Problem statement: Assume $N=\sum_{i=-\infty}^{\infty} \delta_{\Gamma_i}$ is a homogeneous Poisson point process (hPPP) on the real line with positive rate $\lambda$. Let $N$ be independent of $(U_n)_{n\in\mathbb{Z}}$ which are i.i.d. uniform on the interval $\langle 0,1\rangle$. Now consider the process $N'=\sum_{\Gamma_i >0} \delta_{\Gamma_i} \boldsymbol{\mathbb{1}}_{U_i < e^{-\Gamma_i}}$ which is also a Poisson process, but on $\langle 0,\infty\rangle$. Calculate $\mathbb{E} N' (\langle 0,a\rangle)$ for $a>0$.
My attempt: I will denote with $\mathcal{L}$ the Lebesgue measure, with $f_U$ the pdf of the uniform random variable on $\langle 0,1 \rangle$ and with $f_Y$ the pdf od $\Gamma_i$. Since $N$ is a hPPP with rate $\lambda$, then its intensity measure is $\mu = \lambda \mathcal{L}$. Since $N'$ is a thinning of $N$, then it is a PPP with intensity measure $\mu' = \lambda \mathcal{L}\mathbb{P}(C)$, where $C=\{ U_i < e^{-\Gamma_i} \}$. It follows that
$$ \begin{align*} \mathbb{E} N' (\langle 0,a\rangle) &= \mu' (\langle 0,a\rangle) \\ &= \lambda a \mathbb{P} (U_i < e^{-\Gamma_i}) \\ &=\lambda a \int_0^{\infty} \int_0^{e^{-y}} f_U (u) f_Y(y) \, \mathrm{d}u \, \mathrm{d}y \\ &= \lambda a \int_0^{\infty} e^{-y} f_Y(y) \, \mathrm{d}y \end{align*} $$
This is where I am stuck because I feel I need to have a closed formula for the expectation, but I'm not sure where to go from here. Are my calculations even correct? How to proceed?
We note the following
This is essentially a restatement of Poisson thinning, hence we skip the proof. Now note that
\begin{align*} N'([0, a]) &= \sum_{i\in\mathbb{Z}} \mathbf{1}_{\{ \Gamma_i \in (0, a], \, U_i < e^{-\Gamma_i} \}} \\ &= \int_{\mathcal{S}} \mathbf{1}_{\{ x \in (0, a], \, u < e^{-x} \}} \, M(\mathrm{d}x\mathrm{d}u). \end{align*}
So by taking expectation, we get
\begin{align*} \mathbb{E}[N'([0, a])] &= \int_{\mathcal{S}} \mathbf{1}_{\{ x \in (0, a], \, u < e^{-x} \}} \, \lambda\mathrm{d}x\mathrm{d}u \\ &= \lambda \int_{0}^{a} \biggl( \int_{0}^{e^{-x}} 1 \, \mathrm{d}u \biggr) \, \mathrm{d}x \\ &= \lambda \int_{0}^{a} e^{-x} \, \mathrm{d}x \\ &= \lambda (1 - e^{-a}). \end{align*}
Regarding OP's approach, I can see that the equality doesn't hold:
$$\mu' (\langle 0,a\rangle) = \lambda a \mathbb{P} (U_i < e^{-\Gamma_i}) $$
I guess OP tried to use $\Gamma_i$ to denote a typical Poisson point here, but this is already problematic. For a random point to qualify as "a typical point" of a homogeneous process on $\mathbb{R}$, it would be a "uniformly chosen point in $\mathbb{R}$", which simply doesn't make sense.
On the other hand, since we are restricting the range to $[0,a]$, it now makes sense talk about a typical Poisson point on $[0, a]$, which is simply a uniformly chosen point in $[0, a]$. So by letting $\Gamma \sim \operatorname{Uniform}([0, a])$, we get
\begin{align*} \mu' (\langle 0,a\rangle) &= \lambda a \mathbb{P} (U < e^{-\Gamma}) \\ &= \lambda a \int_{0}^{a} \int_{0}^{1} \mathbf{1}_{\{ u < e^{-x}\}} \, \mathrm{d}u \frac{\mathrm{d}x}{a}, \end{align*}
which is essentially the same as what is obtained in my answer.