How this exponential family is rewritten using dominating measure The exponential family is given by :
$$p(x;\theta)=\exp\left\{C(x)+\theta^{i}F_{i}(x)-\psi(\theta)\right\}\hspace{3cm} (1)$$ the $n$ functions $F_{1}(x),...,F_{n}(x)$ are random variables. Let $x_i=F_{i}(x)$.
Suppose we define the probability density function on the $n$-dimensional random variable $x=[x_i]$ with respect to the dominating measure
$$d\mu(x)=\exp\{C(x)\}dx\hspace{3cm}(2)$$
Then we may rewrite equation $(1)$ as
$$p(x;\theta)=\exp\{\theta^{i}x_{i}-\psi(\theta)\}\hspace{3cm} (3)$$
Can how we wrote $(3)$ ??
 A: Let $P$ be a probability measure charging $(\mathbb{R}^n,\mathscr{B}(\mathbb{R}^n))$ with density $fg$ wrt the Lebesgue measure $dx$ given by two nonnegative measurable functions $f$, $g$. Then we may equivalently write
$$P(A)=\int_Afg dx=\int_Afd\mu,A \in \mathscr{B}(\mathbb{R}^n)$$
so that $P$ has density $f$ wrt the measure $d\mu$.
To see this: since $g$ is nonnegative measurable, $\mu(A):=\int_A gdx$ is a measure on $(\mathbb{R}^n,\mathscr{B}(\mathbb{R}^n))$. To prove the above, we need to prove the second equality: (1) for indicators, (2) for nonnegative measurable functions $f$. So $\int_A\mathbf{1}_Bgdx=\int_{A\cap B}gdx=\mu(A\cap B)=\int_A\mathbf{1}_Bd\mu$ for any $B \in \mathscr{B}(\mathbb{R}^n)$. Now since $f$ can be approximated by an increasing sequence of nonnegative simple functions $f_n$ we get by monotone convergence/Beppo Levi
$$\begin{aligned}\int_Afgdx&=\sup_{n \in \mathbb{N}}\int_A\overbrace{\sum_{k\leq N_n}\phi_{k,n}\mathbf{1}_{B_{k,n}}}^{=f_n}\,gdx=\\
&=\sup_{n \in \mathbb{N}}\sum_{k\leq N_n}\phi_{k,n}\int_{A\cap B_{k,n}}gdx=\\
&=\sup_{n \in \mathbb{N}}\sum_{k\leq N_n}\phi_{k,n}\int_{A\cap B_{k,n}}d\mu=\\
&=\sup_{n \in \mathbb{N}}\int_A f_n d\mu=\\
&=\int_Afd\mu\end{aligned}$$
Since $A,f$ are arbitrary, the result follows.
