I understand that a simple function $s:\mathbb{R}^2 \to \mathbb{R}$ is any function which assumes only a finite number of distinct values. It can also be written as a linear combination of indicator functions $s= \sum_{k=1}^N \alpha_k f_{A_k}$ with disjoint sets $A_k$ and distinct $\alpha_k \in \mathbb{R}$. I'm trying to prove that a simple function, defined above, is measurable if and only if all of the sets $A_k$ are measurable. Since I am still somewhat new to these types of proofs, I am having difficulty being rigorous, though I have a general idea of how to explain this.
[Math] Show that a simple function is measurable if its parts are all measurable
measure-theoryreal-analysis
Best Answer
Assume that $A_1,\dots,A_N$ are measurable. Let $B$ a Borel subset of $\Bbb R$ that do not contain zero. Let $I:=\{k\in [N],\alpha_k\in B\}$ (this set may be empty). Then $s^{-1}(B)=\bigcup_{k\in I}A_k$. Indeed, let $k\in I$; then $\alpha_k\in B$. If $x$ belongs to $A_k$, then $s(x)=\alpha_k\in B$ hence for each $k\in I$, $A_k\subset s^{-1}(B)$. Conversely, if $x\in s^{-1}(B)$, then $s(x)\in B$. Since $s$ can only take the values $\alpha_l$ for some $1\leqslant l\leqslant n$ and $s(x)\in B$, necessarily $s(x)=\alpha_k$ for some $k$ in $I$. Since the $\alpha_k$ are distinct, we necessarily have $ x\in A_k$. If $B$ contains zero, write $B=\left(B\setminus \{0\}\right)\cup\{0\}$ and notice that $s^{-1} \left(\{0\}\right)= \mathbb R^2\setminus\bigcup_{k=1}^{N}A_k$.
So, if the $A_k$ are all measurable, so will be $s$.
Now, assume $s$ measurable, and let $k\in [N]$. Then $s^{-1}(\{\alpha_k\})$ is measurable, as so is $\{\alpha_k\}$. But $s^{-1}(\{\alpha_k\})=A_k$, which concludes the proof.