# 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. - How does one define E(X) when X(B) is infinite for some relatively compact B? (Not sure the hefty bounty now offered is appropriate for a question whose answer is probably a variation on "Check the definitions of the objects involved".) – Did Dec 8 '12 at 8:34 @Matt N. Thanks for the bounty. I thought about it myself but you came ahead. I will clearify the question (and think about it again) when I have time. – AndreasS Dec 9 '12 at 20:49 Ok : ) And no problem. – Rudy the Reindeer Dec 9 '12 at 20:50 @AndreasS Does this answer answer your question? – Rudy the Reindeer Dec 15 '12 at 7:50 @MattN. Not quite yet. My big problem lies in the second part, which hasn't been adressed yet – AndreasS Dec 16 '12 at 15:06 ## 1 Answer 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). - Thanks for the clarification for other members (I should have done that...). I've come more or less to the same conclusions as you. However, my main problem is the second part, in which I don't see a connection to the presentation$X=\sum{\lambda_n X_n}$... So thank you for your further edits. – AndreasS Dec 16 '12 at 15:09 I'm having great difficulty interpreting a lot of it; it's just a lot of definitions to me, and without an intuition to guide me, the work is a lot harder. My intuition from undergrad stats tells me that$\mathbb E[X]=\int_{-\infty}^\infty x d{\bf P}[X=x]$means that$\mathbb E[X]$can be represented as a limit of a sum of RVs with$\lambda_n=x_n\,dx$. I have no idea how to formalize that concept in this setup, though. – Mario Carneiro Dec 16 '12 at 15:56 I'm finally done with the second part! That problem is a doozy. Note that for the second part, they are still using the definition of$X$as a sum, so you only need to show that [$E[X]\in{\cal M}(E)$] + [$X$is a sum] implies [$X\$ is a random measure] (note the way the problem is used in the proof on the next page). –  Mario Carneiro Dec 16 '12 at 17:26
Ah, I see. It was just too nice a statement to be true in that generality (i.e. when X was of an arbitrary form...) Thanks a lot for the effort! –  AndreasS Dec 16 '12 at 21:25
@MattN. Score! That was totally worth learning measure theory. :) –  Mario Carneiro Dec 16 '12 at 21:43