Can the step-functions be chosen monotonically? Assume we are on the interval $[a,b]$. Assume we have a partition of the interval $\{a=x_0<x_1<x_2,<\ldots x_{n+1}\}$. A step-function is a function $f$ where there exists $c_0,c_1,\ldots ,c_n$ such that
$$f(x)=c_0I_{x_0}(x)+\sum\limits_{i=0}^nc_iI_{(x_i,x_{i+1}]}(x).$$
Assume that $g\in L^1([a,b])$, $g\ge0$. Then there exists a sequence of step-functions such that they converge to $g$ in $L^1([a,b])$. Can this sequence be chosen so that they converge monotonically?
 A: Apostol discussed this in his Mathematical Analysis.
Let's generalize a bit. Let $I$ be an interval. A function $s:I\to\mathbb {R} $ is said to be step function on $I$ if there is a closed interval $[a, b] \subseteq I$ such that restriction of $s$ to $[a, b] $ is a step function on $[a, b] $ (see definition in your question) and further $s(x) =0$ if $x\in I\setminus [a, b] $.
Apostol then proves the following deep result:

Theorem 1: Let $\{s_n\} $ be an increasing sequence of step functions on an interval $I$ such that $\int_I s_n$ is bounded above. Then the sequence $\{s_n\} $ converges to a limit function $f$ almost everywhere on $I$.

A function like $f$ in above result is said to be an upper function on $I$ and we define $$\int_I f=\lim_{n\to\infty} \int_I s_n\tag{1}$$
Next Apostol shows that if $f$ is any Riemann integrable function on $[a, b] $ then $f$ is also an upper function on $[a, b] $ (this involves approximating $f$ by sequence of step functions corresponding to lower Darboux sums on a uniform partition into $2^n$ subintervals). And then Apostol remarks that there are upper functions $f$ defined on an interval $I$ such that $-f$ is not an upper function on $I$ thereby showing that class of upper functions is larger than the class of Riemann integrable functions.
A function $f:I\to\mathbb {R} $ is said to be Lebesgue integrable on $I$ (written $f\in L(I)$) if we can write $f=u-v$ where $u, v$ are upper functions on $I$. The decomposition into $u, v$ is not unique and we define $$\int_I f=\int_I u-\int_I v\tag{2}$$

The answer to your question can be given now based on the example of an upper function $f$ such that $-f$ is not an upper function. Apostol does provide it in an exercise.
Let $I=[0, 1]$ and $$r_1,r_2,\dots,r_n,\dots$$ be the rationals in $I$ and $I_n=[r_n-4^{-n},r_n+4^{-n}]\cap I$. Let $f(x) =1$ if $x$ lies in some $I_n$ otherwise $f(x) =0$. Let $f_n(x) =1$ if $x\in I_n$ and $f_n(x) =0$ otherwise and $$s_n=\max(f_1,f_2,\dots,f_n)$$ then $s_n$ is an increasing sequence of step functions on $I$ and $s_n(x) \to f(x) $ almost everywhere on $I$ which shows that $f$ is an upper function on $I$. And $$\int_I f\leq \sum l(I_n) \leq\frac{2}{3}\tag{3}$$ Next Apostol says that if $s$ is any step function on $I$ such that $s(x) \leq - f(x) $ then $s(x) \leq - 1$ almost everywhere on $I$ and thus $\int_I s\leq - 1$. If $-f$ were an upper function on $I$ then we would have $\int_I (-f) \leq - 1$ which contradicts $(3)$.
The function $g$ defined by $g(x) =2-f(x)$ is positive on $I$ and clearly belongs to $L^1(I)$ but it can not be represented as the limit of an increasing sequence of step functions on $I$.
However your question does not specifically ask for increasing sequence but rather a monotone sequence. It should be obvious that the function $g$ above can be represented as limit of a decreasing sequence of step functions on $I$.
A: Let's consider measurable mapping $Y$ of $(\Omega, A)$ into $(\mathbb{R}, \mathfrak{B}(\mathbb{R}))$ and construct
$$Y_n=\sum_{q=1}^{n2^n}\frac{q-1}{2^n}\cdot \mathbb{1}_{\{q-1 \leqslant Y \cdot 2^n < q \}}+n\mathbb{1}_{\{ Y \geqslant n\}}$$
where $\mathbb{1}_B$ is indicator of $B$.
Then $Y_n$ is increasing sequence which converges to $Y$.
Taken from Jacques Neveu  "Mathematical Foundations Of The calculus Of probability", 1965, page 34.
Addition 1.
Let's consider now approximation using step function. As above we have approximation with simple functions, then it's enough to take case, when $f$ is characteristic function of $E$, $f=\chi_E $. We use, that if set $E$ have finite measure, then for $\forall \epsilon >0$ there exists a finite union of closed disjoint segments $F = \cup_{j=1}^{N}Q_j$, such that $m(E \Delta F)<\epsilon$. Therefore $f(x)=\sum_{j=1}^{N}\chi_{Q_j}(x)$, except on a set of measure $<\epsilon$. So for $\forall k \geqslant 1$ we have step function $\psi_k(x)$, which is not equal $f(x)$ on set with measure $<2^{-k}$. So we obtain step functions $\psi_k \to f$ except on some set with measure $0$.
A: 
I shall prove that if $f$ is a positive measurable function on $(E,\mathcal{E},\mu)$, then there exists an increasing sequence $(f_n)$ of step functions which converges towards $f$.

