Lemma 2.2 on Poisson Random Measures from Kyprianou's Introductory Lectures on Fluctuations of Levy Processes I have several problems to understand the proof of the Lemma 2.2 (page 37) from Kyprianou's
"Introductory Lectures On Fluctuations Levy Processes with Applications". We assume:
Let $Z_i: \Omega \to \mathbb{R} \backslash \{0\}, i \ge 1$ be a sequence
of independent and identically distributed (iid)
random variables with common distribution function $F$. Further let
$T_i: \Omega \to \mathbb{N} \cup \infty, i \ge 0$ be the arrival times
in the real valued Poisson process $N =  \{N_t : t \ge 0 \}$ with
parameter $\lambda >0$.
Suppose we fix an element $A \in 
\mathcal{B}[0, \infty) \otimes \mathcal{B}(\mathbb{R} \backslash \{0\}) $
in the product Borel sigma algebra. Define the new random variable
$$ N(A):= \#\{i \ge 0: (T_i, Z_i) \in A \} 
= \sum_{i=0}^{\infty} 1_{(T_i, Z_i) \in A}. $$
where $\sum_{i=0}^{\infty} 1_{(T_i, Z_i) \in A}$ is the sum of indicator functions
$1_{(T_i, Z_i) \in A}$.
I not understand an argument in the proof of following lemma:

Lemma 2.2. Let $k \in \mathbb{N}, k \le 1$. If $A_1, ...,A_k$ are disjoint sets in
product Borel algebra
$\mathcal{B}[0, \infty) \otimes \mathcal{B}(\mathbb{R} \backslash \{0\}) $
then $N(A_1), ...,N(A_k)$ are mutually independent and Poisson distributed
with parameters $\lambda_i := \lambda \int_{A_i} at \otimes F(dx)$, respectively.


Proof. First recall a classic result concerning the Poisson process $\{N_t : t \ge 0\}$.
That is, the law of $\{T_1, ..., T_n \}$ conditional on the event $\{N_t = n \}$
is the same as the law of an ordered independent sample of size $n$ from the
uniform distribution on $[0, t]$. (Exercise 2.2 has the full details).


This latter fact together
with the fact that the variables $\{Z_i : i = 1, ..., k\}$ are independent and
identically distributed with common law $F$ implies that conditional on $\{N_t = n \}$,
the joint law of the pairs $\{(T_i, Z_i) : i = 1, ..., k \}$ is that of $n$
independent bivariate random variables with common distribution
$t^{-1}ds \otimes F(dx)$ on $[0, t] \otimes \mathbb{R}$
ordered in time. In particular, for any
$A \in \mathcal{B}[0, t] \otimes \mathcal{B}(\mathbb{R} \backslash \{0\})$,
the random variable $N(A)$ conditional on the event $\{N_t = n \}$ is a Binomial
random variable $B(n,p)$ with probability of success given by
$p= \int_A t^{-1}ds \otimes F(dx)$. ... The rest of the proof is fine

Question: Why
$N(A)$ conditional on the event $\{N_t = n \}$ is a Binomial
random variable $B(n,p)$. Equivalently, why
$$  P(N(A)=k \ \vert \ N_t=n)= \binom{n}{k}p^k(1-p)^{n-k}? $$
Let's try to recapitulate what happens before. By quoted exercise 2.2
the distribution of $\{T_1, ..., T_n \}$ conditional on the event $\{N_t = n \}$
has the same ditribution as an ordered independent sample of size $n$ from the
uniform distribution on $[0, t]$.
That means that for every set $E \in \mathcal{B}([0, \infty)^n)$, we have
$$ P((T_1, ..., T_n) \in E \ \vert \ N_t = n) = \int_E \frac{n!}{t^n} \cdot 1_{(0 \le t_1 \le ... \le t_n \le t)} dt_1 \cdot ... \cdot dt_n $$
where the right hand side should represent the distribution of an ordered independent
sample of size $n$
taken from the uniform distribution on $[0, t]$. It seems that that the factor
$n!$ catches all possible orderings of an arbitrary $n$-tuple
$(t_1,..., t_n)$.
What I not understand at that point is what is the diffrence between the right
hand side distribution and the usual distribution of the NOT ordered
product of independent and identically distributed
random variables $t_i: \Omega \to [0, \infty)$ with
uniform distribution on $[0, t]$.
