Let $M$ denote the set of measurable subsets of $\mathbb{R}$. Show
that there exists a non-measurable function $g:\mathbb{R}\to \mathbb{R}$ such that $g^{-1}(\{x\}) \in M$ for every $x \in
> \mathbb{R}$.Hint: Construct an injective function $g$ such that $\{x \in \mathbb{R}: g(x) > 0\}$ is a non-measurable set.
I am trying to figure out how to show the result using a construction satisfying the hint. My idea was:
Let $E$ be a non-measurable subset of $(0,\infty)$. Define $g$ as follows:
$$g(x)=\begin{cases}
x & x \in E \\
-x & x \in [0,\infty) \setminus E \\
x & x \in (-\infty,0) \end{cases}.$$
Then $g^{-1}(\{ x \})$ is measurable for each singleton (because it is finite (either 1 or 2 values)) but $g^{-1}((0,\infty)) = \{x \in \mathbb{R} : g(x) > 0 \}=E$, which is not measurable.
However, this is obviously not quite injective. How can I modify this to get an injective function?
Best Answer
Let $f: \mathbb{R} \to (0,\infty)$ be an injective function, e.g. $f(x) = \exp(x)$. Now define
$$g(x) := \begin{cases} x, & x \in E \\ -f(x), & x \notin E. \end{cases}$$
As $f$ is injective and takes only strictly positive values, it is not difficult to see that $g$ is injective but not measurable (as $\{g>0\} = E$).