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$A\subset\mathbb R^n$ is measurable if for all $B\subset \mathbb R^n$

$$m^*(B)=m^*(B\cap A)+m^*(B\setminus A)$$

This is just a definition, and it basically says that the outer measure can be partitioned into the outer measure of the intersection of set $A$ with any other set, and the outer measure of the intersection of that set with the complement of $A.$

But why is it defined as such?

One thing it achieves is countable additivity: given two disjoint Lebesgue measurable sets $E$ and $F:$

$$\begin{align}m^*(E\cap F)&= m^*\left((E\cap F) \cap E \right ) + m^*\left((E\cap F) \cap E^c \right )\\[2ex] &=m^*(E )+ m^*(F) \end{align}$$

And relatedly, how does it connect with the alternative definition:

For all $\epsilon>0$ there exist an open set $G$ and a closed set $F$ such that $F\subset A\subset G$ and $m^*(G\setminus F)<\epsilon$ ?

Antoni Parellada
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    The is the Caratheodory criterion, which is (to my mind) one of the cleverest and most non-obvious results in analysis (though, in retrospect, it is exactly the right definition). Basically, the idea is that a set is measurable if, no matter how to carve it up, the measure doesn't change. By looking at the intersections with $B$ and $B^\complement$ for arbitrary sets $B$, we are carving up $A$ in as many ways as is possible. You might think about what this has to do with disjoint additivity (which, depending on how the axioms are stated, is either an axiom or theorem). – Xander Henderson Mar 12 '18 at 23:26
  • @peterag. Right! The A and B in the def'n must be reversed. – DanielWainfleet Mar 13 '18 at 06:35
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    See https://math.stackexchange.com/a/667529/442

    I cannot mark this as a duplicate, since the poster there never accepted any answer.

     – GEdgar Mar 13 '18 at 11:01
  • @DanielWainfleet the post has been updated, so I deleted my comment... – peter a g Mar 13 '18 at 13:23
  • If I were teaching this I would prefer not to take this as the def'n but as a result following from a more intuitive def'n in terms of open covers. – DanielWainfleet Mar 14 '18 at 06:57

1 Answers1

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Lebesgue's outer measure $m^*$ is constructed as an attempt to extend the notion of length of an interval, in a coherent way, to all of $\mathcal P(\mathbb R)$.

But then Vitali sets exist, which show that $m^*$ is cannot be a measure. Caratheodory's criterion gives you a family of subsets $\mathcal M$ that satisfies three things:

  • restricted to $\mathcal M$, the outer measure $m^*$ is a measure

  • $\mathcal M$ is a $\sigma$-algebra

  • $\mathcal M$ contains all open sets, and thus the Borel $\sigma$-algebra

The three points together give us a big enough domain (bigger than the Borel $\sigma$-algebra) where the idea of $m^*$ works and actually provides a measure. e: the Lebesgue measure.

On the question about how Caratheodory came up with such an idea, I cannot really say. With the criterion already in place, one may think that the motivation is to take sets where $m^*$ is additive.

Martin Argerami
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