I'm trying to follow the proof of basu's theorem given by the book Mathematical statistics in Dr. Jun Shao's book:
Theorem: Let $V$ and $T$ be two statistics of $X$ from a population $P\in \textbf{P}$. If $V$ is an ancillary statistic and $T$ is boundedly complete and sufficient for $P\in \textbf{P}$, then $V$ and $T$ are independent w.r.t for any $P\in \textbf{P}$.
Proof:
$\textbf{Part 1:}$Let $B$ be any event in the range of $V$. Since $V$ is ancillary $P(V^{-1}(B))$ is a constant.
My understanding of why is this true. An ancillary statistic's distribution is independent of the population $P$. While the measure from the space $(X,\sigma(X),P)$ is used to measure the subset of $X$ that satisfies the event $V(B)$ since it does not depend on the population then $P(V^{-1}(B))=c$ must evaluate to the same constant $c$ for all $P\in \textbf{P}$ if not then dependency follows.
$\textbf{Part 2:}$Since $T$ is sufficient, $E[I_{(B)}(V)|T]$ is a function of $T$ independent of $P$.
This just follows from the definition of sufficiency.
$\textbf{Part 3:}$Since $E\{E[I_{(B)}(V)|T]-P(V^{-1}(B))\}=0$ , for all $P\in\textbf{P}$ then $P(V^{-1}(B))=E[I_{(B)}(V)|T]=P((V^{-1}(B))|T)$ by boundedly completeness.
This follows because $$E_T\{E[I_{(B)}(V)|T]-P(V^{-1}(B))\}=E_T[E[I_{(B)}(V)|T]]-E_T[P(V^{-1}(B)]=E[I_{(B)}(V)]-P(V^{-1}(B))=0$$
Because $P(V^{-1}(B))$ is a constant and the law of total expectation. Then by completeness $E[I_{(B)}(V)|T]=P(V^{-1}(B))$ must hold, but why must $P(V^{-1}(B)|T)$ also be equal?
$\textbf{Part 4:}$ Let $A$ be an event in the range of $T$. Then $$P(T^{-1}(A)\cap V^{-1}(B))=E\{E[I_{A}(T)I_{B}(V)|T]\}$$
My understanding is that $E[I_{A}(T)I_{B}(V)]=P(T^{-1}(A)\cap V^{-1}(B))$ this can be shown by direct calculation.
The step $P(T^{-1}(A)\cap V^{-1}(B))=E\{E[I_{A}(T)I_{B}(V)|T]\}$ follows from the law of total expectation.
$\textbf{Part 5:}$ $$E\{E[I_{A}(T)I_{B}(V)|T]\}=E\{I_{A}(T)E[I_{B}(V)|T]\}=E\{I_{A}(T)P(V^{-1}(B))$$
This follows because $I_{A}(T)$ is a constant and the second equation follows from part 3.
$\textbf{Part 6:}$ $E[I_{A}(T)P(V^{-1}(B))]=P(T^{-1}(A))(P(V^{-1}(B))$
This follows because $P(V^{-1}(B))$ is a constant and the expectation is taken over $T$ for the final part.
Is my understanding correct? Sorry if it was a long post.