$f\in W^{1,p}((0,1))$ is absolute continuous This is an exercise in Evans's PDE book:

$f\in W^{1,p}((0,1))$ is absolute continuous where $1\leq p < \infty $.


Try : By definition of sobolev space, $f$ has weak derivative $f'$ 
So $$\ f(x)= f(0) +\int_0^x f'(t)dt \tag{*}$$ 
By fundamental theorem of calculus, $f$ is absolutely continuous. 
But I cannot show ($\ast $). How can we complete the proof ?
 A: Approximate $f$ by smooth functions $f_n$ convergent to $f$ in $W^{(1,p)}$. Then, in particular, $f_n'\to f'$ in $L^p$ and $f_n(0)\to f(0)$ by the Sobolev embedding theorem. $(*)$ holds for $f_n$, and it remains to pass to the limit as $n\to\infty$. (Note that
$$
\int_0^x f'(t)dt=\langle \chi_x,f'\rangle,
$$
where $\chi_x$ is the characteristic function of the interval $[0,x]$ and lies in $L^{p'}$,
$1/p+1/p'=1$.) This proves $(*)$ for $f$.
A: The context of the exercise is the very very beginning, i.e., there is only a definition of Sobolev space $W^{1,p}(0,1)$ in terms of weak derivatives and Lebesgue space $L^p(0,1)$, while there is nothing else, e.g., there are no Sobolev embedding theorems, there are no approximations by smooth functions. Of course, one could try to establish first the seems-to-be-required approximation by smooth functions in conjunction with the embedding $W^{1,p}\hookrightarrow C$, jumping somehow over the representation $(\ast)$. This would be an extremely complicated approach to a rather trivial but still most fundamental exercise.  Under the circumstances, the only known correct simple approach to proving $(\ast)$ reduces to just two elementary exercises, being by itself most instructive for beginners. First, taking
$$
g(x)\overset{\rm def}{=}\int\limits_0^x f'(t)\,dt,
$$ 
find its weak derivative $g'$ using the standard definition of a weak derivative, namely,
$$
\begin{align}
\int\limits_0^1 g(t)\varphi'(t)\,dt=
\int\limits_0^1 \varphi'(t)\,dt\int\limits_0^t f'(s)\,ds=
\int\limits_0^1 f'(s)\,ds\int\limits_t^1 \varphi'(t)\,dt\\
=\int\limits_0^1 f'(s)\bigl(\varphi(1)-\varphi(s)\bigr)\,ds
=-\int\limits_0^1 f'(s)\varphi(s)\,ds\quad\forall\,\varphi\in C_0^{\infty}(0,1),
\end{align}
$$
whence follows that $\,g'=f'\,$ or $\,(f-g)'=0\,$ a.e. on $\,(0,1)$. Second, taking a function $h\in L^1_{loc}(0,1)$ that does possess a weak derivative $\,h'=0\,$ a.e. on $\,(0,1)$, prove that there is some constant $C$ such that $\,h=C\,$ a.e. on $\,(0,1)$. To achieve this, start with the definition of $\,h'=0\,$, i.e.,
$$
\int\limits_0^1 h(x)\varphi'(x)\,dx=0\quad\forall\,\varphi\in C_0^{\infty}(0,1).\tag{1}
$$
Choosing a function $\eta\in C_0^{\infty}(0,1)$ such that $\int_0^1 \eta(x)\,dx=1$, which is easy to construct explicily, consider a function
$$
\Phi(x)=\psi(x)-\eta(x)\int\limits_0^1\psi(s)\,ds
$$
with an arbitrary given $\psi\in C_0^{\infty}(0,1)$. It is clear that
$\Phi\in C_0^{\infty}(0,1)$ and $\int_0^1 \Phi(x)\,dx=0$, hence a test function 
$$
\varphi(x)\overset{\rm def}{=}\int\limits_0^x\Phi(s)\,ds\in C_0^{\infty}(0,1).\tag{2}
$$
Now, substituting $(2)$ in $(1)$ results in the identity
$$
\int\limits_0^1 h(x)\psi(x)\,dx=\int\limits_0^1 C\psi(x)\,dx\quad
\forall\,\psi\in C_0^{\infty}(0,1)
$$
with the constant
$$
C\overset{\rm def}{=}\int\limits_0^1 h(x)\eta(x)\,dx.
$$
Therefore, the locally integrable function $\,h=C\,$, and hence $\,f=C+g\,$ a.e. on 
$\,(0,1)$. Since $g$ is continuous on $[0,1]$, the function $f$ can be redefined on a subset of Lebesgue measure zero on $[0,1]$ to become everywhewe continuous, in wich case 
$C=f(0)$ with $f$ satifying $(\ast)$. Interestingly enough, Paul Du Bois-Reymond was the first to have done the second exercise yet in 1879 for a function $h\in C(0,1)$, i.e., was the first to find a weak solution $h\in C(0,1)$ of the simplest differential equation $h'=0$. This remarkable but little known fact was dicovered by Jean Dieudonné (see pages 221-222 in his History of Functional Analysis). 
