Distribution of a random measure is determined by the characteristic function

Question

I ham trying to understand a proof from a book I am reading. It says the proof follows directly from the prior theorem and I just can't see that.

Let $X$ be a random measure on a locally finite, seperable topological Room $E$.

Theorem: The distribution $P_X$ of a random measure X is defined by both the values of the Laplace-Transformation $L_X(f), f \in C^+_c(E)$ and the values of the characteristic function $\phi_X(f), f \in C_c(E)$ with $$L_X(f) = \mathbb{E}[exp(-\int f dX)]$$ $$\phi_X(f) = \mathbb{E}[exp(i\int f dX)]$$ Proof: This follows directly from the prior theorem and the Laplace-transformation/the characteristic function being unique.

I am sitting here for some while now and not seeing it. The prior theorem is:

The distribution $P_X$ of a random measure X is distinctly determined by the distribution of $$((I_{f_1}, .., I_{f_n}): n \in \mathbb{N}; f_1, ... ,f_n \in C_c^+(E))$$ as well as by the distribution of $$((I_{A_1}, ..., I_{A_n}): n \in \mathbb{N}; A_1, ..., A_n \in B_b(E), A_i \cap A_j = \emptyset \text{ for } i \neq j)$$

With

• $I_f(\mu) = \int f d\mu$
• $I_A(\mu) = \mu(A)$
• $B_b$ the relative compact borel sets (their closure is compact)

Any help, tips, whatsoever are highly appreciated!

Answer 1

Argue something like this: Suppose $\phi_X(f) = \phi_{X'}(f)$ for every $f$. Pick sets $A_1,\ldots,A_n$ in $B_b$ and constants $t_1,\ldots,t_n$ and put $f=\sum t_i \chi_{A_i}$, where $\chi_{A_i}$ is the indicator function of set $A_i$ (taking value $1$ on $A_i$ and $0$ elsewhere). By assumption, $$\mathbb{E}[\exp(i\int f\, dX)] = \mathbb{E}[\exp(i\int f\, dX')]\tag{*} $$ But $\int f\,dX=\int\sum t_i\chi_{A_i}\,dX = \sum t_i\int \chi_{A_i}\,dX=\sum t_i X(A_i)$. The equality (*) holds for all choices of $t_1,\ldots,t_n$, so by uniqueness of characteristic functions in $\mathbb R^n$, the collection $(X(A_1),\ldots,X(A_n))$ has the same joint distribution as the collection $(X'(A_1),\ldots,X'(A_n))$. Now apply the second half of the prior theorem.

Answer 2

I found the solution:

If $f \in C_c(E)$, then $\forall t \in \mathbb{R}: t * f \in C_c$. For $f_1, ... f_n \in C_c(E), t \in \mathbb{R^n}$ set $f = (f_1, ..., f_n)$

$$\phi_X(t \cdot f) = \mathbb{E}[\exp(i t \cdot I_f(X))] = \int_E \exp(it \cdot x) d\mathbb{P}_{I_f \circ X}$$

From the uniqueness of the characteristic function $\mathbb{P}_{I_f \circ X}$ is uniquly defined through this. From the prior theorem we know that because we chose $(f_1, ..., f_n)$ arbitary this uniquly defines the distribution $\mathbb{P}_{X}$