# Approximation of non-measurable functions

Let $I = [0,1]$, $\phi, \psi: I \rightarrow \mathbb{R}$ functions. Does $\phi$ exist such that any $\psi$ adhering to the following inequality is non-measurable: $$\sup_{x \in I} |\phi(x) - \psi(x)| \le 1?$$ Assuming so, is it possible to replace 1 with $C$ representing a non-negative constant (and are there any bounds on what value $C$ can take)?

In particular, taking $\psi = \phi$, this criterion should force $\phi$ to be non-measurable itself. Besides risking this observation, I am at a loss for establishing such a $\phi$. I was trying to work with characteristic functions (i.e.,for non-measurable sets) since these can afford surprisingly-useful examples; nothing crystallized, though, so perhaps I went in a bad direction. I hope that someone visiting the site could assist me here. The second question is more of a personal question/ "philosophical" question for edification. Thanks in advance for any assistance you can give (it seems really curious that measure theory could yield a function with such a property)!

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Let $E$ be your favourite non-measurable set. Take $\phi=(2C+1)\;1_E$. Then, if $|\psi-\phi|\leq C$, we have $$\psi^{-1}(C,\infty)=E,$$ so $\psi$ is non-measurable. The number $C$ can then be any positive constant.
Let $X = B([0,1])$ be the Banach space of all bounded real-valued functions on $[0,1]$, equipped with the supremum norm $||f|| = \sup_{t \in [0,1]} |f(t)|$. The measurable functions $M$ are a closed subspace of $X$, since a uniform limit of measurable functions is measurable. Hence the set of non-measurable functions is open. If $f$ is non-measurable, then $d(f,M) := \inf_{g \in M} ||f-g|| > 0$. Since $M$ is a vector space and the norm is homogeneous, we have $d(af,M) = a d(f,M)$ for any constant $a > 0$. So taking any $a > C d(f,M)$, we have $d(af,M) > C$. So $af$ has the property you want.
This shows that not only does such a non-measurable function exist for each $C$, but that any non-measurable function can be scaled to have this property.