# Lebesgue non-measurable function

Can we give an example of Lebesgue non-measurable function, for which set $\{x: f(x)=C\}~\forall C\in\mathbb{R}$ is measurable? Thanks.

• You can get such a functions by adding pretty much any analytic function to a non-measurable function. – leftaroundabout Apr 16 '15 at 17:13

Let $S$ a non-measurable subset of $]0,+\infty[$. Define $$g(x)=\begin{cases} x\text{ if } x\in S\\-x\text{ if } x\notin S\end{cases}$$

$g^{-1}(y)$ is finite $\forall y\in \mathbb{R}$, but $\{ g\geq 0\}\setminus\ ]-\infty,0]=S$ is not measurable.

• What does $]0, +\infty]$ denote? – David Zhang Apr 16 '15 at 19:37
• @DavidZhang Editing right... NOW! – user228113 Apr 16 '15 at 19:39
• Huh. I'm not sure what $]0, +\infty[$ denotes either. Is it the complement of $[0, +\infty]$, perhaps? – David Zhang Apr 16 '15 at 19:43
• I believe $]$ and $[$ are notations of open intervals. It's uncommon, but, for example, I've seen it in Brown's Topology and Groupoids. – user83387 Apr 16 '15 at 19:45
• Its much more common in continental Europe. – Calvin Khor Apr 16 '15 at 23:03

Take $V$ to be a non-measurable set on $[0,1]$, and consider on $[0,1]$ the function $$f(x) := x \mathbf{1}_V(x) + (-10 -x)\mathbf{1}_{[0,1]\setminus V}(x)$$ where $\mathbf{1}_V$ is the indicator function of $V$, $$\mathbf{1}_V(x) := \begin{cases} 1 & x∈ V \\ 0 & x \notin V \end{cases}$$Then the preimage of any $C∈ℝ$ is a singleton or empty and hence measurable.

Its also not hard to use this idea to make a non-measurable function on $ℝ$, also satisfying your criterion: $$f(x) := e^x \mathbf{1}_V(x) + (-10 -e^x)\mathbf{1}_{ℝ \setminus V}(x)$$

• Excuse me, what does $\mathbf{1}_V(x)$ mean? – Wanksta Apr 16 '15 at 15:49
• 1 if x is in V, 0 otherwise. – Paul Apr 16 '15 at 15:52
• @olegas You'll often see it called the indicator function or characteristic function of the set $V$, though the first name is preferred since "characteristic function" has different meanings in different fields. – David Zhang Apr 16 '15 at 19:36