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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 ?

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  • $\begingroup$ 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. $\endgroup$ – Jakobian Jul 3 '19 at 21:36
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    $\begingroup$ $(1)$ 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. ] $\endgroup$ – MathematicsStudent1122 Jul 3 '19 at 21:37
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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.

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