In this case we should have for every $E \in \mathcal{B}([0, \infty)^n)$
$$ P_{(t_1,..., t_n)}(E) = \int_{pr_1(E)} \frac{dt_1}{t^{-1}} \cdot ...
  \int_{pr_n(E)} \frac{dt_n}{t} = \prod_{i=1}^n 
\int_{pr_i(E)} \frac{dt_i}{t} $$
where $pr_i:[0, \infty)^n \to [0, \infty) $ is the canonical projection function onto
$i$-th coordinate.
So the first problem is that I not understand why we need to introduce this
ordering $t_1 \le ... \le t_n$ of the $t_i$'s at
the right side of the equation above. Shouldn't the equality
$$   \prod_{i=1}^n 
\int_{pr_i(E)} \frac{dt_i}{t} =             
\int_E \frac{n!}{t^n} 1_{(0 \le t_1 \le ... \le t_n \le t)} dt_1 \cdot ... \cdot dt_n  $$
always hold?
Next we consider the common distribution of pairs
$\{(T_i, Z_i) : i = 1, ..., k \}$ conditional on $\{N_t = n \}$. Taking any
$E \in (\mathcal{B}[0, \infty) \otimes \mathcal{B}(\mathbb{R} \backslash \{0\}))^n)$
the proof states that
$$  P(((T_1,Z_1), ..., (T_n, Z_n)) \in E \vert N_t = n) = 
\int_E \frac{n!}{t^n} 1_{(0 \le t_1 \le ... \le t_n \le t)} \cdot (dt_1 \otimes F(dx_1)) \cdot ...
\cdot ( dt_n \otimes F(dx_n)) $$
Now the question is why this impies that $N(A)$ conditional on the event $\{N_t = n \}$
(so in other words the distribution $P_{N(A)}( \bullet \ \vert \ N_t = n)$) is a Binomial
random variable $B(n,p)$. Is it obvious to see?
 A: For your first question, the two are actually different. Note that we actually want to work with $E = \times_{i=1}^n E_i \in (\mathcal{B}([0,t])\otimes\mathcal{B}(\mathbb{R}\setminus\{0\}))^n$ because that is the collection of events on which we get the desired result by conditioning on $N_t$ (look at the last paragraph of the proof you copied and notice what space $A$ lies in).
An ordered uniform distribution is simply a uniform distribution conditioned on the event $\{0\leq T_1 \leq T_2 \leq \cdots \leq T_n \leq t\}$. So if we can write $E = \times_{i=1}^n E_i$, then it is not necessarily true that
$$\prod_{i=1}^n \int_{E_i}\frac{dt_i}{t} = \int_E\frac{n!}{t^n}1_{\{0\leq t_1 \leq t_2 \leq \cdots \leq t_n\leq t\}}\,dt_1\cdots dt_n,$$
to see why this is the case consider when $n = 2$, $E_1 = \{T_1 > 5\}$ and $E_2 = \{T_2 < 2\}$ for $t = 10$. Then the left-hand side of the expression above is equal to $\frac{1}{10}$ while the right side is $0$. The reason we are working with the ordered uniform distribution instead of the ordinary uniform distribution is simply because we implicitly assumed at the beginning that $ 0\leq T_1 \leq T_2 \leq \cdots$, and conditioning on $N_t =n$ requires $T_n \leq t$.
For your second question, first note that for $A \in \mathcal{B}([0,t])\otimes\mathcal{B}(\mathbb{R}\setminus\{0\})$,
$$N(A) = \left|\{i \leq N_t: (T_i,Z_i) \in A\}\right|.$$
Now, working with ordered random variables is tricky, so let's find a way to rewrite the distribution of $((T_i,Z_i))_{i=1,\dots,n}$ in terms of unordered, i.i.d. random variables. Let $\mathcal{E}$ be the set of permutations, $\sigma$, of $\{1,\dots,n\}$. Let $E$ be defined as above. The computation below is complicated, but I'll walk through it after the computation.
