)f(x)$$
If $P_{\theta}(X=x) = g_{\theta}(t)f(x)$,
$$P_{\theta}(T=t) = \sum_{x: T(x)=t}{P_{\theta}(X=x)} = g_{\theta}(t)\sum_{x: T(x)=t}{f(x)}$$
Hence,
$$P_{\theta}(X=x|T=t) = \frac{P_{\theta}(X=x,T=t)}{P_{\theta}(T=t)} = \frac{P_{\theta}(X=x)}{P_{\theta}(T=t)} = \frac{g_{\theta}(t)f(x)}{g_{\theta}(t)\sum_{y: T(y)=t}{f(y)}} = \frac{f(x)}{\sum_{y: T(y)=t}{f(y)}}$$
which does not depend on $\theta$.
This idea is also more or less the one the book you posted was trying to follow.
