Question: Given any bounded set $E$ in $\mathbb{R}^{N}$, is there a general way to choose a sequence $(G\_{n})$ of open sets such that $\chi\_{G\_{n}}\downarrow\chi\_{E}$ pointwise?

Here $\chi$ denotes the characteristic/indicator function on $E$.

One may think of $G\_{n}:=E+B(0,1/n)$, but we have instead that $\chi\_{G\_{n}}\downarrow\chi\_{\overline{E}}$ pointwise.

So if $E$ is closed, then it follows.

**As @TheoBendit has noted in the comment box that it is what $G\_{\delta}$ set is meant to be.**

So the answer is no.

Let me loose a little bit. What if I require only that $\chi\_{G\_{n}}\downarrow\chi\_{E}$ pointwise almost everywhere? 

One may think of remove the boundary $\partial E$ of the set $E$, but note that the boundary need no to be of measure zero, for example, $\partial(\mathbb{Q}\cap[0,1])=[0,1]$.

I also hope to have something like $\chi\_{G\_{n}}\downarrow\chi\_{E}$ pointwise quasi-everywhere with respect to $\text{Cap}\_{\alpha,s}$.

Here I introduce briefly the definition of $\text{Cap}\_{\alpha,s}$.

For compact sets $K$, we let $\text{Cap}\_{\alpha,s}(K)=\inf\{\|\varphi\|\_{W^{\alpha,s}}^{s}:\varphi\geq 1~\text{on}~K\}$.

For open sets $G$, we let $\text{Cap}\_{\alpha,s}(G)=\sup\{\text{Cap}\_{\alpha,s}(K): K~\text{is compact in}~G\}$.

For arbitrary sets $E$, we let $\text{Cap}\_{\alpha,s}(E)=\inf\{\text{Cap}\_{\alpha,s}(G): G~\text{is open that contains}~E\}$.

Answer: 
Let $\mu$ denote Lebesgue measure on $\Bbb R^N. $

If $E$ is $\mu$-measurable then there exists a sequence $(G\_n)\_n$ of open sets such that $E\subset \cap\_nG\_n$ and $\mu(\,(\cap\_nG\_n)\setminus E\,)=0.$ 

Let $H\_n=\cap \{G\_j:j\le n\}.$

If $p\in E$ then $\chi\_{H\_n}(p)=1=\chi\_E(p)$ for all $n.$

If $p\not \in \cap\_n H\_n$ then $\{n:p\in H\_n\}$ is finite. So for all but finitely many $n$ we have $\chi\_{H\_n} (p)=0=\chi\_E(p).$

So if $\chi\_{H\_n}(p)$ does not converge to $\chi\_E(p)$ then $p\in (\cap\_nH\_n)\setminus E, $which is a set of measure $0.$
For arbitrary bounded set $E$, then $\text{Cap}(E)<\infty$ and hence by outer regularity, we have $G\_{n}\supseteq E$ such that $\text{Cap}(E)\leq\text{Cap}(G\_{n})<\text{Cap}(E)+1/n$ for each $n=1,2,...$

Then $E\subseteq\displaystyle\bigcap\_{n}G\_{n}$. But $\text{Cap}\left(\displaystyle\bigcap\_{n}G\_{n}-E\right)+\text{Cap}\left(E\right)\geq\text{Cap}\left(\displaystyle\bigcap\_{n}G\_{n}\right)$, so the trick by @DanielWainfleet may fail for capacity case.

Nevertheless, if $\text{Cap}(E)=0$, then it follows:

Choosing decreasing $G\_{n}\supseteq E$ such that $\text{Cap}(G\_{n})<1/n$, then $\text{Cap}\left(\displaystyle\bigcap\_{n}G\_{n}-E\right)\leq\text{Cap}\left(\displaystyle\bigcap\_{n}G\_{n}\right)\leq\text{Cap}(G\_{n})<1/n$ for each $n=1,2,...$, so in this case, $\text{Cap}\left(\displaystyle\bigcap\_{n}G\_{n}-E\right)=0$. 

The rest follows by the trick of @DanielWainfleet.