Trying to prove "If the expectation of the random variable is a measure, then the random variable is a random measure." I try to solve the exercise 24.1.1 in Klenke's book on probability.
The first part I got but I have a problem with:

If $X$ is a random variable with values in $(\cal\tilde{M}$$(E),\cal{B}($$\cal\tilde{M}$$(E)))$ such that $\mathbb{E}[X] \in \cal{M}$$(E)$, then $X$ is a random measure.

Where $E$ is a seperable metric space, $\cal\tilde{M}$$(E)$ is the space of all signed Radon measures on $E$ together with the vague topology and $\cal{M}$$(E)$ is the space of all Radon measures on $E$. 
Here my thoughts so far:
If $X$ was constructed as in the first part of the exercise, then it would easily follow as $\mathbb{E}[X] \in \cal{M}$$(E)$ implies $\mathbb{P}(X(B)<\infty)=1$ for every $B$ relatively compact. 
So if I was able to decompose any such $X$ into the form $X=\sum_{n =1}^{\infty}{\lambda_n X_n}$ given in the first part, then I would get the claim. However, I don't think this is possible as X is a priori a signed measure.
Thanks for any hints.
 A: Unrolling the definitions:


*

*A class of sets ${\cal A}\subseteq2^\Omega$ is called a $\sigma$-algebra if $\Omega\in{\cal A}$ and ${\cal A}$ is closed under complements and countable unions.

*A pair $(\Omega,{\cal A})$ consisting of a nonempty set $\Omega$ and a $\sigma$-algebra ${\cal A}\subseteq 2^\Omega$ is called a measurable space. A triple $(\Omega,{\cal A},\mu)$ is called a probability space if $(\Omega,{\cal A})$ is a measurable space
and if $\mu$ is a measure on ${\cal A}$ with $\mu(\Omega)=1$.

*A map $X:\Omega\to\Omega'$ is called ${\cal A}–{\cal A}'$-measurable (or, briefly, measurable) if $X^{−1}({\cal A}) := \{X^{−1}(A) : A\in{\cal A}\} \subseteq {\cal A}$.

*Let $(\Omega',{\cal A}')$ be a measurable space and $(\Omega,{\cal A},{\bf P})$ a probability space and let $X :\Omega\to\Omega'$ be ${\cal A}–{\cal A}'$-measurable. Then $X$ is called a random variable with values in $(\Omega',{\cal A}')$.

*A $\sigma$-finite measure $\mu$ on $(E,{\cal E})$ is called a Radon measure if for any point $x\in E$, there exists an open neighborhood $U\ni x$ such that $\mu(U)<\infty$, and $$\mu(A)=\sup\{\mu(K):K\subseteq A\mbox{ is compact}\}\quad \forall A\in{\cal E},$$ and let ${\cal M}(E):=\{\mbox{Radon measures on }(E,{\cal E})\}$.

*Let ${\cal B}_b(E) = \{B\in {\cal B}(E) : B\mbox{ is relatively compact}\}$.

*Denote by ${\mathbb M} = \sigma(\{I_A : A\in{\cal B}_b(E)\})$ the smallest $\sigma$-algebra on
${\cal M}(E)$ with respect to which all maps $I_A : \mu\mapsto\mu(A),\ A\in{\cal B}_b(E)$, are measurable.

*Let ${\cal\widetilde M}(E)$ be the space of all measures on $E$ endowed with the $\sigma$-algebra $\widetilde{\mathbb M} = \sigma(I_A : A\in {\cal B}_b(E))$.

*A random measure on $E$ is a random variable $X$ on some probability space $(\Omega,{\cal A},{\bf P})$ with values in $({\cal\widetilde M}(E),\widetilde{\mathbb M})$ and with ${\bf P}[X\in {\cal M}(E)] = 1$.

*Let $X_1,X_2,\dots$ be random measures on $E$ and $\lambda_1,\lambda_2,\dots\in[0,\infty)$. Define $X:=\sum_{n=1}^\infty\lambda_nX_n$. Show that $X$ is a random measure iff we have ${\bf P}[X(B)<\infty]=1$ for all $B\in{\cal B}_b(E)$. Infer that if $X$ is a random variable with values in $({\cal\widetilde M}(E),\widetilde{\mathbb M})$ and ${\bf E}[X]\in{\cal M}(E)$, then $X$ is a random measure.


If this seems a bit gratuitous, understand that this is mostly for my own sanity in parsing the bevy of definitions involved here, and also so that new readers won't also have to consult the book as I did to understand the problem statement.
(a) Since we know $X=\sum_{n=1}^\infty\lambda_nX_n$ with each $X_i$ a random measure, let ${\cal C}=\bigcap_{n=1}^\infty X_n^{-1}({\cal M}(E))$. Then given $B\in{\cal C}$ with $X(B)=\mu$, we can write $\mu=\sum_{n=1}^\infty\lambda_n\mu_n$ where $\mu_n\in{\cal M}(E)$ so that $$\mu(A)=\sum_{n=1}^\infty\lambda_n\mu_n(A)=\sum_{n=1}^\infty\lambda_n\sup\{\mu_n(K):K\subseteq A\mbox{ compact}\}$$
$$=\sup\left\{\sum_{n=1}^\infty\lambda_n\mu_n(K):K\subseteq A\mbox{ compact}\right\};$$
thus $\mu$ is inner regular. Also, because $E$ is separable, there is an open set $x\in U\in{\cal B}_b(E)$, so $(\forall B\in{\cal B}_b(E))\mu(B)<\infty$ would be a sufficient condition for $\mu(U)<\infty$ so that $\mu$ is locally finite. Thus $\mu\in{\cal M}(E)$, so $X({\cal C})\cap\{\mu:\mu({\cal B}_b(E))<\infty\}\subseteq{\cal M}(E)$. But ${\cal C}$ is a countable intersection of sets of probability $1$ and ${\bf P}[X({\cal B}_b(E))<\infty]=1$, so the intersection is of probability $1$ as well as any superset thereof; thus ${\bf P}(X^{-1}({\cal M}(E)))=1$, i.e. ${\bf P}[X\in {\cal M}(E)] = 1$ and $X$ is a random measure.
(b) Now we wish to show that ${\bf E}[X]\in{\cal M}(E)$ implies $X$ is a random measure. Because $E[X]$ is locally finite, there is a $U(x)\ni x$ for any $x\in E$ such that $E[X](U)<\infty$. Since $E$ is separable, it has a countable base, and given any $B\in{\cal B}_b(E)$, we have $B\subseteq\bigcup_{n=1}^\infty U(x_n)$. Since $B$ is compact, there is a finite subcovering, and therefore $|E[X](B)|\leq\sum_{i=1}^n|E[X](U_i)|<\infty$. Thus
$${\bf E}[X](B):={\bf E}[X(B)]:=\int X(B)\,d{\bf P}<\infty\Rightarrow{\bf P}[X(B)<\infty]=1.$$
Therefore, since $X=\sum_{i=1}^\infty\lambda_i X_i$, $X$ is a random measure by part (a).
