Non-measurable bijection

Question

I wanted to find a non-measurable bijection $f: \Bbb R \to \Bbb R$. Of course it must be non-continuous so I thought of taking a Vitali set $V\subset(0,1)$ and defining $$f(x)=\cases{ x & if $x\leq 0$ \\ x & if $x \in (0,1) \cap V^C$ \\ x+1 & if $x \in V$ \\x+1 & if $x\geq 1$} $$ We can verify that $f^{-1}((1,2))=V$ so $f$ is not measurable but it is a bijection. However, I know that some of you usually come up with simpler/easier/creative/ more beautiful counterexamples so I wanted to find out what other examples of such functions can be constructed. (Of course I think we can have uncountably many other examples by taking a continuous bijection $f$ and shifting a non measurable part of the image by a constant upwards...)

Answer 1

You can simplify by taking $f(x)=x+1$ if $x \in V$ and $f(x)=x-1$ if $x \in V+1$, with $f(x)=x$ if $x \notin V\cup(V+1)$.