Very simple question on sufficiency In the factorization theorem for checking if a statistic is sufficient
$f(\mathbf{x};\theta)=g(T(\mathbf{x});\theta)h(\mathbf{x})$. 
Could $h(\mathbf{x})=1$? I know this factorisation most probably is not very useful at all, but it's just a curiosity. 
 A: The answer is yes and here is a simple example where that happens. Consider a sequence of iid Bernouli trials with probability $\theta$. The joint pdf of the experiment then is
$$  f \left( \mathbf{x}; \theta \right) = \theta ^{\sum_{i=1}^n x_i} \left( 1 - \theta \right)^{n-\sum _{i=1}^n x_i } = \left( \frac{\theta}{1-\theta} \right)^{\sum_{i=1}^n x_i} \left(1-\theta \right) ^{n} $$
And it becomes clear that the experiment depends on $\mathbf{x}$ only through $\sum_{i=1}^n x_i$, hence this must be the sufficient statistic. This is often good enough but we also need to identify  the function $h \left( \mathbf{x} \right)$. Since this function cannot depend on $\theta$ however, it has to be a constant function in the present experiment. So we take $h \left( \mathbf{x} \right) = 1$ and we're done.
A: In a formal (mathematical) way, the answer to your question is always yes, even when imposing the additional constraint that $g$ is the density of $T(x)$: all you need is to chose the dominating measure [against which the density $f(\mathbf{x};\theta)=g(T(\mathbf{x});\theta)h(\mathbf{x})$ is defined] so that $h(x)=1$. 
In other words, you can always make the (factor) function $h$ part of the dominating measure instead of being part of the density function.
