Let \(f\in L^1(\RR^n)\) and \(\varepsilon> 0\), and let \(B_N\coloneqq\theset{x\in\RR^n\suchthat\abs{x}<N}\).
Small tails. There exists \(N\in\NN\) such that \(\int_{\RR^n\setminus B_N} \abs{f} < \varepsilon\).
Absolute continuity. There exists \(\delta>0\) such that \(\int_E \abs{f} < \varepsilon\) for every Lebesgue measurable \(E\subseteq\RR^n\) with \(m(E) < \delta\).
For small tails, \(\abs{f}\chi_{\RR^n\setminus B_N}\to 0\) pointwise as \(N\to\infty\) and is dominated by \(\abs{f}\in L^1\), so \(\int_{\RR^n\setminus B_N}\abs{f}\to 0\) by dominated convergence.
For absolute continuity, put \(f_M\coloneqq\min(\abs{f}, M)\) for \(M\in\NN\). By monotone convergence \(\int(\abs{f} - f_M)\to 0\), so there is \(M\) with \(\int(\abs{f} - f_M)<\varepsilon/2\). With \(\delta\coloneqq\varepsilon/(2M)\) and \(m(E)<\delta\), \[ \int_E\abs{f}\leq\int(\abs{f} - f_M) + \int_E f_M < \frac{\varepsilon}{2} + M\,m(E) < \varepsilon. \]