Can I Prove that a finite to one function from a space with a reasonable topology to the discrete topology is always non measurable? Specifically, I am trying to prove that if I take $\Bbb R^m$ with the standard measure and $P(\Bbb R)$ with the discrete measure, so every set in $P(P(\Bbb R))$ is open and closed, $f:\Bbb R^m -> P(\Bbb R)$, and $f$ is finite to one, then $f$ is not measurable. But, I this would apply to any domain set with non-measurable sets and any range set with the discrete topology.
I've seen this question: Can a function be measurable but not with respect to a finer topology?
It seems to show that if $f$ is 1 to 1, then this is true. It feels like I should be able to send this to finite to 1, but I haven't been able to do that yet.
Any help with the general or specific question would be great. Thanks in advance.
PS, this is my first ever question, so if there's a way to link this question to that one or something, do let me know.
 A: Let the ordinal $c$ be the cardinal of $\Bbb R,$ which is also the cardinal of $\Bbb R^m $ and the cardinal of the set of all closed uncountable subsets of $\Bbb R^m$ and the cardinal of each uncountable closed subset of $\Bbb R^m.$
Let $F=\{f^{-1}\{x\}\}: \emptyset \ne x\in P(\Bbb R)\}.$ Then $F$ is a pair-wise disjoint family of finite sets and $\cup F=\Bbb R^m$.
Let $\{C_a:a<c\}$ be the set of all uncountable closed subsets of $\Bbb R^m.$ 
Define $\{A(d):d<c\}$ and $\{B(d):d<c\}$ as follows, recursively: 
For $a<c$ suppose that $\{A(d):d<a\}$ and $\{B(d):d<a\}$ are subsets of $F.$ Then the cardinals of $\cup\{A(d):d<a\}$ and $\cup \{B(d):d<a\}$ are each at most $|a\times \omega|,$ which is $<c.$ So we may take distinct $$x_a,y_a \in C_a\setminus ((\cup\{A(d):d<a\})\cup (\cup \{B(d):d<a\}))$$ such that $f(x_a)\ne f(y_a).$
Then define $A(a)=\{f^{-1}\{f(x_a\}\}\cup  \{A(d):d<a\}$ and $B(a)=\{f^{-1}\{f(y_a)\}\}\cup \{B(d):d<a\}.$ 
Now let $A=\cup \{A(a):a<c\}$ and $B=\cup \{B(a):a<c\}.$ Then $A,B$ are disjoint.
We have $A=f^{-1}\{f(y):y\in A\}.$
But $A$ is not Lebesgue-measurable because for every $C_a$ we have $x_a\in A\cap C_a$ but $y_a\in B\cap C_a\subset (R^m\setminus A)\cap C_a.$
Note: Lebesgue measure $m$ is inner-regular. If $A$ were measurable then $m(A)=\sup\{m(C): C=\overline C\subset A\}=0$ (because closed subsets of $A$ are countable sets) but also we would have $m(\Bbb R^m\setminus A)=0$ (because closed subsets that are disjoint from  $A$ are countable sets.)
Footnote: The idea is an adaptation of a (somewhat similar, but simpler) construction that shows there exists a non-measurable subset of $\Bbb R.$
