We know that a function from $\mathbb R$ to $\mathbb R$ is continuous iff the graph can be drawn without lifting the pen. I want to know if there is a similar intuitive characterisation for measurable sets in particular Lebesgue measurable set. I mean how, by just looking at a function can I determine whether it is Lebesgue measurable or not. I know this is a vague question but it would help me a lot to visualize.
2026-03-30 00:17:54.1774829874
Criterion for a function to be Lebegsue measurable.
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I'm uncertain how you could easily visualize this, as just about every set we end up working with in practice is measurable. We have to go through quite a bit of effort using the axiom of choice somewhere to produce non-measurable sets.
I find this equivalent definition to be a far better visualization of what it means for a set to be measurable $E$ is measurable if and only if it can be used to parition all sets and preserve the outer measure additively: $$\forall A\subseteq \mathbb{R}, m^*(A)=m^*(A\cap E)+m^*(A\cap E^c) $$ wiki
Basically this says that your set can't be used to split up any other set in a way to make additivity of outer measure break down. Again, just about every set not specifically constructed to be a counterexample will do, and every counterexample constructed will be one that was developed inherently with Axiom of Choice.
As a side note, just having the axiom of choice being false is not quite enough to have a math system with no nonmeasurable sets of real numbers, we actually have that
"ZF+all subsets of Real numbers are measurable" is equiconsistent with "ZF+ not AC+ there exists an inaccessible cardinal"
where inaccessible cardinals are the "smallest"* of the "Large cardinal hypothesis", sets too big to build straight from ZF. In this case, it'd be a set too big to be built from smaller sets using only union, cartesian products, and power sets.
(* on smallest, there is one thing that is technically smaller, but according to my grad school set theory teacher, just about nobody uses "worldly" cardinals, which would be smaller)