If $f$ is integrable in $[a-1,b+1]$, then prove that $\lim \limits_{h\to 0} \ \int_{a}^{b}|f(x+h)-f(x)| ~ dx=0$ I'd really appreciate your help with the following problem: 

Let $f$ be a integrable  function in $[a-1,b+1]$. I need to prove that:
  $$\lim_{h\to 0} \int_{a}^{b} |f(x+h)-f(x)|dx=0 .$$

Basically I need to show that $\forall  \varepsilon  \exists  \delta .|h|< \delta \to \left|\int_{a}^{b}|f(x+h)-f(x)|dx\right| < \varepsilon $.
In addition, I know that because of $f$ being integrable I can write that
$\sum \limits_{i=0}^{n-1}(\sup f-\inf f) \Lambda x_{i} < \varepsilon$ for a division of $[a,b]$.
I tried to define $g=|f(x+h)-f(x)|$ and work with that, but it didn't lead me to any smart conclusion. Also I tried to put the limit inside of the integral and to use the derivative, but I'm not sure when I can do that.
Thank you for the help!
 A: This can be proven by approximating $f$ by a continuous function $g$ so that $\int_a^b|f(x)-g(x)|\mathrm{d}x<\epsilon$. Since $g$ is continuous on a compact set, it is uniformly continuous, so there is a $\delta>0$ so that $|g(x+\delta)-g(x)|<\frac{\epsilon}{b-a}$. Therefore,
$$
\begin{align}
\int_a^b|f(x+\delta)-f(x)|\mathrm{d}x
&\le\int_a^b|f(x+\delta)-g(x+\delta)|\mathrm{d}x\\
&\qquad+\int_a^b|g(x+\delta)-g(x)|\mathrm{d}x\\
&\qquad+\int_a^b|f(x)-g(x)|\mathrm{d}x\\
&\le\epsilon+\frac{\epsilon}{b-a}(b-a)+\epsilon\\
&=3\epsilon
\end{align}
$$
A: Let $\varepsilon > 0$ and take a partition of $[a-1,b+1]$ for which $\sum (\sup f - \inf f)\Lambda x_i < \varepsilon(b-a)$.
Now, since $f$ is integrable in $[a-1,b+1]$ let $M$ be the essential supremum of $f$ there (assuming we're talking about proper integrals). You get that any $|h|<1$ satisfies that $|f(x+h)-f(x)|<2M$.
Take $\delta>0$ such that $\delta$ is less than the minimal length of an interval in the partition and less than $\varepsilon$, for a given interval $I$ take the part of $[a,b]$ where $x\in I$ yet $x+h \not \in I$, this has length of at most $\delta$, which is smaller than $\varepsilon$, and the same goes for tha part of $[a,b]$ where $x\notin I$ yet $x+h \in I$.
You get that integrating over all of these parts is bound from above by $4Mn\varepsilon$ where $n$ is the number of intervals in the partition.
Integrating over all the rest is bound by $2\varepsilon(b-a)$.
All in all the value of the integral is bound by a function of $\varepsilon$, which concludes the proof.
