This depends on precisely what you meant by "approximated by". If you meant the exact same thing as in the first paragraph, i.e., there exists a sequence $S_n = \sum_{i = 1}^{k_n} c_{(n, i)}1_{[a_{(n, i)}, b_{(n, i)}]}$ s.t. $S_n \nearrow f$. (For simplicity, I'll assume $c_{(n, i)} > 0$ for all $n, i$ - this doesn't really matter since we're approximating nonnegative functions anyway. I'll also assume $a_{(n, i)} < b_{(n, i)}$ for all $n, i$ - I'll only be dealing with a.e. approximations, so a degenerate interval can simply be ignored.) Even if $\nearrow$ is only a.e., i.e., $S_n \leq S_m$ a.e. whenever $n \leq m$ and $f = \lim_n S_n$ a.e., you cannot get all measurable $f \geq 0$ from this. For example, let $A$ be any closed nowhere dense set with positive measure (say, a fat Cantor set), then $1_A$ cannot be so approximated. Indeed, if $S = \sum_{i = 1}^k c_i1_{[a_i, b_i]} \leq 1_A$, then $1_{[a_i, b_i]} \leq 1_A$ a.e., so the open set $(a_i, b_i) \subset A$ a.e., i.e., $(a_i, b_i) \setminus A$ is null. As $A$ is closed, $(a_i, b_i) \setminus A$ is open. The only null open set is the empty set, so $(a_i, b_i) \subset A$ fully - but this is impossible as $A$ has no interior.
In fact, the following is a characterization of all $f$ that can be approximated this way:
Proposition: $f \geq 0$ can be approximated as above iff for any $C \geq 0$, the sets $\{x: f(x) > C\}$ is almost open. (Here, by a set $A$ being almost open, I meant there exists an open set $O$ s.t. $O \triangle A$ is null.)
Proof: ($\Rightarrow$) Since we're only dealing with a.e. approximations, we may write $S_n = \sum_{i = 1}^{k_n} c_{(n, i)}1_{(a_{(n, i)}, b_{(n, i)})}$ and assume the intervals involved are disjoint. As $S_n \nearrow f$, it is then easy to see that, a.e.,
$$\{x: f(x) > C\} = \bigcup_{n = 1}^\infty \bigcup_{i: c_{(n, i)} > C} (a_{(n, i)}, b_{(n, i)})$$
Thus, $\{x: f(x) > C\}$ is almost open.
($\Leftarrow$) Since we're only dealing with a.e. convergence, instead of Lebesgue measure, I'll change to an equivalent probability measure $\mu$, say, the Gaussian measure. I'll prove then there is an increasing sequence $S_n$ s.t. $S_n \to f$ in measure. Then a subsequence would convergence to $f$ a.e. (See here.) Recall also the topology of convergence in measure is metrizable, so we may choose a metric $\rho$ generating it. $\rho$ may be chosen in such a way so that, if $f_1 \leq f_2 \leq f_3$ a.e., then $\rho(f_2, f_3) \leq \rho(f_1, f_3)$. (See here.)
Now, for each $k$, we define,
$$T_k' = \sum_{i = 0}^{k2^k - 1} 2^{-k} 1_{\{x: f(x) > i2^{-k}\}}$$
One may check that $T_k' \nearrow f$ everywhere. By assumption on $f$, $\{x: f(x) > i2^{-k}\}$ is almost open for all $k, i$, so we may replace these sets by open sets $O_{(k, i)}$, then,
$$T_k = \sum_{i = 0}^{k2^k - 1} 2^{-k} 1_{O_{(k, i)}}$$
and $T_k \nearrow f$ a.e. By Egorov's theorem, we have $\rho(T_k, f) \to 0$.
Now, every open set is a disjoint union of countably many open intervals (namely the connected components of the said open set). Up to a.e., we may take these intervals to be bounded. Thus, for each $k, i$, we may write,
$$O_{(k, i)} = \bigcup_{j = 1}^\infty (a_{(k, i, j)}, b_{(k, i, j)})$$
a.e., where the intervals on the right are disjoint from each other. But this means, for each $k$, if,
$$G_{(k, N)} = \sum_{i = 0}^{k2^k - 1} \sum_{j = 1}^N 2^{-k} 1_{[a_{(k, i, j)}, b_{(k, i, j)}]}$$
Then $G_{(k, N)} \nearrow T_k$ a.e. as $N \to \infty$. Again, by Egorov's theorem, $\rho(G_{(k, N)}, T_k) \to 0$ as $N \to \infty$.
