Following the previous question posted by me, an analogue question is the following:
Suppose that $f\geq 0$ is measurable. Is there a sequence of simple functions $(s_{n})$ such that $s_{n}\downarrow f$ pointwise almost everywhere, where the simple functions $s_{n}$ are of the form \begin{align*} s_{n}=\sum_{\text{finite sum}}a_{i}\chi_{G_{i}}, \end{align*} where $G_{i}$ is open?
Following the nice trick by @zhw. one writes \begin{align*} f=\sum_{n}a_{n}\chi_{A_{n}}~~~~\text{a.e.}, \end{align*} where each $A_{n}$ is bounded measurable and $a_{n}>0$. Following the trick by @DanielWainfleet in this post, for each measurable $A_{n}$, one can find a decreasing sequence $(G_{k,n})_{k=1}^{\infty}$ of open sets such that $E\subseteq G_{k,n}$ and that $\chi_{G_{k,n}}\downarrow\chi_{A_{n}}$ a.e. as $k\rightarrow\infty$.
And again by @zhw.'s trick, one writes \begin{align*} \chi_{A_{n}}=\chi_{G_{1,n}}-\sum_{k=2}^{\infty}\left(\chi_{G_{k-1,n}}-\chi_{G_{k,n}}\right). \end{align*} Hence, \begin{align*} f=\sum_{n}a_{n}\left(\chi_{G_{1,n}}-\sum_{k=2}^{\infty}\left(\chi_{G_{k-1,n}}-\chi_{G_{k,n}}\right)\right), \end{align*} but how do we arrange the terms in the sum to achieve the goal?