The theorem says that continuity, bounded variation (BV), and # (Property N) imply absolute continuity. Here's how it fails if we have two of the properties:
BV & Property N
Let $f(x) = \operatorname{sign} x$. The range is finite, so $f(E)$ always has zero Lebesgue measure. But $f$ is not absolutely continuous.
Continuity & Property N
Let $f(x) = x\sin (1/x)$, $f(0)=0$. Property $N$ follows from the fact that $f$ is locally Lipschitz in $\mathbb{R} \setminus\{0\}$. But $f$ is not BV, so is not absolutely continuous.
Continuity & BV
Let $f$ be the Cantor function.
You can construct $f$ as follows: let $m$ denote Lebesgue measure. Once $m(A)=0$, we can choose for each $i=1,2,\cdots$, a sequence of intervals $I_{ij}$ such that $$A\subset \cup_{j=1}^\infty I_{ij},\ \sum_{j=1}^\infty m(I_{ij})\le \frac{1}{2^i}. \tag{1}$$
Note that each $x$ belongs to infinitely many intervals $I_{ij}$. Define $f:[0,1]\to\mathbb{R}$ by $$f(x)=\sum _{i,j=1}^\infty m(I_{ij}\cap [0,x]).$$
Note that $f$ is increasing. Also $f$ satisfies $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\infty,\ \forall x\in A.$$
Indeed, as $x$ belongs to infinitely many intervals $I_{ij}$, we can assume without of generality that $x$ belongs to $J_k=(x-a_k,x+a_k)$ for all $k=1,2,\cdots$, that each $J_k=I_{ij}$ for some $i,j$ and $a_k\to 0$. Note that
\begin{eqnarray}
\frac{f(x-h)-f(x)}{h} &=& \frac{1}{h}\sum _{i,j=1}^\infty m(I_{ij}\cap [x-h,x]) \nonumber \\
&\ge& \frac{1}{h}\sum _{k=1}^\infty m(J_k\cap [x-h,x]) \nonumber \\
&\ge& \frac{1}{h}(Nh+\sum _{k=N+1}^\infty a_k),
\end{eqnarray}
where $N$ depending on $h$ is choosen in such a way that $h\le a_N$. As $h\to 0^+$, we see from the previous inequality that the left derivative is infinity. With an similar reasoning, you can conclude that the right derivative is also infinity.
To prove that $f$ is absolutely continuous, note that $$\sum _{k=1}^N |f(x_{k+1})-f(x_k)|=\sum _{k=1}^n \sum_{i,j=1}^\infty m(I_{ij}\cap[x_k,x_{k+1}]). \tag{2}$$
Once $(1)$ is satisfied, you conclude from $(2)$ that $f$ is absolutely continuous. To get a strictly increasing functions, just consider $g(x)=x+f(x)$.
Note: The original construction is due to professor Porter and it is contained here.
Best Answer
Let $A\subset[0,1]$ be a closed set of positive measure that doesn't contain any interval (this, for instance). Then the function $\phi(x)=d(x,A)$ is Lipshchitz, hence absolutely continuous, and $\{\phi=0\}=A$ doesn't contain any interval.