The compound Poisson distribution is defined as(see Levy processes and infinitely divisible distributions page: 18):

Let $c>0$ and $\sigma$ be a measure on $\mathbb{R}$ with $\sigma(\{0\})=0$, a compound distribution $\mu$ on $\mathbb{R}$ is called a compound Poisson if its characteristic function is

$$ \Phi_{\mu}(t) = e^{c(\Phi_\mu - 1)} $$

My question is: Why we need the condition $\sigma(\{0\})=0$?

Similarly, in the book Probability: a comprehensive course (page: 333) stated the following theorem:

A probability measure $\mu$ on $\mathbb{R}$ is infinitely divisible if and only if there exists a sequence of finite measures $(\nu_n)_{n\in\mathbb{N}}$ on $\mathbb{R} - \{0\}$ such that $CPoi_{\nu_n} \rightarrow \mu(n\rightarrow\infty)$, where $CPoi_{\nu_n}$ denote the compound distribution with intensity $\nu_n$.

My question is: Why $\nu_n$ is defined on $\mathbb{R}-\{0\}$ other than $\mathbb{R}$?

The compound Poisson distribution is defined as(see Levy processes and infinitely divisible distributions page: 18):

Let $c>0$ and $\sigma$ be a measure on $\mathbb{R}$ with $\sigma(\{0\})=0$, a compound distribution $\mu$ on $\mathbb{R}$ is called a compound Poisson if its characteristic function is

$$ \Phi_{\mu}(t) = e^{c(\Phi_\sigma - 1)} $$

My question is: Why we need the condition $\sigma(\{0\})=0$?

Similarly, in the book Probability: a comprehensive course (page: 333) stated the following theorem:

A probability measure $\mu$ on $\mathbb{R}$ is infinitely divisible if and only if there exists a sequence of finite measures $(\nu_n)_{n\in\mathbb{N}}$ on $\mathbb{R} - \{0\}$ such that $CPoi_{\nu_n} \rightarrow \mu(n\rightarrow\infty)$, where $CPoi_{\nu_n}$ denote the compound distribution with intensity $\nu_n$.

My question is: Why $\nu_n$ is defined on $\mathbb{R}-\{0\}$ other than $\mathbb{R}$?