Proving that some set is dense in $L^\infty(\Omega)$ Let $\Omega$ be an open subset of $\mathbb R^n$ and define $$S:= \big\{ s: \Omega\subseteq \mathbb R^n \to \mathbb C | s \textrm{ is measurable and } s(\Omega) \textrm{ is a finite set} \big\}. $$ 
I have to prove that $S$ is a dense subset of the Banach space $L^\infty(\Omega)$ of essentially bounded functions.
My thoughts: I have already proved that for any measurable function $\phi\colon \Omega \to \mathbb C$ there exists a sequence of simple functions $\big( \vartheta_k\colon \Omega \to \mathbb C \big)_{k\in \mathbb N}$ such that $0\leq |\vartheta_1|\leq |\vartheta_2|\leq \cdots\leq |\phi|$ and $\lim\limits_{k\to \infty} \vartheta_k(x)=\phi(x)$ for each $x\in \Omega$. It is easy to see that each $\vartheta_k$ belongs to $S$. 
So, if we take $f\in [\phi]$ for some $[\phi]\in L^\infty(\Omega)$ (let's say that we take $ f=\phi$), there does exists a sequence of functions $\vartheta_k\colon \Omega \to \mathbb C$ as above. 
I have troubles to show that $\lim\limits_{k\to \infty}\lVert \phi-\vartheta_k \rVert_{\infty}=0$, as far as we only have pointwise convergence. It must be easy to proceed at this point but I just don't see how right now. How would you conclude?, do you see another way to solve the problem?
Thanks in advance!
 A: You can do separately for the real and imaginary part. So let $f \in L^{\infty}(\Omega)$, real valued. Consider $M>0$ so that $ \{x\in \Omega\ | \ |f(x)| > M\}$ has measure zero.  Now, for every $k\ge 1$ natural, consider the function $f_k$ that equals $\frac{l M}{2^k}$ on the set $\{x \in \Omega \ | \ \frac{l M}{2^k} \le f(x) < \frac{(l+1)M}{2^k} \}$ for every $l$ integer, $-2^k \le l \le 2^k$. On the set $\{x \ | |f(x)|>M\}$ define $f_k$ to be $0$ ( any other value works). One notices that $f_k$ is increasing and $0\le f-f_k < 2^{-k}$ for every $k\ge 1$ on the set $\{x \ | \ -M \le f(x) \le M\}$. Moreover, each $f_k$ is measurable and takes finitely many values. 
A: Suppose $f:\Omega\to \mathbb C$ is measurable, with $|f|\le M $ on $\Omega.$ Let $\epsilon>0.$ Partition $[-M,M]^2$ into finitely many pairwise disjoint Borel sets $E_1,\dots E_n,$ with $\text { diam } E_k < \epsilon$ for every $k.$ (Slap an $\epsilon/2$ grid on $[-M,M]^2$ and a choice of appropriate $E_k$'s reveals itself.) Then each $f^{-1}(E_k)$ is measurable, and $\cup \,f^{-1}(E_k) = \Omega.$ Select any points $z_k\in E_k, k = 1,\dots ,n.$ Set $s=\sum_{k=1}^{n}z_k\chi_{f^{-1}(E_k)}.$ Then $s$ is measurable, takes on only the values $z_1,\dots ,z_n,$ and $|s-f|<\epsilon$ pointwise everywhere on $\Omega.$ Hence $\|s-f\|_\infty\le\epsilon.$
