prove that $u$ is equal a.e. to an absolutely continuous function Prove that if $n=1$ and $u\in W^{1,p}(0,1) $ for some $1\leq p&lt;\infty$, then $u$ is equal a.e. to an absolutely continuous function,and $u'$ (which exists a.e.) belongs to $L^{p}(0,1)$.

My idea:due to $(0,1)$ is a bounded and we can think that $u$ has a compact support,if $u\in W^{1,p}((0,1))$,then the $u^{\varepsilon}=\eta_{\varepsilon}*u$ is a smooth function with
$ u^{\varepsilon}\rightrightarrows u $ when $\varepsilon\to 0^{+}$,and $\|Du^{\varepsilon}\|_{p}\leq \|Du\|_{p} $,then for $x,y\in (0,1),x\neq y$, we have
\begin{align*}
 u^{\varepsilon}(x)-u^{\varepsilon}(y)&amp;= \int_{0}^{1} (x-y)\cdot (Du^{\varepsilon}(tx+(1-t)y))dt 
\end{align*}
then we have
$$ |u^{\varepsilon}(x)-u^{\varepsilon}(y)|\leq \|Du^{\varepsilon}\|_{p}|x-y|\leq \|Du\|_{p}|x-y| $$
Let $\varepsilon\to 0$,we get
$$ |u(x)-u(y)|\leq \|Du\|_{p}|x-y| $$
Another approach without convolutions:

Since $Du$ is an $L^p(0,1)$ function, hence $L^1(0,1)$ in particular, for each $x \in (0,1)$ the integral $w(x) = \int_0^x Du(t)\,dt$ makes sense.  By the fundamental theorem of calculus for Lebesgue integrals, $w$ is absolutely continuous and $w' = Du$ almost everywhere.  Intuitively, we should expect that $w$ differs from $u$ by a constant.  To find the constant, set $c = \int_0^1 (u(t) - w(t))\,dt$.  Then set $v = w+c$, which is again absolutely continuous and has $v' = Du$ almost everywhere.  Now for any test function $\varphi \in C^\infty_c((0,1))$, use the definition of weak derivative and integration by parts to show that $\int_0^1 (u(t)-v(t)) \varphi(t)\,dt = 0$.  Conclude that $u-v = 0$ almost everywhere.
Yes, this is one way to do it. I would suggest you to read this book, chapter 8 for a complete introduction on 1 dimensional Sobolev space.

Also, this note collects more results, but without proof, of Sobolev function properties in one dimensions.