If $f$ has derivative in $L^p(\mathbb{R})$, then it is absolutely continuous in any bounded interval

Question

Let $f\in L^p(\mathbb{R})$, with derivative in $L^p(\mathbb{R})$. That is, there exists $h\in L^p(\mathbb{R})$ such that: $$ \lim_{y\to 0} \int_{\mathbb{R}} \left| \frac{f(x+y)-f(x)}{y}- h(x)\right|^p\; dx =0 $$ Prove that $f$ is absolutely continuous in any bounded interval.

Any suggestion??

Answer 1

Let us assume that our bounded interval is $I$ and $(x_k,y_k)$ is a finite sequence of pairwise disjoint sub-intervals of $I$, fulfilling $\sum_k|y_k-x_k|=\delta$. We have $$ f(y_k)-f(x_k) = \int_{x_k}^{y_k}h(x)\,dx $$ hence by the triangle inequality and Holder's inequality $$ \sum_k\left|f(y_k)-f(x_k)\right|\leq \int_{\bigcup_k(x_k,y_k)}|h(x)|\,dx\leq \|h\|_p\,\delta^{1-\frac{1}{p}} $$ and as long as $\delta\to 0$ we have that the LHS tends to zero as well.
This proves the absolute continuity of $f$.