This question was mistakenly asked in $\mathrm{lim}_{h\rightarrow 0} \int_0^h |f(x+h)-f(x)| dx=0$ almost everywhere. But it turned out that the person was only asking for $$\lim_{h \to 0} \int_0^h |f(x+t)-f(x)| dx = 0$$ which is easy to verify. But is the statement true if you repalce $t$ by $h$?
I had the following idea to proof it: If $f \in L^1(\mathbb{R})$ then there exists a sequence of continuous function $f_n \to f$. Now
$$ \int_0^h |f(x+h)-f(x)| dx \leq \int_o^h |f(x+h)-f(x) - f_n(x+h) + f_n(x)| dx \\ + \int_0^h |f_n(x+h)-f_n(x)| dx $$
now for every $n$ we have
$$ \lim_{h \to 0} \int_0^h |f_n(x+h)-f_n(x)| dx = 0 $$
and for every $h$ we have
$$ \lim_{n \to \infty} \int_o^h |f(x+h)-f(x) - f_n(x+h) + f_n(x)| dx \to 0 $$
But I was not able to conclude
$$ \lim_{h \to 0}\int_0^h |f(x+h)-f(x)| dx = 0$$
Am I missing anything, or is this not possible?
If $f$ is integrable (at least on a neighborhood of $0$), then we have $$ \int_0^h | f(x+h) - f(x) | dx \le \int_0^h |f(x+h)| dx + \int_0^h |f(x)| dx = \int_h^{2h} |f(y)| dy + \int_0^h |f(x)| dx \to 0 $$ for $h\to 0$.