For $f\in L^1(\RR^d)$, $\int_{\abs{x}\ge r}\abs{f}\to0$ as $r\to\infty$.Suggest a slogan
Let \(d\geq1\), let \(m\) be Lebesgue measure on \(\RR^d\), and let \(f\colon\RR^d\to\CC\) be integrable with respect to \(m\). For \(r>0\), let \(B_r(0)\coloneqq\theset{x\in\RR^d\suchthat\abs{x}<r}\). Then \[ \lim_{r\to\infty}\int_{\RR^d\setminus B_r(0)}\abs{f}\,dm=0. \] Equivalently, for every \(\varepsilon>0\) there exists \(r>0\) such that \[ \int_{\RR^d\setminus B_r(0)}\abs{f}\,dm<\varepsilon. \]