Why every injective function satisfies the hypothesis?

Question

Here is the question I am trying to solve:

Suppose $f$ is a real-valued function on $\mathbb R$ such that $f^{-1}(c)$ is measurable for each $c.$ Is $f$ necessarily measurable?

Here is the hint I found here Must $f$ be measurable if each $f^{-1}(c)$ is? :

"Hint: Every injective function satisfies the hypothesis. You can take any nonmeasurable set, map it injectively into $(0,\infty)$, and map its complement injectively into $(-\infty,0)$ (for example using $e^x$ for a simple formula)."

My questions are:

1- why every injective function satisfies the hypothesis?

2- And why if I took any nonmeasurable set and map it injectively into $(0,\infty)$, and and map its complement injectively into $(-\infty,0)$ I will get the counter example I want?

Could someone help me answer these questions please?

Answer 1

If $f$ is injective then $f^{-1}(c)$ is either empty (the measure is $0$) or one point (the measure is $0$).

Take a non-measurable subset $T$ of the interval $(0,1)$. Then $S=T\cup [1,2]$ is non-measurable and of cardinality continuum. The latter means there is a bijection $f$ between $S$ and $(0,+\infty)$. Take a bijection $g$ between $\mathbb{R}\setminus S$ and $(-\infty, 0]$. Then the function $h(x)$ which is defined as $g^{-1}(x)$ if $x\le 0$ and $f^{-1}(x)$ if $x>0$, is injective, and so is $h^{-1}$. But $h^{-1}(0,+\infty)=S$ is non-measurable, hence $h^{-1}$ is a function from $\mathbb R$ onto itself such that the preimage of every point is of measure $0$ by the function itself is non-measurable.

Answer 2

For the first point, if $f$ is injective, then the preimage of each $c$ is either $\emptyset$ or a singleton, which are both measurables sets.

For the second one: take an unmeasurable set $A$, hence $A$ is uncountable (any countable set is measurable since it has measure equal to 0). On the other hand, its complement $A^{c}$ is uncountable, too: otherwise $\mu(A^{c})=0<+\infty$, hence $\mu(A)=\mu(\mathbb R)-\mu(A^{c})=+\infty-0=+\infty$, which contradicts the assumption on $A$.

Hence, because of the cardinality, you can build a bijection $A\to(0,+\infty)$ and another one $A^{c}\to(-\infty,0)$, that is you built an injective function $f\colon\mathbb R\to \mathbb R$. It is injective, hence it satisfies the hypothesis $f^{-1}(c)$ measurable, $\forall c\in\mathbb R$. On the other hand, $(0,+\infty)$ is measurable, but $f^{-1}(0,+\infty)=A$ is not by hypothesis, that is $f$ is not measurable.