Does there exists a non-measurable function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that for every function $g : \mathbb{R} \rightarrow \mathbb{R}$ with $$ |g(x)-f(x)|<1 $$ for all $x \in \mathbb{R}$, then $g$ is also non-measurable?
I am thinking about function $f$ such that $f(x)=|x|+100$ for $x$ in any non-measurable subset of $\mathbb{R}$ and $f(x)=|x|-100$ for $x$ in any measurable set of $\mathbb{R}$. But I don't know whether this works or not since measurable set and non-measurable can intersect. Thanks.
Note : the measure here is Lebesgue measure
Your basic idea works. Fix a non-measurable subset $A \subseteq \mathbb{R}$ and define $f = 3 1_A$. If $g$ is measurable then $g^{-1}(2, \infty)$ is measurable. If $|f(x) - g(x)| < 1$ for every $x$ then $g(x) > 2$ iff $x \in A$ so that $A$ is measurable which is a contradiction.