Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)!

share|cite|improve this question

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.

share|cite|improve this answer
+1 for letting us choose our favorite non-measurable set – leo Mar 3 '12 at 0:37

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.