$$P(((T_1,Z_1),\cdots,(T_n,Z_n))\in E|N_t = n) = \int_E \frac{n!}{t^n} 1_{\{0\leq t_1 \leq t_2 \leq \cdots \leq t_n\leq t\}}(dt_1\otimes F(dx_1))\cdots (dt_n\otimes F(dx_n)) \\= \frac{n!}{t^n} \int_E 1_{\{0\leq t_1 \leq t_2 \leq \cdots \leq t_n\leq t\}}(dt_1\otimes F(dx_1))\cdots (dt_n\otimes F(dx_n)) \\= \sum_{\sigma \in \mathcal{E}} \frac{1}{t^n}\int_E 1_{\{0\leq t_{\sigma(1)} \leq t_{\sigma(2)} \leq \cdots \leq t_{\sigma(n)}\leq t\}}(dt_1\otimes F(dx_1))\cdots (dt_n\otimes F(dx_n))\\= \frac{1}{t^n}\int_E \sum_{\sigma \in \mathcal{E}} 1_{\{0\leq t_{\sigma(1)} \leq t_{\sigma(2)} \leq \cdots \leq t_{\sigma(n)}\leq t\}}(dt_1\otimes F(dx_1))\cdots (dt_n\otimes F(dx_n))\\= \frac{1}{t^n}\int_E (dt_1\otimes F(dx_1))\cdots (dt_n\otimes F(dx_n))\\=\prod_{i=1}^n \frac{1}{t} \int_{E_i} dt_i\otimes F(dx_i)\\= \prod_{i=1}^n P((U_i,Z_i) \in E_i),$$
where $\{U_i\}_{i=1,\dots,n}$ is a collection of i.i.d. uniform$[0,t]$ random variables that are independent of $\{Z_i\}_{i\in\mathbb{N}}$. What did I do there? In the first equality, I just wrote the probability as an integral and in the second equality I removed some constants from the integral. The third equality is the tricky part. Notice that $\otimes_{i=1}^n (dt_i\otimes F(dx_i))$ is a measure describing the distribution of i.i.d. random variables. Note also that permuting i.i.d. random variables does not change their joint distribution. So, for any permutation $\sigma \in \mathcal{E}$, $\int_{E} 1_{\{0\leq t_1\leq \cdots\leq t_n\leq t\}} \otimes_{i=1}^n(dt_i\otimes F(dx_i)) = \int_{E} 1_{\{0\leq t_{\sigma(1)}\leq \cdots\leq t_{\sigma(n)}\leq t\}} \otimes_{i=1}^n(dt_i\otimes F(dx_i))$. After that it's just simple algebra.
Note that $N(A) = k$ if and only if there exist exactly $k$ values of $i$ such that $(T_i,Z_i) \in A$. However, we can use the calculation above to write this in terms of the i.i.d. random variables $(U_i,Z_i)$ which are in $A$ with probability $p = \frac{1}{t}\int_Adt_1\otimes F(dx_1)$. Let $\mathscr{E}$ be the set of subsets of size $k$ in $\{1,\dots,n\}$ and note that $|\mathscr{E}| = \binom{n}{k}$. Then,
$$P(N(A) = k|N_t = n) = \sum_{I \in \mathscr{E}} P(((T_1,Z_1),\dots,(T_n,Z_n)) \in (\times_{i \in I} A)\times(\times_{i\notin I} A^c))\\=\sum_{I \in \mathscr{E}} \left(\prod_{i \in I} P((U_i,Z_i) \in A)\right)\left(\prod_{i \notin I} P((U_i,Z_i) \notin A)\right) \\= \binom{n}{k}p^k(1-p)^k,$$
as desired. Hope that helps! Feel free to ask if you have any questions, there are some messy computations up there. I tried to make the individual steps and overall intuition as simple as possible.
Edit: I completely forgot to give intuition why we get this. As you can see above, the only reason we work with ordered uniform random variables is because in the problem statement, we assumed that $\{T_i\}$ were ordered so we really have no choice. We can't allow $T_2 < T_1$. The trick is to randomly reorder $\{T_i\}$ so that the resulting collection of random variables is i.i.d.. To do that, let $\sigma$ be a permutation of $\{1,\dots,n\}$ chosen uniformly at random. Then we show that $U_i = T_{\sigma(i)}$ forms an i.i.d. sequence. This implies that $\{(U_i,Z_{\sigma(i)})\}$ (which is equal in distribution to $\{(U_i,Z_i)\}$ which I wrote above) are i.i.d., and $N(A)$ counts the number of $\{(T_i,Z_i)\} = \{U_{\sigma^{-1}(i)},Z_i\}$ in $A$. That is, you have $n$ Bernoulli random variables that sum to $k$ and are equal to 1 with probability $p$. Thus, you end up with a Binomial$(n,p)$ random variable.