Now, for each $n > 0$, first choose $k$ large enough s.t. $\rho(T_k, f) < \frac{1}{n}$, then choose $N$ large enough s.t. $\rho(G_{(k, N)}, T_k) < \frac{1}{n}$. Let $S_n' = G_{(k, N)}$. Then $S_n'$ is of the desired form and $\rho(S_n', f) < \frac{2}{n} \to 0$, so $S_n' \to f$ in measure. The only issue is that $S_n'$ may not be increasing. This, however, can be rectified by defining,
$$S_n = \max(S_1', \cdots, S_n')$$
Note that the maximum of finitely many simple functions of the desired form is again, up to a.e., a simple function of the desired form. Since all $S_n' \leq f$ a.e., we have $S_n' \leq S_n \leq f$ a.e., so $\rho(S_n, f) \leq \rho(S_n', f) \to 0$, i.e., $S_n \to f$ in measure. As $S_n$ is clearly increasing, this proves the claim. $\square$
Remark: There is a slightly simpler characterization: $f$ satisfies the properties in the proposition above iff $f$ equals a.e. to a lower semicontinuous function. Indeed, if $f$ equals a.e. to a lower semicontinuous function $g$, then $\{x: f(x) > C\}$ equals $\{x: g(x) > C\}$ a.e. and the latter set is open. Conversely, if $\{x: f(x) > C\}$ is almost open for all $C \geq 0$, then as we have seen, we have a sequence $T_k = \sum_{i = 0}^{k2^k - 1} 2^{-k} 1_{O_{(k, i)}}$ satisfying $T_k \nearrow f$ a.e. and with all $O_{(k, i)}$ open. But then all $T_k$ are lower semicontinuous, so $f = \sup_k T_k$ a.e. and the RHS, being the supremum of lower semicontinuous functions, is lower semicontinuous.
Now, the result is different if by "approximated by", you meant something weaker - say, there exists a sequence $S_n = \sum_{i = 1}^{k_n} c_{(n, i)}1_{[a_{(n, i)}, b_{(n, i)}]}$ s.t. $S_n \to f$ a.e., without assuming the sequence is increasing. In this case, it is actually the case that every measurable function can be approximated:
Proposition: Every measurable $f \geq 0$ can be approximated as above.
Proof: The proof is actually somewhat similar. Again, instead of Lebesgue measure, I'll change to an equivalent probability measure $\mu$. And again, I'll prove then there is a sequence $S_n$ of the desired form s.t. $S_n \to f$ in measure. Then a subsequence would convergence to $f$ a.e. Again, let $\rho$ metrizes the topology of convergence in measure. Let $T_k = \sum_{i = 1}^{m_k} c_{(k, i)} 1_{A_{(k, i)}}$ be any sequence of nonnegative simple functions that converge to $f$ a.e. For each $k, i$, by outer regularity of $\mu$, there is a decreasing sequence of open sets $(O_{(k, i, j)})_j$ s.t. $A_{(k, i)} = \cap_{j = 1}^\infty O_{(k, i, j)}$ a.e. But then, for each $k$, if,
$$G_{(k, j)} = \sum_{i = 1}^{m_k} c_{(k, i)} 1_{O_{(k, i, j)}}$$
We have $G_{(k, j)} \to T_k$ a.e. as $j \to \infty$. Now, for each $k, i, j$, as we have seen before, we may write,
$$O_{(k, i, j)} = \bigcup_{l = 1}^\infty (a_{(k, i, j, l)}, b_{(k, i, j, l)})$$
a.e., where the intervals on the right are disjoint from each other. But then, for each $k, j$, if,
$$H_{(k, j, N)} = \sum_{i = 1}^{m_k} \sum_{l = 1}^N c_{(k, i)} 1_{[a_{(k, i, j, l)}, b_{(k, i, j, l)}]}$$
Then $H_{(k, j, N)} \to G_{(k, j)}$ a.e. as $N \to \infty$. From this point we apply Egorov's theorem repetitively: For each $n$, first choose $k$ large enough s.t. $\rho(T_k, f) < \frac{1}{n}$, then choose $j$ large enough s.t. $\rho(G_{(k, j)}, T_k) < \frac{1}{n}$, and finally choose $N$ large enough s.t. $\rho(H_{(k, j, N)}, G_{(k, j)}) < \frac{1}{n}$. Let $S_n = H_{(k, j, N)}$. Then $\rho(S_n, f) < \frac{3}{n} \to 0$, so $S_n \to f$ in measure, as desired. $\square$
