Let $(\Omega, \mathcal F, \mu)$ be a measure space. A function $f$ is called $\mu$-simple if $f= \sum_{k=1}^n a_k 1_{A_k}$ where $(a_k)_{k=1}^n \subset \mathbb R_{\neq 0}$ and $(A_k)_{k=1}^n$ is a sequence of pairwise disjoint sets in $\mathcal F$ with finite measures. I'm trying to prove below result, i.e.,
Theorem . If $\mu(\Omega) < \infty$, then the space of $\mu$-simple functions is dense in $L^\infty (\Omega)$.
Could you have a check on my below attempt?
Proof Fix $f \in L^\infty (\Omega)$. Then there is a finite $M>0$ such that $|f| \le M$ a.e. Fix $\varepsilon >0$. We will find a $\mu$-simple $g$ such that $\|f-g\|_\infty \le \varepsilon$. Let $N:= \lceil 2M /\varepsilon \rceil$. Let $$ A_k := \{-M + k\varepsilon \le f < -M +(k+1) \varepsilon\} \in \mathcal F \quad \forall k \in \{0, 1, \ldots, N\}. $$
Clearly, $\Omega \setminus \bigcup_{k=1}^{N} A_k$ is a $\mu$-null set. Because $\mu$ is finite, $\mu(A_k) < \infty$ for all $1 \le k \le N$. Let $$ g := \sum_{k=1}^{N} (-M + k\varepsilon) 1_{A_k}. $$
Clearly, $g$ satisfies our requirement. This completes the proof.