Suppose that $f\geq 0$ is bounded measurable and compactly supported. Is there a sequence of simple functions $(s_{n})$ such that $s_{n}\uparrow f$ pointwise almost everywhere, where the simple functions $s_{n}$ are of the form \begin{align*} s_{n}=\sum_{\text{finite sum}}a_{i}\chi_{K_{i}}, \end{align*} where $K_{i}$ is compact?
I have posted this question yesterday, but the wording may be somehow vague that make the question itself unfathomable and I have discarded the post and make a new one here, now it should be clearly phrased and understandable.
Anyway, for $0\leq f<M$, traditionally one may take \begin{align*} s_{n}=\sum_{k=1}^{2^{n}M-1}\dfrac{k}{2^{n}}\chi_{f^{-1}[k2^{-n},(k+1)2^{-n})}, \end{align*} but the difficulty is that, $f^{-1}[k2^{-n},(k+1)2^{-n})$ need no to be compact.
One may think of the inner regularity of the measurable set $f^{-1}[k2^{-n},(k+1)2^{-n})$ to approximate it by a compact set, say, for fixed $k,n$, a compact set $K_{k,n,m}$ is such that \begin{align*} \left|f^{-1}[k2^{-n},(k+1)2^{-n})-K_{k,n,m}\right|<\dfrac{1}{m}, \end{align*} so we end up with a double indices sequence $(t_{n,m})$ such that \begin{align*} t_{n,m}=\sum_{k=1}^{2^{n}M-1}\dfrac{k}{2^{n}}\chi_{K_{k,n,m}}. \end{align*} The difficulty would be, how do we enumerate $t_{n,m}$ into a single index sequence such that it is increasing?
The $K_{k,n,m}$ can be selected such that for each fixed $k,n$, $K_{k,n,m}\subseteq K_{k,n,m+1}$, but this does not entail the double indices sequence $(t_{n,m})$ can be reduced to an increasing single index sequence. A naive trial would be considering the diagonal $(t_{n,n})$, but this does not work.
If $f:\mathbb R\to [0,\infty)$ is measurable, then as you know, there is a sequence of simple functions $0\le s_1 \le s_2 \le \cdots$ that converge to $f$ a.e. Setting $s_0=0,$ it follows that
$$f =\sum_{n=1}^{\infty}(s_n-s_{n-1}) \,\,\,\text {a.e.}$$
Now each $s_n-s_{n-1}$ is a nonnegative simple function. So we can write
$$s_n-s_{n-1}= \sum_{m=1}^{m_n} a_{nm}\chi_{E_{nm}}.$$
Here all $a_{mn}\ge 0$ and the sets $E_{nm}$ are all measurable of course. Therefore
$$f= \sum_{n=1}^{\infty}\sum_{m=1}^{m_n} a_{nm}\chi_{E_{nm}}\,\,\text {a.e.}$$
Every summand in the above double sum is nonnegative, so we can arrange this sum in linear form to get
$$\tag 1 f= \sum_{j=1}^{\infty} b_{j}\chi_{F_{j}}\,\,\text {a.e.},$$
where all $b_j\ge 0.$
Lemma: If $F\subset \mathbb R$ is measurable, then there are pairwise disjoint compact sets $K_1,K_2,\dots \subset F$ such that $m(F \setminus (\cup_{n=1}^\infty K_n))=0.$ Hence $\chi_F = \sum_{n=1}^{\infty}\chi_{K_n}$ a.e.
I'll sketch a proof of the lemma at the end. Assuming it holds, apply it to each $F_j$ in $(1).$ After linear ordering of the resulting double sum, we then have nonnegative $c_n$ and compact sets $K_1,K_2,\dots $ such that
$$f=\sum_{n=1}^{\infty}c_n\chi_{K_n} \,\,\text {a.e.}.$$
To finish, simply let $S_n=\sum_{m=1}^{n}c_m\chi_{K_m}.$ The sequence $S_n$ solves the problem.
Proof sketch of lemma: Since we can write $F$ as a countable union of disjoint measurable sets of finite measure, we can assume $m(F)<\infty.$ Then, much like Daniel Fischer mentioned in a comment, there is a compact $K_1\subset F$ such that $m(F\setminus K_1)<1.$ Then there is a compact $K_2 \subset F\setminus K_1$ such that $m(F\setminus (K_1\cup K_2))<1/2.$ Continue this process to obtain disjoint compact sets $K_n \subset F$ such that $m(F\setminus (K_1\cup\cdots\cup K_n))<1/n$ for each $n.$ It follows that $m(F\setminus (\cup_{n=1}^\infty K_n)) =0.$