Proving an equality involving a Lebesgue integrable function on R.

Question

I'm having trouble with the following equality. I'm not even sure how to begin. Please help.

Let $f$ be a real-valued, Lebesgue integrable function on $\mathbb{R}$. Prove that

$$ \lim_{t \to 0} \int_{\mathbb{R}} \lvert\, f(x+t) - f(x)\rvert\, \mathrm{d}x = 0 $$

Answer 1

Hint. You need to combine the following facts:

a. If $\int_{\mathbb R}\lvert f\rvert<\infty$, then for every $\varepsilon>0$, there exist $R>0$, such that $$ \int_{\lvert x\rvert> R}\lvert f\rvert<\varepsilon. $$

b. If $f\in L^1[a,b]$, then for every $\varepsilon>0$, there exist $g\in C[a,b]$, such that $$ \int_a^b\lvert\, f-g\rvert<\varepsilon. $$

c. If $g\in C[a,b]$, then $g$ is uniformly continuous, and hence, $$ \lim_{h\to 0}\int_{a+h}^{b-h}\lvert g(t+h)-g(t)\rvert\,dt=0. $$

Thus $$ \int_{\mathbb R}\lvert f(x+t)-f(x)\rvert\,dx\le 2\varepsilon+\int_{-R}^R\lvert f(x+t)-f(x)\rvert\,dx\le 4\varepsilon+\int_{-R+h}^{R-h}\lvert g(x+t)-g(x)\rvert\,dx. $$