I have a question about the p.d.f. of exponential family. Suppose $(X,\mathcal{F})$ is a measurable space and $\{F\_{\theta},\theta\in \Theta\}$ is a distribution family on $(X,\mathcal{F})$. When $\{F\_{\theta},\theta\in \Theta\}$ is dominated by a $\sigma$ finite measure $\mu$, $\{F\_{\theta},\theta\in \Theta\}$ is a exponential family if and only if
$$
\frac{d F\_\theta}{d\mu}=\exp\{\eta(\theta)^\top T(x)-\xi(\theta)\}h(x),\forall \theta\in\Theta,
$$
where $h(x)$ is non-negative measurable which is stated in many references like Lehman and Casella, Theory of Estimation.


It is also said that $h(x)$ is not important because it can be absorbed in $d\mu$(so we can even let $h(x)=1$). Let $\lambda(A)=\int\_Ah(x)d\mu,\forall A\in\mathcal{F}$,then
$$
\frac{d F\_\theta}{d\lambda}=\exp\{\eta(\theta)^\top T(x)-\xi(\theta)\}.
$$


**I am confused with the formula $\frac{d F\_\theta}{d\lambda}$ above**. I know that since $F\_\theta\ll \lambda\ll\mu$, we can derive that ($\mu$-a.e.)
$$
\begin{aligned}
\frac{d F\_\theta}{d\mu}&=\frac{d F\_\theta}{d\lambda}\frac{d \lambda}{d\mu},
\\
\exp\{\eta(\theta)^\top T(x)-\xi(\theta)\}h(x)&=\frac{d F\_\theta}{d\lambda}h(x).
\end{aligned}
$$
But we don't know wheather $h(x)>0$ $\mu-$ a.e. or not, how can we get
$$
\frac{d F\_\theta}{d\lambda}=\exp\{\eta(\theta)^\top T(x)-\xi(\theta)\}?
$$
Should we emphasize $h(x)>0$ in the definition? Or maybe we just consider the set $\{x\in X:h(x)>0\}$? If so, how to interpret that $h(x)$ can be set $1$? Does this problem correspond with the support of a distribution?


Any help would be appreciated.



p.d.f. of exponential family
$\newcommand\la\lambda$Let $F:=F\_\theta$ and $R:=\exp\{\eta(\theta)^\top T-\xi(\theta)\}$, so that
$$\frac{dF}{d\mu}=Rh,$$
which means that
$$\int g\,dF=\int gRh\,d\mu \tag{1}\label{1}$$
for all measurable functions $g$ such that either one (or, equivalently, both) of the latter two integrals exist (say in $[-\infty,\infty]$).


In a standard manner, we can show that the condition that $\la(A)=\int\_A h\,d\mu$ for all measurable $A$ is equivalent to the condition that
$$\int f\,d\la=\int fh\,d\mu \tag{2}\label{2}$$
for all measurable functions $f$ such that either one (or, equivalently, both) of the latter two integrals exist in $[-\infty,\infty]$.


Letting now $f=gR$ in \eqref{2}, from \eqref{1} we get
$$\int gR\,d\la=\int g\,dF$$
for all measurable functions $g$ such that either one (or, equivalently, both) of the latter two integrals exist in $[-\infty,\infty]$. This means
$$\frac{dF}{d\la}=R,$$
as desired.




---


(The set $\{x\colon h(x)>0\}$ plays no role in this consideration -- since we only multiply here, and never divide.)


