Definition of non measurable sets

Question

A set $A\subset \mathbb{R}^n$ is called measurable if $\forall B \subset \mathbb{R}^n \colon \mu(B) = \mu(B \cap A) + \mu(B\backslash A)$, where $\mu$ is the $n$-dimensional Lebesgue measure. A set $A \subset \mathbb{R}^n$ is therefore not measurable if $\exists B \subset \mathbb{R}^n \colon \mu(B) < \mu(B \cap A) + \mu(B\backslash A)$, since $\mu$ is monotone with respect to inclusion. The name ''non measurable'' suggests that we can't ''compute'' the volume of a set. This suggestion also makes sense if we talk about the vitali set $P$. There one shows that $\mu(P)= \mu([0,1])=1$ and $\mu(P)=0$ or $\mu(P)=\infty$. Therefor the vitali set has no reasonable volume. Now, from this observation that there are sets which we can't assign a volume (otherwise we have a contradiction), how does one come up with the standard definition of measurability which I presented above?

Answer 1

Statement of this in a notation I prefer ...

A set $A\subset \mathbb{R}^n$ is called $\mu^*$-measurable iff $\forall B \subset \mathbb{R}^n \colon \mu^*(B) = \mu^*(B \cap A) + \mu^*(B\backslash A)$, where $\mu^*$ is an outer measure.

This definition is due to Carathéodory.

Carathéodory's reasoning for the definition is known (see his personal notes given in his collected works). For Lebesgue measure in $\mathbb R$, the condition for measurability given by Lebesgue could be stated

A set $A\subset \mathbb{R}$ is called Lebesgue measurable iff for all bounded intervals $B \subset \mathbb{R}$ we have $\mu^*(B) = \mu^*(B \cap A) +\mu^*(B\backslash A)$, where $\mu^*$ is Lebesgue outer measure.

Carathéodory was trying to generalize to outer measures $\mu^*$ in $\mathbb R^n$ where all intervals have infinite outer measure, so Lebesgue's formulation was useless. He came up with the idea to use all subsets, not just the intervals. In cases where bounded intervals have finite measure, we can show Carathéodory's definition agrees with Lebesgue's.

An example, already given by Carathéodory, is: "arc length" measure in $\mathbb R^n$, with $n \ge 2$.

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In Lebesgue's case, when our set $A$ is bounded, we take a large "interval" (in $n>2$ a hypercube) $B \supseteq A$. Then we define inner measure $\mu_*(A) = \mu^*(B) - \mu^*(B\setminus A)$. The equation $\mu^*(B) = \mu^*(B \cap A) +\mu^*(B\backslash A)$ then tells us exactly that the inner measure of $A$ equals the outer measure of $A$. This is the concept Lebesgue started with.