Let $H$ be a $\mathbb R$-Hilbert space, $\mu$ be a finite measure on $\mathcal B(H)$ with $\mu(\{0\})=0$ and $(L_t)_{t\ge0}$ be a $H$-valued càdlàg Lévy process on a probability space $(\Omega,\mathcal A,\operatorname P)$ with $$L_t=\sum_{i=1}^{N_t}Y_i\;\;\;\text{for all }t\ge0\tag1$$ for some $H$-valued independent identically distributed process $(Y_n)_{n\in\mathbb N}$ on $(\Omega,\mathcal A,\operatorname P)$ with $Y_1\sim\lambda^{-1}\mu$ for some $\lambda>0$ and some càdlàg Poisson process $(N_t)_{t\ge0}$ on $(\Omega,\mathcal A,\operatorname P)$.

Let $t\ge0$ and $B\in\mathcal B(H\setminus\{0\})$. I would like to show that $$\left|\left\{s\in[0,t]:\Delta L_s\in B\right\}\right|=\sum_{i=1}^{N_t}1_B(Y_i),\tag2$$ where $$\Delta L_s:=L_s-L_{s-}=L_s-\lim_{r\to s-}L_r\;\;\;\text{for }s\ge0.$$ How can we do that?

*Remark*: Since the measure is not involved in $(2)$ it might be unimportant for the claim, but we may note that $$Z_n:=\sum_{i=1}^nY_i\;\;\;\text{for }n\in\mathbb N$$ is a time-homogenous Markov chain and hence $$L_t=Z_{N_t}\;\;\;\text{for all }t\ge0$$ is a time-homogeneous Markov process.