I have been given these definitions in my statistical inference class:
Let $(X_1,...,X_n)$ be a simple random sampling of $X\rightarrow\{P_\theta:\theta \in \Theta\}$ and $T\equiv T(X_1,...,X_n)$ a statistic. We say that
$T$ is sufficient for $\theta$ if the conditional distribution of $(X_1,...,X_n)$ to each value of $T$ is independent of $\theta$ ($\theta$ does not appear).
$T$ is complete if for every unidimensional measurable function $g$ $$ E_\theta[g(T)]=0 \ \forall \theta \in \Theta \Rightarrow P_\theta[g(T)=0]=1 \ \forall \theta \in \Theta $$
I'm interested in knowing under what functions sufficiency and completeness are preserved. I already know that if $T$ is sufficient for $\theta$ and $T=f(U)$ then U is sufficient for $\theta$, and therefore, if $f$ is biyective, $T'=f(T)$ is also sufficient. However, we haven't seen any properties for completeness and I have done the following:
Let $T$ be a complete statistic, and $T'=f(T)$ with $f$ measurable. Then $T'$ is complete. I proceed as follows:
Let $g$ be an unidimensional and measurable function such that $E_\theta[g(T')]=0 \ \forall \theta \in \Theta$. Then $E_\theta[g(f(T))]=0 \ \forall \theta \in \Theta$ and by completeness of $T$, $P_\theta[g(f(T))=0]=1 \ \forall \theta \in \Theta$, so $P_\theta[g(T')=0]=1 \ \forall \theta \in \Theta$ and $T'$ is complete. I wasn't told what "measurable" means in the definitions, but I guess it is borel measurable (so that $g\circ f$ is borel measurable and the previous line makes sense).
Is this last result correct? If $T$ is sufficient and complete and $f$ is biyective and measurable, can I conclude that $T'=f(T)$ is also sufficient and complete?