Let $f:\Bbb R\to \Bbb R$ be an $L^1$ function and $f_\epsilon(x):=f(x+\epsilon)$, $\mu$ is the Lebesgue measure, prove that $$\lim_{\epsilon\to 0}\int|f_\epsilon-f|\mathrm d\mu=0.$$
I tried to first deal with the simplest case, that is, when $f=\chi_E$ for some measurable subset $E$. I even managed to reduce these conditions further to stronger ones: now I need only to show that for $G\in G_\delta,\mu(G)<\infty$, $\mu(G\setminus (G-\epsilon))+\mu((G-\epsilon)\setminus G))\to 0$ as $\epsilon\to 0$, but this looks quite troublesome.
Am I on the right track? (I guess so coz once I'm done with the characteristic functions I can manage to prove the desired general cases.). If not, is there any other, preferably more elegant and direct approach to this problem? Thanks!
Hint: $C_c(\Bbb R)$ (the space of continuous functions of compact support) is $\| \cdot \|_1$-dense in $L^1(\Bbb R)$. Try to prove it for functions in this space. Given that, proceed as follows:
Let $f \in L^1(\Bbb R)$ and $\eta > 0$. Then, there is $\varphi_{\eta} \in C_c(\Bbb R)$, such that $\| f - \varphi_{\eta} \|_1 < \eta /3$. Denote by $\tau_{\epsilon}g(x) = g(x + \epsilon)$. Then,
$$\|f - \tau_{\epsilon}f\|_1 \le \|f - \varphi_{\eta} \|_1 + \|\varphi_{\eta} - \tau_{\epsilon} \varphi_{\eta} \|_1 + \|\tau_{\epsilon} \varphi_{\eta} - \tau_{\epsilon} f \|_1$$
now, note that $\|\tau_{\epsilon} \varphi_{\eta} - \tau_{\epsilon} f \|_1 = \|f - \varphi_{\eta}\|_1$, and use the first result to conclude