If $E$ is finite-dimensional and $\mu$ finite show that $\mathcal S(X,\mu,E)$ is dense in $\mathcal L_\infty(X,\mu,E)$
This is an exercise of the book Analysis III of Amann and Escher. Here $\mathcal S(X,\mu,E)$ is the space of $\mu$-simple functions and $\mathcal L_\infty(X,\mu,E)$ is the space with the seminorm $\|{\cdot}\|_\infty$ of functions essentially bounded, with $E$ a finite-dimensional Banach space.
I give an attempt of a proof. Can you check the correctness of the proof or comment in my mistake? Thank you.
Proof:
Set $A_{j,n}:=\{x\in X:|f(x)|\in[\|f\|_\infty j2^{-n},\|f\|_\infty(j+1)2^{-n})\}$ for $j=0,\ldots, 2^n-1$. Then clearly each $A_{j,n}$ is measurable, $A_{j,n}\cap A_{k,n}=\emptyset$ when $j\neq k$ and $X=\bigcup_{j=0}^{2^n-1}A_{j,n}$.
Note that $f(A_{j,n})$ is relatively compact because $E\cong\Bbb K^n$ (for $\Bbb K\in\{\Bbb R,\Bbb C\}$) for some $n\in\Bbb N$. Then there is a finite open cover of $f(A_{j,n})$ of open balls of arbitrarily small radius.
Then let $B_{1,n},\ldots,B_{m,n}$ a finite open cover of open balls of radius $2^{-n-1}$ of $f(A_{j,n})$. Then we set $V_{k,n}:=(B_{k,n}\setminus(\bigcup_{h=1}^{k-1}B_{h,n}))\cap f(A_{j,n})$, so the $V_{k,n}$ are a partition of borel sets of $f(A_{j,n})$, so clearly the sets $R_{k,n}:=f^{-1}(V_{k,n})\cap A_{j,n}$ are measurable and a partition of $A_{j,n}$. Finally choose some $d_k\in V_{k,n}$ for each $k$ and set $g_{j,n}(x):=\sum_{k=1}^m\chi_{R_{k,n}}(x)d_k$.
Then, by construction, $g_{j,n}$ is $\mu$-simple and also is $g_n:=\sum_{j=0}^{2^n-1}g_{j,n}$. More over, it have the property that $|f(x)-g_n(x)|<2^{-n}$ for each $x\in X$, so we can conclude that $\mathcal S(X,\mu, E)$ is dense in $\mathcal L_\infty(X,\mu,E)$.$\Box$
You use the fact that E has finite dimension when you say ''$f(A_{j,n}$ is relatively compact " . In a finite-dimensional space, it is true that a bounded set is relatively compact because it's closure is bounded and closed thus compact. But in a infinite-dimensional space, a Riesz's theorem claims that all the balls are not compact. For this reason if $E$ is not finite-dimensional then $f(A_{j,n})$ is bounded but not anything more and you can't find a finite open cover.