I have to prove that a Borel measure $\mu$ on $\mathbb{R}^+$ such that $$ \int_0^\infty (z\wedge z^2)\mu(dz)<\infty $$ is a Levy measure.
MY ATTEMPT:
We need to check that $\int_0^\infty(1\wedge z^2)\mu(dz)<\infty$; in particular we have: $$ \begin{split} \int_0^\infty(1\wedge z^2)\mu(dz) & =\int_0^1(1\wedge z^2)\mu(dz)+\int_1^\infty(1\wedge z^2)\mu(dz)\\ & \leq \int_0^1\mu(dz)+\int_1^\infty(z\wedge z^2)\mu(dz)\\ & =\mu([0,1))+\int_1^\infty(z\wedge z^2)\mu(dz)<\infty \end{split} $$ Does it make sense?
Your argument is not quite correct as Lévy measures may be infinite on $(0,1]$.
Notice that for $x>0$ $$1\wedge x^2 \leq \big(x\wedge x^2\big)\mathbb{1}_{(0,1]}(x) + \big(x\wedge x^2\big)\mathbb{1}_{(1,\infty)}(x)=x\wedge x^2 $$