To this end, let us first notice that $g$ can always be written in the form
$$f(x) = \sup\{\alpha \in \mathbb{Q}_+ : f(x) \ge \alpha\} = \sup_{\alpha\in\mathbb{Q}_+} \alpha I_{\{f \ge \alpha\}}(x). $$
So, $f$ is the supremum of a countable family of step functions (namely a constant times the indicator function of a measurable set), of the form $\alpha I_{\{f \ge \alpha\}}$. Thus, in particular $g$ is the supremum of a sequence $(g_n)$ of step functions. In order to get an increasing sequence, let us set
$$ f_n = \max\{g_1,\dots,g_n\} $$
Since $f = \sup_n g_n$, it is also $f = \sup_n f_n = \lim_n f_n$.
A: The following answer has some overlap with the accepted answer. But I think it's a shorter proof and worth posting.
I'll work on $[0,1].$ Let $U\subset (0,1)$ be open and dense, with $m(U)<1.$ (An appropriate union of open intervals centered at the rationals in $(0,1)$ will do.) Define $g=0$ on $U,$ $g=1$ on $[0,1]\setminus U.$
Lemma: If $s$ is a step function such that $s\le g$ a.e., then $s\le 0$ a.e.
Proof: Let $I_k=(x_{k-1},x_k]$ be any of the intervals corresponding to $s,$ with $s=c_k$ on $I_k.$ By the density of $U,$ $U\cap I_k$ is a set of positive measure. Since $g=0$ on $U$ and $s\le g$ a.e., we have $c_k\le 0.$ Thus $s\le 0$ on $I_k.$ Since $I_k$ was any of the intervals, it follows that $s\le 0$ a.e. in $[0,1].$
Now suppose we have an increasing sequence of step functions $s_n$ that converge to $g$ in $L^1.$ Then some subsequence $s_{n_k}$ also converges pointwise to $g$ a.e. We then have each $s_{n_k}\le g$ a.e. By the lemma, $s_{n_k}\le 0$ a.e. for each $k.$ Thus
$$\tag 1 \int_0^1 |g-s_{n_k}|\ge \int_0^1 |g| = m([0,1]\setminus U)>0$$
for all $k.$ Hence $s_{n_k}$ does not converge in $L^1$ to $g,$ contradiction.
If we instead consider $h=1-g,$ exactly the same ideas show there is no decreasing sequence of step functions $s_n$ that converge to $h$ in $L^1.$ Finally, if we want one example $f$ for which no monotonic sequence of step functions converges to $f$ in $L^1,$ let $f=g$ on $[0,1],$ where $g$ is as above, and on $(1,2]$ let $f$ be the $h$ above, translated one unit to the right.
