Can the measurable mapping in the definition of complete statistics depend on sample size? 
*

*The definition of complete statistics is from
http://en.wikipedia.org/wiki/Completeness_(statistics)#Definition

The statistic $s$ is said to be complete for the distribution of $X$ if
  for every measurable function $g$ (which must be independent of $θ$) the
  following implication holds: 
$E(g(s(X))) = 0$ for all $θ$ implies that $P_θ(g(s(X)) = 0) = 1$ for all $θ$.

Let the codomain of statistic $s$ be $\mathbb R^m$, then is $g$ a
measurable mapping from $\mathbb R^m$ to $\mathbb R$?
Since $g$ acts on the codomain of $s$, does $g$ not know the sample
size $n$ of $X = \{X_1, \dots, X_n\}$? Only $s$ acts on the sample
$X$,so does only $s$ know the sample size of $X$?
But in a solution to the problem 6.15 in Casella and Berger's
Statistical Inference,
when proving a statistic is not complete, $g$ is chosen to depend on
$n$.  
what is $g$ actually? A measurable mapping from $\mathbb R^m$ to
$\mathbb R$ which doesn't know the sample size, or something like a
statistic which knows  the sample size? 
Or does the statistic $s$ also output the sample size $n$ of its
input $X$? I.e. the codomain of a statistic $s$ is $\mathbb R^m
\times \mathbb N$? So that $g$ is defined on $\mathbb R^m \times
\mathbb N$, i.e. $g$ get to know the sample size from the output of
statistic $s$?

*In general (going beyond the concept of complete statistics), when
talking about a mapping $g$ on a statistic $s(X)$, i.e. $g(s(X))$,
do we always assume $g$ knows the sample size of $X$, i.e. is the
sample size of $X$ always an input to $g$?
Even further do we assume $g$ know the entire input sample $X$ (not just its size $n$) to the
statistic $s$? e.g. $s(X) = \sum_i X_i$, and does it make sense that
$g(s(X)) = (\sum_i X_i) + (\sum_i X_i^2)$?
See also https://math.stackexchange.com/questions/918632/in-composition-of-two-mappings-can-the-outer-mapping-access-the-arguments-of-th
Thanks.
 A: Things are clearer if one thinks of completeness as a property of a parametric family of distributions
$$
\mathcal F = \{F_\theta: \theta \in \Theta\},
$$
where $F_\theta$ is a probability distribution on (say) $\mathbb R^m$. Then $\mathcal F$ is complete if 
$$
\int g(x) \ dF_\theta(x) = 0 \iff F_\theta(g(x) = 0) = 1 \mbox{ for all $\theta$}.
$$
When one talks about a statistic $s(X)$ being sufficient, where $X$ is an $\mathbb R^n$-valued random vector and $X \sim F^n_\theta$ (i.e. $X_i \stackrel{iid}{\sim} F_\theta$) for some unknown $\theta$, what one really means is that the parametric family $\{G_\theta: \theta \in \Theta\}$ is complete where $G_\theta$ is the distribution of $s(X)$ when $X \sim F^n_\theta$. 
As such, then, the notion of completeness has nothing to do with the sample size and the function $s(X)$ needed to get a complete statistic might change with the sample size. For example if $X_i \stackrel{iid}{\sim} N(\mu, 1)$ then $s(X) = \sum_{i=1}^n X_i$ depends on the sample size. What the solution to 6.15 shows is that, irrespective of the value of $n$, the parametric family of distributions of the statistic $(\bar X, S^2)$ does not correspond to a complete family.
A different $g$ is required for each value of $n$, but for each value of $n$ we know such a function exists so the statistic is not complete for any sample size. 
The function $g$ is just a mapping from the space $\mathbb R^m$ to the real numbers. $g$ doesn't "know" anything other than its input. But, again, the function $g$ needed to show incompleteness depends on the distribution of $s(X)$ and $s(X)$ itself may depend on the sample size. Indeed, $s: \mathbb R^n \to \mathbb R^m$ so $s$ itself obviously must depend on the sample size. 
