Everyone might get used to the following question in the context when $f$ is assumed to be Lebesgue integrable function. In this problem, we give new hypotheses
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a non-negative continuous function satisfying $\int_{\mathbb{R}}f(x)dx<\infty.$ Then prove that $$\lim_{h\rightarrow 0}\int_{\mathbb{R}}|f(x+h)-f(x)|dx=0.$$
I have been thinking about mimicking the idea of proof from $f$ being Lebesgue integrable function. To me, it does not seem work successfully. Without applying knowledge from Lebesgue integration, how does everyone tackle the problem?
The finitenss of the improper Riemann integral $\int_{-\infty}^{\infty} f(x)dx$ implies that given $\epsilon >0$ we can find $\Delta$ such that $\int_{-\infty}^{-\Delta} f(x)dx <\epsilon$ and $\int_{\Delta}^{\infty} f(x)dx<\epsilon$. For $|h| <1$ the function $f(x+h)-f(x)$ vanishes for $|x| >\Delta +1$. Hence $\int_{\mathbb R} |f(x+h)-f(x)|dx=\int_{-\Delta-1}^{\Delta +1} |f(x+h)-f(x)|dx$. By uniform continuity of $f$ ( on $[-\Delta -2, \Delta +2]$) we can choose $\delta >0$ such that $|h|<\delta$ implies $|f(x+h)-f(x)|<\frac {\epsilon} {2\Delta +2}$ whenever $x \in [-\Delta -1,\Delta +1]$. Hence $\int_{\mathbb R} |f(x+h)-f(x)|dx<3\epsilon$ for $|h| <\delta$.