Problem in derivation of Poisson Process I have a problem in understanding the derivation of the  Poisson Process in the second edition of Klenke "Probability Theory". At the end of page 124 (screenshot), he explains the meaning of P5 that is $\limsup_{\epsilon\to 0}P[N_\epsilon\geq 2]=0$, by defining 
$$\lambda=\limsup_{\epsilon\to 0}P[N_\epsilon\geq 2].$$
And writes: For any $n\in\mathbb{N}$ and $\epsilon>0$, we have
$$
P[N_{2^{-n}}\geq 2]\geq\lfloor\frac{2^{-n}}{\epsilon}\rfloor P[N_{\epsilon}\geq2]-\lfloor\frac{2^{-n}}{\epsilon}\rfloor^{2}P[N_{\epsilon}\geq2]^{2}.
$$
Hence, 
$$2^n P[N_{2^{-n}}\geq 2]\geq \lambda-2^{-n}\lambda^2\xrightarrow{n\to\infty} \lambda.$$
I cannot fully derive none of the above inequalities. 
For the first one this is what I did:
Let $R_{n,\epsilon}\triangleq\lfloor\frac{2^{-n}}{\epsilon}\rfloor$
and $\underline{R}_{n,\epsilon}\triangleq2^{-n}\mod\epsilon$, so
that $2^{-n}=R_{n,\epsilon}\epsilon+\underline{R}_{n,\epsilon}$,
with $0\leq\underline{R}_{n,\epsilon}<\epsilon.$ We use the first
three proprieties P1, P2, and P3 to derive the above relation as follows,
$$\begin{align*}
P[N_{2^{-n}}\geq2]= & P\left[N_{\bigcup_{r=0}^{R_{n,\epsilon}-1}(r\epsilon,(r+1)\epsilon]\cup(R_{n,\epsilon}\epsilon,2^{-n}]}\geq2\right]=P\left[\sum_{r=0}^{R_{n,\epsilon}-1}N_{(r\epsilon,(r+1)\epsilon]}+N_{(R_{n,\epsilon}\epsilon,2^{-n}]}\geq2\right]\\
\geq & P\left[\bigcup_{r=0}^{R_{n,\epsilon}-1}\left\{ N_{(r\epsilon,(r+1)\epsilon]}\geq2\right\} \right]=R_{n,\epsilon}P\left[N_{(r\epsilon,(r+1)\epsilon]}\geq2\right]=\lfloor\frac{2^{-n}}{\epsilon}\rfloor P\left[N_{\epsilon}\geq2\right]
\end{align*},$$
Which only gives the first term of the inequality but not the second squared term. Given the second line we do not really need it as it goes to 0 by taking $n$ to infinity, but I am afraid I might have done a mistake in the above result. 
I appreciate if someone could verify it and also explains how the second relation after the "hence" comes about (I can see that but not exactly).
 A: For notational convenience, write $A_r := \left\{ N_{(r\epsilon,(r+1)\epsilon]}\geq2\right\} $. The argument for the first inequality goes like this:
$$
P(N_{2^{-n}}\ge2)\stackrel{(1)}\ge 
P(\bigcup_r A_r)\stackrel{(2)}\ge \sum_r P(A_r) - \sum_{r<s}P(A_r\cap A_s)
$$
In the above, (1) follows because the event $\bigcup_rA_r$ implies the event $\{N_{2^{-n}}\ge2\}$. Inequality (2) is a Bonferroni inequality, namely the one obtained by truncating inclusion-exclusion at two terms. Next, replace $\sum_r P(A_r)$ with $R_{n,\epsilon}P(A_0)$ as you've done. Simplify $\sum_{r<s}P(A_r\cap A_s)$ to $\sum_{r<s}P(A_r)P(A_s)=\sum_{r<s}P(A_0)^2$ by properties P3 and P2; and clearly $\sum_{r<s}P(A_0)^2$ is bounded above by $R_{N,\epsilon}^2$ times $P(A_0)^2$, which completes the proof.
As for the second inequality, you multiply the first inequality through by $2^n$ (for fixed $n$), then apply $\limsup$, which preserves the inequality. Then bound the RHS of the result using the inequality $$\limsup (a_\epsilon-b_\epsilon)\ge \limsup a_\epsilon - \limsup b_\epsilon,\tag3$$ (which follows from $\limsup (x_\epsilon + y_\epsilon)\le \limsup x_\epsilon + \limsup y_\epsilon$). Finally use P5 to argue that the limsups on the RHS of (3) are equal to the stated values.
For example, to see why
$$\limsup_\epsilon 2^n\epsilon\lfloor 2^{-n}/\epsilon\rfloor \frac1\epsilon P(N_\epsilon\ge2)=\lambda,$$
you can use 
 the inequality $x-1<\lfloor x\rfloor\le x$ to argue that $2^n\epsilon\lfloor 2^{-n}/\epsilon\rfloor\to1$ as $\epsilon\to0$. (Remember $n$ is still fixed.)
