$g$ is Lebesgue integrable, $\int |g(x + \epsilon) - g(x)|\,dx \to 0$ as $\epsilon \to 0$? Suppose that $g$ is Lebesgue integrable. How do I see that$$\int |g(x + \epsilon) - g(x)|\,dx \to 0$$as $\epsilon \to 0$?
 A: Hint 1: prove it is true when $g$ is continuous.
Hint 2: show that $$\int |g(x+\epsilon) - g(x)| \, dx \le 2 \|g - \phi\|_{L^1} + \int |\phi(x+\epsilon) - \phi(x)| \, dx$$
whenever $\phi \in L^1$. 
Hint 3: try approximating $g$ by a continuous function $\phi$. 
A: First, prove this is true when $g$ is a step function, meaning that $g=\sum_{i=1}^n c_i\chi_{(a_i,b_i)}$ is a simple function which is constant on several open intervals $(a_i,b_i)$. 
Then, use the fact that step functions are dense in $L_1(\mathbb R)$. This is true since simple functions $f=\sum_i c_i \chi_{E_i}$ are dense, and for any measurable $E$ there exists a finite union of open intervals $A$ for which $\mu(E\Delta A)<\epsilon$, where $\Delta$ is the symmetric difference of sets, $A\Delta B=(A\setminus B)\cup (B\setminus A)$. This last fact is proven in Folland, Proposition 1.20, for example.

In more detail: 
For any function $f$, let $f_\epsilon$ denote the function $f_\epsilon(x)=f(x+\epsilon)$. Given $\delta>0$, choose a step function $\phi$ so that $\|g-\phi\|_{L^1}<\delta/2$. This implies $\|g_\epsilon-\phi_\epsilon\|_{L^1}<\delta/2$ as well. Since $\phi$ is a step function, it is easy to prove $\lim_{n\to\infty}\|\phi_\epsilon - \phi\|_{L^1}=0$. Then,
$$
\limsup_{n\to\infty} \|g_\epsilon - g\|_{L^1}\le \limsup_{n\to\infty}\|g_\epsilon-\phi_\epsilon\|_{L^1}+\|\phi_\epsilon - \phi\|_{L^1}+\|\phi-g\|_{L^1}=\delta/2+0+\delta/2
$$
This holds for all $\delta>0$, completing the proof.
