# Application of the Superposition Theorem for Poisson point processes

The problem below is Exercise 3.8 of Last and Penrose Lectures on Poisson Processes. I have been thinking for the better part of the day about it but couldn't write down even one insightful line. I am especially puzzled about the connection with the Superposition theorem.

Problem: Let $\lambda$ be an $s$-finite measure on $( \mathbb{X}, \mathcal{X})$ and let $\eta$ be a Poisson point process with intensity measure $\lambda$. Let $f:\boldsymbol{N} \to \mathbb{R}_+$ measurable, be such that $\mathbb{E}(f(\eta)) < \infty$ and further let $\eta'$ be a Poisson point process with intensity measure $\lambda'$ such that $\lambda= \lambda' + \nu$ where $\nu$ is a finite measure.

Apply the Superposition Theorem in order to show that $\mathbb{E}(f ( \eta')) < \infty$

Note: $\boldsymbol{N}$ denotes the set of point measures

My approach: Since the Superposition theorem is mentioned in the exercise my only idea was to introduce another Poisson point process $X$ with intensity $\nu$ that is independent of $\eta'$, by the Superposition theorem I could then conclude that $X + \eta'$ is a Poisson point process with intensity $\lambda' + \nu = \lambda$.

Update: I have managed to write down some additional lines, namely
Let $\eta \sim$PPP($\lambda$) and $\eta' \sim$ PPP($\lambda'$), choose $X \sim$PPP($\nu$) such that $X$ is independent of $\eta'$, by the Superposition theorem for PPP we obtain that $$\eta' + X \sim \text{PPP}(\lambda' + \nu)= \text{PPP}(\lambda) \sim \eta$$ In particular for $f: \boldsymbol{N} \to \mathbb{R}_+$ measurable we have that $$\mathbb{E}(f( \eta' + X)) = \mathbb{E}(f ( \eta)) < \infty$$ Certainly we have $\eta' \leq \eta' + X$, I need to rigorously justify why $$\mathbb{E}(f ( \eta')) \leq \mathbb{E}(f( \eta' +X)) \tag{*}$$ The reason for this must be connected to the fact that we have $\lambda = \lambda' + \nu$ where both $\lambda, \lambda'$ are $s$-finite and $\nu$ is finite. Since $\nu$ is a finite measure I can say that $\lambda' \leq \lambda' + \nu$ and on (*) the integral on the RHS measures all elements of $\boldsymbol{N}$ that get measured on the LHS but finitely many more because $\nu$ is finite, thus * holds.

Questions: Is my reasoning correct? Is there a way how to rigorously justify * ?

As declared in the OP we let $\eta \sim$PPP($\lambda$) and $\eta' \sim$ PPP( $\lambda'$) where $\lambda, \lambda'$ are $\sigma$-finite measures. We then choose $X \sim$PPP($\nu$) where $\nu$ a finite measure such that $X$ is independent of $\eta'$.
By the Superposition theorem it follows that $\eta' + X \sim$ PPP( $\lambda' + \nu$)=PPP($\lambda) \sim \eta$ and in particular we have for $f: \boldsymbol{N} \to \mathbb{R}_+$ measurable that $\mathbb{E}(f( \eta' + X))= \mathbb{E}(f( \eta))< \infty$. Furthermore we have $$\infty > \mathbb{E}(f( \eta' + X)) \geq \mathbb{E}(f( \eta') 1_{X=0})\overset{1}=\mathbb{E}(f( \eta')) \mathbb{P}(X=0) \overset{2}= \mathbb{E}(f( \eta')) e^{- \lambda} \\ \overset{3}\implies \mathbb{E}(f( \eta'))< \infty$$
1) $X$ is independent of $\eta'$
2) $X \sim$PPP($\lambda)$
3) $\lambda$ is a finite measure and thus $e^{-\lambda}>0$