Let \(f\colon\RR^n\to\CC\) be uniformly continuous with \(f\in L^1(\RR^n)\). Then \(f(x) \to 0\) as \(\abs{x}\to \infty\).
Suppose not. Then there exist \(\varepsilon>0\) and points \(x_k\in\RR^n\) with \(\abs{x_k}\to\infty\) and \(\abs{f(x_k)}\geq\varepsilon\) for all \(k\). By uniform continuity, choose \(\delta>0\) such that \(\abs{x-y}<\delta\) implies \(\abs{f(x)-f(y)}<\varepsilon/2\); then \(\abs{f}\geq\varepsilon/2\) on each ball \(B(x_k,\delta)\). Passing to a subsequence, we may assume \(\abs{x_{k+1}}\geq\abs{x_k}+2\delta\), so the balls \(B(x_k,\delta)\) are pairwise disjoint. Hence \[ \int_{\RR^n}\abs{f}\geq\sum_{k}\int_{B(x_k,\delta)}\abs{f}\geq\sum_k\frac{\varepsilon}{2}\vol\qty{B(0,\delta)}=\infty, \] contradicting \(f\in L^1(\RR^n)\).