Equivalence of a Levy measure

Question

We know that a measure $\nu$ is a Levy measure if: \begin{equation}\label{1}\tag{1} \nu(\{0\}),\quad \int_{|x|<1}|x|^{2} \nu (dx) <\infty, \quad \int_{|x| \geq 1} \nu (dx)<\infty. \end{equation} It straightforward to show that \begin{equation}\tag{*} \int \frac{|x|^2}{1+ |x|^2} \nu (dx) < \infty \end{equation} But how to show that (*) implies (\ref{1}) ?

Answer 1

For $|x| \leq 1$, $$|x|^2 - \frac{|x|^2}{1 + |x|^2} = \frac{|x|^4}{1 + |x|^2} \leq \frac{|x|^2}{1 + |x|^2} \in L^1(\nu).$$ So the first integral is finite. For $|x| \geq 1$, $$1 - \frac{|x|^2}{1 + |x|^2} = \frac{1}{1 + |x|^2} \leq \frac{|x|^2}{1 + |x|^2} \in L^1(\nu).$$ So the second integral is finite.

However, we might have $\nu(\{0\}) = \infty$. For example, let $\nu(\{0\}) = \infty$, $\nu(\{0\}^c) = 0$.