For every \(f \in L^{1}(\RR^n)\), \[ \lim_{h \to 0} \int_{\RR^n}\abs{f(x+h)-f(x)}\,dx=0 . \]
Let \(\varepsilon>0\). Since compactly supported continuous functions are dense in \(L^1(\RR^n)\), choose \(g\in C_c(\RR^n)\) with \(\norm{f-g}_{L^1}<\varepsilon/3\). By translation invariance of Lebesgue measure, \(\int\abs{f(x+h)-g(x+h)}\,dx=\norm{f-g}_{L^1}<\varepsilon/3\) for every \(h\). The function \(g\) is uniformly continuous and, for \(\abs{h}\le1\), \(x\mapsto g(x+h)-g(x)\) vanishes outside the compact set \(K=\supp g+\overline{B(0,1)}\), so \(\int\abs{g(x+h)-g(x)}\,dx\le \vol(K)\sup_x\abs{g(x+h)-g(x)}<\varepsilon/3\) for \(\abs{h}\) small enough. The triangle inequality then gives \(\int\abs{f(x+h)-f(x)}\,dx<\varepsilon\) for \(\abs{h}\) small enough.