# About functions on non-measurable sets

Let $$E$$ be a Lebesgue measurable subset of $$\mathbb{R}$$. We say that a function $$f:E\longrightarrow \mathbb{R}$$ is Lebesgue measurable if, for every $$\alpha\in \mathbb{R}$$, the set $$\{x\in E: f(x)>\alpha\}$$ is Lebesgue measurable (equivalently, "$$>$$" can be replaced by "$$\geq$$", "$$<$$", or, "$$\leq$$").

Now we consider a subset $$E$$ of $$\mathbb{R}$$, which is NOT Lebesgue measurable. The following questions seem to be interesting:

(1) Is it possible to define/have a function $$f:E\longrightarrow \mathbb{R}$$ such that, for every $$\alpha\in \mathbb{R}$$, the set $$\{x\in E: f(x)>\alpha\}$$ is Lebesgue measurable ?

(2) Is it possible to define/have a continuous function $$f:E\longrightarrow \mathbb{R}$$ such that, for every $$\alpha\in \mathbb{R}$$, the set $$\{x\in E: f(x)>\alpha\}$$ is Lebesgue measurable ?

Does anybody have answers/counterexamples concerning these (possible) pathologies? Or, these questions make sense ?

• Sets which are not Lebesgue measurable are quite pathological. Even finding such a set requires axiom of choice. I wouldn't worry about them so much. – Jakobian Jul 3 '19 at 21:36
• $(1)$ is false, because $$E = \bigcup_{k \in \mathbb{Z}} \{x \in E : f(x)>k\}$$ and the countable union of measurable sets is measurable, contradiction. ] – MathematicsStudent1122 Jul 3 '19 at 21:37

No, (1) (and hence (2) as well) cannot happen. Remember that the measurable sets form a $$\sigma$$-algebra: in particular, the union of countably many measurable sets is measurable.
Suppose $$E,f$$ are as in (1). We have $$E=\bigcup_{\alpha\in\mathbb{R}}\{x\in E: f(x)>\alpha\}.$$ But every real number is bigger than some rational, so this is the same as $$\bigcup_{\alpha\in\mathbb{Q}}\{x\in E: f(x)>\alpha\}.$$ By assumption, this latter set is a countable union of measurable sets (since $$\mathbb{Q}$$ is countable), so $$E$$ was measurable to begin with.
• We could also have used $$\mathbb{Z}$$ in place of $$\mathbb{Q}$$, and it's certainly easier to see that $$\mathbb{Z}$$ is countable; but using $$\mathbb{Q}$$ as an "$$\mathbb{R}$$-substitute" is so useful that I want to do it here too even if it's a bit silly. The specific property that makes $$\mathbb{Q}$$ really useful is that it's dense in $$\mathbb{R}$$; obviously $$\mathbb{Z}$$ isn't. Of course that's unnecessary here, but it gets used extensively down the road.
Incidentally, I've phrased this as a proof by contradiction, but that's really unnecessary: what we're actually doing is showing that for any $$E,f$$, if $$\{x\in E: f(x)>\alpha\}$$ is measurable for each $$\alpha\in\mathbb{R}$$ then $$E$$ is a union of countably many measurable sets and hence measurable. This is a completely direct proof.