If $\int_{\mathbb{R}} \min(1,x^2)\nu(dx)<\infty$ then $\nu(A)< \infty$ for all A if $0$ is in the interior of $A^c$

Question

I am currently reading "Fluctuations of Lévy Processes with Applications" by Kyprianou, where he mentions (p.36 at the top) that if $\nu$ is a measure on $\mathbb{R}\setminus\{0\}$ satisfying $\int_{\mathbb{R}} \min(1,x^2)\nu(dx)<\infty$ then it implies that $\nu(A)< \infty$ for all Borel sets A if $0$ is in the interior of $A^c$. However I don't see why this is true ; any help is much appreciated.

Answer 1

By assumption $A\cap\{x\in\mathbb{R}:|x|<\varepsilon\}=\emptyset$ for $\varepsilon>0$ small enough. Hence \begin{align} \infty>\int_A1\wedge x^2\,dx&=\int_{A\cap\{x:|x|<\varepsilon\}}1\wedge x^2\,\nu(dx)+\int_{A\cap\{x:|x|\geq\varepsilon\}}1\wedge x^2\,\nu(dx)\\ &\geq 0 + (1\wedge \varepsilon^2)\int_{A\cap\{x:|x|\geq\varepsilon\}}\,\nu(dx)\\ &=\big(1\wedge \varepsilon^2\big)\nu(A\cap\{x:|x|\geq\varepsilon\})\\ &=\big(1\wedge\varepsilon^2\big)\nu(A) \end{align}