Measurable function simple --> measurable extension

Question

If $(X, \mathbb A)$ is a measurable space and $M \subset X$ arbitrary and $\mathbb A|_M := \{ A \cap M : A \in \mathbb A\}$. Let's define a measurable function $f: (M, \mathbb A_M) \to (\mathbb R, B(\mathbb R))$.

If we define this $f$ as being simple, why does it then admits a measurable extension?

Answer 1

The image of simple function $f$ is finite. Let's say it is $\{a_1,\dots,a_n\}$ where the $a_i\in\mathbb R$ are distinct.

Then the preimage of a Borel set wrt $f$ can be written as $\bigcup_{i=1}^n f^{-1}(\{a_i\})=\bigcup_{i=1}^n(A_i\cap M)$ where $A_i\in\mathbb A$ for $i=1,\dots,n$ such that the sets $M\cap A_i$ are disjoint and cover $M$.

Now define $B_i\subseteq X$ by $B_1=X\setminus (A_2\cup\cdots\cup A_n)$ and $B_i=A_i\setminus(A_1\cup\cdots\cup A_{i-1})$ for $i=2,\dots,n$.

The sets are disjoint with $B_i\cap M=A_i\cap M$ for $i=1,\dots,n$ and cover $X$.

Now we can prescribe a measurable simple function on $X$ that extends $f$ by stating that $x\mapsto a_i$ iff $x\in B_i$.

Answer 2

First step: if $f$ is a multiple of an indicator function, then it admits a measurable extension.

Second step: if $f,g : (M, \mathbb A_M) \to (\mathbb R, B(\mathbb R))$ admit measurable extensions, then so does $f+g$.