Well-definedness of the characteristic function of a compound Poisson variable I am reading about compound Poisson variables and cannot get through the following statement.
Let $\nu$ be a non-zero finite measure on $\mathbb{R}\setminus\{0\}$. Assume that
$$\int \min\{|u|,1\}\nu(du) < \infty. \tag{$\ast$}$$
We say that the distribution of a random variable $S$ belongs to a class $\mathcal{P}_{1,0}$ if its characteristic function writes as
$$\mathbb{E}e^{itS} = \exp\left\{ \int (e^{itu}-1)\nu(du) \right\}$$
for some measure $\nu$ satisfying $(\ast)$. Note that for any real $t$ the integral on the right-hand side is well defined since $|e^{itu}-1| \leq \min\{|t||u|, 2\}$.
Question 1: It is easy to see that $|e^{itu} -1|$ is bounded by 2. How to derive the other bound $|t||u|$ and what is its significance?
Question 2: How to make use of $(\ast)$ and actually show that the integral is well defined?
$$\int|e^{itu}-1|\nu(du) \leq \int \min\{|t||u|, 2\} \nu(du) \leq \dots$$
Thank you!
Regards,
Ivan

If $a\in\mathbb{R}$ then: $$\left|e^{ia}-1\right|=\left|\cos a-1+i\sin a\right|=\sqrt{\left(\cos a-1\right)^{2}+\sin^{2}a}=\sqrt{2-2\cos a}=\sqrt{2-2\left(1-2\sin^{2}\left(\frac{1}{2}a\right)\right)}=2\left|\sin\left(\frac{1}{2}a\right)\right|\leq\left|a\right|$$
That could help for your first question. 
I don't see its functionality. $\left|e^{itu}-1\right|\leq2$ is enought to ensure that: $$\int\left|e^{itu}-1\right|\nu\left(du\right)\leq2\nu\left(\mathbb{R}\backslash\left\{ 0\right\} \right)<\infty$$
