probability theory - Poisson point process construction

Let $(\mathcal{X},\mathcal{H},\mu)$ be a $\sigma$-finite measure space.

Suppose that $0<\mu(\mathcal{X})<\infty$. Let $(N;X_j:j=1,2,\ldots)$ be independent on a probability space $(\Omega,\mathcal{F},P)$ with each $X_j$ having distribution $\mu/\mu(\mathcal{X})$ on $(\mathcal{X},\mathcal{H})$ and $N$ having a Poisson distribution with mean $\mu(\mathcal{X})$.

Set

$$V(\omega)= (X_j(\omega):j\leq N(\omega))$$
a multi set of members of $\mathcal{X}$.

How can I show that this construction gives a Possion point process $V$ in $(\mathcal{X},\mathcal{H})$ with intensity $\mu$.

Here I note that $V(\omega)=\emptyset$ if $N(\omega)=0$ and I understand that in general, a point process is a random variable $N$ from some probability space $(\Omega,\mathcal{F},P)$ to a space of counting measures on ${\bf R}$, say $(M,\mathcal{M})$. So each $N(\omega)$ is a measure which gives mass to points
$$
\ldots < X_{-2}(\omega) < X_{-1}(\omega) < X_0(\omega) < X_1(\omega) < X_2(\omega) < \ldots
$$
of ${\bf R}$ (here the convention is that $X_0 \leq 0$. The $X_i$ are random variables themselves, called the points of $N$.

The intensity of a point process is defined to be
$$
\lambda_N
= {\bf E}[N(0,1]].
$$

It´s really hard for me this problem, could someone help me pls.

Thanks for your time and help.

Answer

Technically, you should put $M(\omega):=\sum_{v\in V(\omega)}\delta_v=\sum_{j=1}^{N(\omega)}\delta_{X_j(\omega)}$, and this random variable will be a Poisson point process of intensity $\mu$. To prove this, we need to show that if $A_1,\ldots,A_n\in\mathcal H$ are disjoint, then
$$\mathbf P(M(A_i)=k_i,1\le i\le n)=\prod_{i=1}^ne^{-\mu(A_i)}\frac{\mu(A_i)^{k_i}}{k_i!}.$$
Break up the problem into steps. First, set $k:=\sum_{i=1}^nk_i$ and $A:=\bigcup_{i=1}^nA_i$. For $m\ge k$, conditioned on $N=m$, the vector
$$\Big(M(A_1),\ldots,M(A_n),M(\mathcal X\setminus A)\Big)$$
is multinomial with $m$ trials and respective event probabilities $\frac{\mu(A_1)}{\mu(\mathcal X)},\ldots,\frac{\mu(A_n)}{\mu(\mathcal X)},\frac{\mu(\mathcal X\setminus A)}{\mu(\mathcal X)}$. Thus,
$$\mathbf P(M(A_i)=k_i,1\le i\le n|N=m)=\frac{m!}{k_1!k_2!\ldots k_n!(m-k)!}\left(\frac{\mu(A_1)}{\mu(\mathcal X)}\right)^{k_1}\ldots\left(\frac{\mu(A_n)}{\mu(\mathcal X)}\right)^{k_n}\left(\frac{\mu(\mathcal X\setminus A)}{\mu(\mathcal X)}\right)^{m-k}\\
=m!(\mu(\mathcal X))^{-m}\prod_{i=1}^n\frac{\mu(A_i)^{k_i}}{k_i!}\cdot\frac{\mu(\mathcal X\setminus A)^{m-k}}{(m-k)!}.$$
Since $\mathbf P(N=m)=e^{-\mu(\mathcal X)}\frac{\mu(\mathcal X)^m}{m!}=\frac{\mu(\mathcal X)^m}{m!}\prod_{i=1}^ne^{-\mu(A_i)}\cdot e^{-\mu(\mathcal X\setminus A)}$
\begin{align*}

\mathbf P(M(A_i)=k_i,1\le i\le n, N=m) &=\mathbf P(M(A_i)=k_i,1\le i\le n|N=m)\mathbf P(N=m)\\
&=\prod_{i=1}^ne^{-\mu(A_i)}\frac{\mu(A_i)^{k_i}}{k_i!}\cdot e^{-\mu(\mathcal X\setminus A)}\frac{\mu(\mathcal X\setminus A)^{m-k}}{(m-k)!}.
\end{align*}
Now simply sum over all $m\ge k$ and use the fact that $\sum_{m=k}^\infty \frac{x^{m-k}}{(m-k)!}=e^x$.