If $\nu$ is a Levy measure, how to show that $\nu( (-\epsilon,\epsilon)^c ) <\infty$

Question

We know that the Levy measure on borelians is a measure $\nu$ such that \begin{equation} \int \frac{|x|^2}{1+ |x|^2} \nu (dx) < \infty \end{equation} We can show that this is equivalent to: \begin{equation} \nu(\{0\}),\quad \int_{|x|<1}|x|^{2} \nu (dx) <\infty, \quad \int_{|x| \geq 1} \nu (dx)<\infty. \end{equation}

So I am interested to show that $$\nu( (-\epsilon,\epsilon)^c )< \infty, \quad \forall \epsilon, \,\, 0< \epsilon < 1.$$.

Notice that $$\nu( (-\epsilon,\epsilon)^c ) = \int_{[|x|> \epsilon]}\nu(dx)= \int_{\mathbb R} \chi_{[|x|> \epsilon]} \nu(dx) $$

From here, I'm failing to show the integral is finite.

Answer 1

$$\nu((-\epsilon,\epsilon)^{c})= \nu((-1,1)^{c})+\int_{\epsilon \leq |x| < 1}\nu (dx)$$ and $$\int_{\epsilon \leq |x| < 1}\nu (dx)\leq \frac 1 {\epsilon^{2}}\int_{\epsilon \leq |x| < 1}|x|^{2}\nu (dx)\leq \frac 1 {\epsilon^{2}}\int_{|x| < 1}|x|^{2}\nu (dx)$$