I would love to have the following result, however I got no clue if it is even true!
Let $B_n:=\{y:\varepsilon_n<|y|\leq\tilde{\varepsilon}_n\}$ for some sequences $1\geq\tilde{\varepsilon}_1>\varepsilon_1>\tilde{\varepsilon}_2>\varepsilon_2>\dots$ and both $\varepsilon_n,\tilde{\varepsilon}_n\downarrow0$. Hence we have $\mathbb{R}^d\setminus\{0\}\supseteq B_n\to\emptyset$. Let $\nu$ be a Levy measure on $\mathbb{R}^d\setminus\{0\}$, that is a $\sigma$-finite measure s.t. $\int_{\mathbb{R}^d\setminus\{0\}}1\wedge|y|^2\mathrm{d}\nu(y)<\infty$. Then
\begin{align} \int_{B_n}|y|\mathrm{d}\nu(y)\to0 \end{align}
Here I cannot apply monotone convergence, dominated convergence or the limsup-part of Fatous Lemma due to a lack of monotonicity or an integrable bound. Any suggestions how to get convergence here? (Sry for potentially wrong tags!)
Not true.
Take $d=1$ and $d\nu=|y|^{-2}\chi_{\{0<|y|<1\}}dy$. Then $\int_{B_n}|y|d\nu=2\int_{\epsilon_n}^{\tilde{\epsilon_n}}y^{-1}dy=2\ln(\frac{\tilde{\epsilon_n}}{\epsilon_n})$, so to get failure of the claim it suffices to choose $\epsilon_n,\tilde{\epsilon_n}$ such that $\lim_n\frac{\tilde{\epsilon_n}}{\epsilon_n}\neq1$.