Show that if the integral of the "derivative" of a L^1 function is 0, then the function is constant a.e. I'm working on the following problem: 
Let $f\in L^1[a,b]$. Prove that if 
$$\lim\limits_{h\to 0}\frac 1h\int_a^b|f(x+h)-f(x)|dx=0,(*)$$
then there is a constant $c$ such that $f(x)=c$ for almost every $x\in (a,b)$. 
I started by noting that $C_c^\infty([a,b])$ is dense in $L^1([a,b])$, so we can find a sequence $\{f_n\}\subseteq C_c[a,b]$ such that $\|f_n-f\|_{L^1[a,b]}\to 0$ as $n\to \infty$. If we have $(*)$ for $f_n$, where $n$ is sufficiently large, then we can conclude that $f_n$ is constant, and therefore $f$ is constant a.e.. But I got stuck when trying to show $(*)$ for $f_n$, $n$ large. It seems that the usual triangle inequality type argument wouldn't work here since there is a $\frac 1h$ in the front. 
So I was wondering if there is any way to make the above argument work? Or if not, how to solve the problem? Any suggestions/hints would be appreciated!

This is an easy consequence of Lebesgue's Theorem. Let $a<c<d<b$ and assume that $c$ and $d$ are Lebesgue points of $f$. Then $\frac {\int_c^{d} f(x+h)dx -\int_c^{d}f(x)dx} h \to 0$. This gives $\frac {\int_d^{d+h} f(x+h)dx -\int_c^{c+h}f(x)dx} h \to 0$. Since $c$ and $d$ are Lebesgue points this gives $f(c)=f(d)$. Since almost all points are Lebesgue points it follows that $f$ is a.e. constant
