Effect of the measure on exponential families This might be a very naive question. 
Wikipedia describes an exponential family as a distribution 
$$f(x \mid \theta) =  h(x) \exp( - \theta x - A(\theta)),$$
where $$A(\theta) = \log\left(\int h(x) \exp( - \theta x) dx\right).$$
I want to understand the role of the underlying measure or basically the $h(x)$ in the above equation.
For instance, can any function of $\theta$ be written as $A(\theta)$ for some choice of $h(x)$. 
 A: This is neither a complete nor thorough answer, but an illustration instead.
Consider the product space $\{0, 1\}^n$ that serves as sample space for the extraction of $n$ i.i.d. Bernoulli r.v.s with parameter $p\in(0, 1)$
$$
P(X_1=x_1, \ldots, X_n=x_n) =
    p^{x_1+\ldots+x_n}(1-p)^{n-x_1-\ldots-x_n}
$$
and their transformation into the binomial r.v. $X:=X_1+\ldots+X_n \sim\text{Binom}(n, p)$ with
$$
P(X=x) = 
    \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x},
$$
which can be written as member of the exponential family in the form
$$
P(X=x) = 
    \exp\left\{
        x \cdot \log\left(\frac{p}{1-p}\right) +
        n \log(1-p) +
        \log\left(\frac{n!}{x!(n-x)!}\right)
    \right\},
$$
which would give $A(p)=n \log(1-p)$ and $h(x)=\log\left(\frac{n!}{x!(n-x)!}\right)$.
In that example, $h(x)$ accounts for the multiplicity of elements $(x_1, \ldots, x_n)$ of $\{0, 1\}^n$ which yield $x_1+\ldots+x_n=x$
$$
h(x) =
    \log\left(
        \#\{
            (x_1, \ldots, x_n)\in\{0, 1\}^n : 
                x_1+\ldots+x_n=x
        \}
    \right) =
    \log\left(\frac{n!}{x!(n-x)!}\right).
$$
