# Question on Palm distribution in Kallenberg's book

I have a question regarding Kallenberg's "Foundations of Modern Probability" 2nd ed, a statement right after Lemma 11.2: "...$\eta\{0\}=1$ a.s.".

Let me state the setup:

$\xi$: a random measure on $\mathbb{R}^d$

$X$: an $\mathbb{R}^d$-indexed process taking values in a measurable space $S$.

$\mathcal{M}(\mathbb{R}^d)$: the space of locally finite measures on $\mathbb{R}^d$.

($X$,$\xi$) are jointly stationary w.r.t. a shift $\theta_t$, $t\in \mathbb{R}^d$ (i.e. $\theta_t x=x+t$, $x\in \mathbb{R}^d$), namely, $\theta_t(X,\xi):= (\theta_tX,\theta_t\xi)$ has the same distribution as $(X,\xi)$.

Assume $\mathrm{E}\xi=c\lambda^d$, where $\lambda^d$ is the Lebesgue measure and $c\in \mathbb{R}_+$ (this is almost a consequence of stationarity).

For a function $f\ge 0$ on $S^{R^d}\times \mathcal{M}(\mathbb{R}^d)$, define $$Q_{X,\xi} f=\mathrm{E} \int_B f(\theta_s(X,\xi))\xi(ds) / \mathrm{E}\xi B,$$ for some Borel set $B\subset\mathbb{R}^d$ with $\lambda^d B\in \mathbb{R}_+$. By Lemma 11.2 and Theorem 2.6 of Kallenberg, the preceding expression actually does not depend on $B$.

$Q_{X,\xi}$ can be viewed a probability measure on $S^{R^d}\times \mathcal{M}(\mathbb{R}^d)$. Let a random pair $(Y,\eta)$ have distribution $Q_{X,\xi}$.

Then the book claims that "When $\xi$ is a simple point process, then ...and we note that $\eta\{0\}=1$ a.s.". I just do not know why "$\eta\{0\}=1$ a.s.".

Take a measurable $A\in \mathcal{M}(\mathbb{R}^d)$. Then $$Q_{X,\xi}(\eta\in A)= \mathrm{E} \int_{[0,1]^d}1_{\{\theta_s \xi\in A\}}\xi(ds)=\\ \mathrm{E} \int_{[0,1]^d}1_{\{\theta_s \xi\in A, \theta_s\xi\{0\}=1\}}\xi(ds)+\mathrm{E} \int_{[0,1]^d}1_{\{\theta_s \xi\in A, \theta_s\xi\{0\}=0\}}\xi(ds).$$ Since the simple point process $\xi$ satisfies $\xi\{s\}=0$ or $1$, the second term in the preceding line must be $0$ because $\theta_s\xi\{0\}=\xi\{s\}$. Therefore, $$Q_{X,\xi}(\eta\in A)=\mathrm{E} \int_{[0,1]^d}1_{\{\theta_s \xi\in A, \theta_s\xi\{0\}=1\}}\xi(ds) = Q_{X,\xi}(\eta\in A, \eta\{0\}=1).$$