Given that $f$ is Darboux(Riemann) integrable on $[c,d]$ and that $[a,b]\subset [c,d]$ prove that$\lim \limits_{h \to 0} \int \limits_{a}^{b} |f(x+h)-f(x)| dx =0$
My attempt : given $\epsilon>0$ we need to prove that there is a partition of P of $[a,b]$ such that $\sum \Delta x \sup_{x \in [x_i ,x_{i+1}]} |f(x+h)-f(x)| < \epsilon$
My problem is with the term $ \sup_{x \in [x_i ,x_{i+1}]} |f(x+h)-f(x)|$ because the $h>0$ could give the sup(max) value when $x+h > x_{i+1}$ or if $h<0$ then we could have the max value when $x+h < x_i$, so i don't really can bound the $sup$ by $\omega _i$.
Could any one give simple way to solve this question, thanks.
Given $\varepsilon>0$, let $g\in C^0([c,d])$ such that $\int_c^d|f(x)-g(x)|dx<\varepsilon$. Being $g$ continuous on the compact set $[c,d]$, $g$ is uniformly continuous (by Heine-Cantor theorem) and then it is easy to prove the conclusion for $g$ instead of $f$. But then $$\limsup_{h\rightarrow0}\int_a^b|f(x+h)-f(x)|dx\le\\\limsup_{h\rightarrow0}\int_a^b|f(x+h)-g(x+h)|dx+\limsup_{h\rightarrow0}\int_a^b|g(x+h)-g(x)|dx+\limsup_{h\rightarrow0}\int_a^b|f(x)-g(x)|dx\le2\varepsilon.$$ By the arbitrariness of $\varepsilon$, the conclusion follows.