On the continuity of functions in 1D Sobolev spaces In the chapter of the Sobolev spaces (Evans PDE book), he mentions that

[...] if $U$ is an open interval in $\mathbb{R}^1$ then $u \in W^{1,p}(U)$ if and only if $u$ equals a.e. an absolutely continuous function whose ordinary derivative (which exists a.e.) belongs to $L^p(U)$. Such a simple characterization is however only available for $\mathbb{R}^1$[...]

What I am trying to figure out is this: what is the intuition behind this property holding with "iff" in 1D and not in higher dimensions?
 A: $\newcommand{\R}{\mathbb{R}}$
To illustrate the intuition, let me focus on the continuity part of that statement, so 
$$
u \in W^{1,1}(\R)
\quad \Longrightarrow \quad 
u \text{ is continuous.}
$$
The proof goes like this: if $u$ happens to be smooth, we can observe that the norm $\| \nabla u \|_{L^1(\R)}$ controls the modulus of continuity of $u$: 
$$
|u(x)-u(y)| \le \int_x^y |\nabla u(z)| dz.
$$
One can then use this observation to prove continuity in full generality. 
In contrast, this is not the case in higher dimensions. We can still use the fundamental theorem of calculus 
$$
|u(x)-u(y)| = \left| \int_0^1 \nabla u(tx+(1-t)y) \cdot (x-y) dt \right| \le \int_{[x,y]} |\nabla u(z)| d \mathcal{H}^1(z),
$$
thus bounding the difference $|u(x)-u(y)|$ by the integral of $|\nabla u|$ over some segment (wrt. to the $1$-dim Hausdorff measure, not $n$-dim of course). However, this can still be arbitrarily large, even if we assume $\| \nabla u \|_{L^1(\R^n)}$ to be small. 
To phrase it differently, the norm $\| \nabla u \|_{L^1(\R^n)}$ does not control the modulus of continuity, for very simple reasons. 

Example. Let $n \ge 2$ and consider the function 
$$
u(x) = 
\begin{cases}
\log |x| & \text{ in the punctured unit ball,} \\
0 & \text{ elsewhere.}
\end{cases}
$$
It's locally Lipschitz in $\R^n \setminus \{ 0 \}$ and one can check that $u \in W^{1,1}(\R^n)$. At the same time, $u$ cannot be modified (on a set of measure zero) to be continuous, since it's essentially unbounded. 

That being said, there are some analogous results in higher dimensions. For example, 
$$
p>n, \ u \in W^{1,p}(\R^n)
\quad \Longrightarrow \quad 
u \text{ is continuous.}
$$
If $n \ge 2$, the above does not hold in the endpoint case $p=n$, i.e., 
$$
n \ge 2, \ u \in W^{1,n}(\R^n)
\quad \not\Longrightarrow \quad 
u \text{ is continuous.}
$$
You could say that $n=1$ is special in this way. However, the following is true: 
$$
u \in W^{n,1}(\R^n)
\quad \Longrightarrow \quad 
u \text{ is continuous.}
$$

And as far as absolute continuity and classical partial derivatives are concerned, there is also an analogue - any function $u \in W^{1,p}(\R^n)$ can be shown to be absolutely continuous on almost all lines, with the classical directional derivative coinciding a.e. with the distributional one. 
