This week I saw the definition of a measurable set for an outer measure.

Let $\mu^*$ be an outer measure on a set $X$. We call $A \subseteq X$ measurable if $$\mu^*(E) = \mu^*(A\cap E) + \mu^*(A^c\cap E)$$ for every $E \subseteq X$.

This is not the first time I've seen this definition. Unlike most other things in mathematics, over time I have gained absolutely no intuition as to why this is the definition.

The only explanation I've ever seen is that a set is measurable if it 'breaks up' other sets in the way you'd want. I don't really see why this is the motivation though. One reason I am not comfortable with it is that you require a measurable set to break up sets which, according to this definition, are non-measurable; why would you require that? Of course, you can't say what a non-measurable set is without first defining what it means to be measurable so I suppose no matter what your condition is, it will have to apply to all subsets of $X$.

Is there an intuitive way to think about the definition of measurable sets? Is there a good reason why we should use this definition, aside from "it works"?


As mentioned by Yuval in the comments, this question has previously been discussed on MathOverflow. I have replicated the accepted answer by Mark below.

Here is an argument that may give some intuition:

Assume that $m^{*}$ is an outer measure on $X$, and let us assume furthermore that this outer measure is finite:

$m^* (X) < \infty$

Define an "inner measure" $m_*$ on $X$ by

$m_* (E) = m^* (X) - m^* (E^c) $

If $m^*$ was, say, induced from a countably additive measure defined on some algebra of sets in $X$ (like Lebesgue measure is built using the algebra of finite disjoint unions of intervals of the form $(a,b]$), then a subset of $X$ will be measurable in the sense of Caratheodory if and only if its outer measure and inner measure agree.

From this viewpoint, the construction of the measure (as well as the $\sigma$-algebra of measurable sets) is just a generalization of the natural construction of the Riemann integral on $\mathbb{R}^n$ - you try to approximate the area of a bounded set $E$ from the outside by using finitely many rectangles, and similarly from the inside, and the set is "measurable in the sense of Riemann" (or "Jordan measurable") if the best outer approximation of its area agrees with the best inner approximation of its area.

The point here (which often isn't emphasized when Riemann integration is taught for the first time) is that the concept of "inner area" is redundant and can be defined in terms of the outer area just as I did above (you take some rectangle containing the set and consider the outer measure of the complement of the set with respect to this rectangle).

Of course, Caratheodory's construction doesn't require $m^*$ to be finite, but I still think that this gives some decent intuition for the general case (unless you think that the construction of the Riemann integral itself is not intuitive :) ).

  • $\begingroup$ For some students the inner/outer measure seems intuitively natural but the Caratheodory requirement does not. It may appear to be a big leap from the statement that $m^*(X) = m^*(A) + m^*(X\setminus A)$ to the perhaps strange condition that $m^*(T)=m^*(T\cap A)+m^*(T\setminus A)$ that is now required to hold for all $T$ not just for $T=X$. Even if $m^*(X)<\infty$ this is what Carathedory must use for measurability of sets with respect to an arbitrary outer measure. $\endgroup$ – B. S. Thomson Nov 3 '15 at 18:23

Old question here, but it is often asked by students of measure theory.

In our text Real Analysis (Bruckner${}^2$*Thomson) Example 2.28 in Section 2.7 there is a motivation for this that Andy used in his classes.

You would have seen that the inner/outer measure idea was successful for Lebesgue measure on an interval. Certainly if you are hoping for additivity of the measure then that is an obvious idea. Besides, when Lebesgue used it the upper=lower idea was pretty much well-established.

The example Andy gives is of a very simple finite outer measure $\mu^*$ on $X=\{1,2,3\}$ with $\mu^*(X)=2$, $\mu^*(\emptyset)=0$ and otherwise $\mu^*(A)=1$. You can certainly define an inner measure but $\mu_*(A)=\mu^*(A)$ works for all subsets and yet that measure is not even finitely additive. Evidently to spot the sets on which the measure will be additive you need to test more than just whether $\mu^*(A)+\mu^*(X\setminus A)=\mu^*(X)$. Caratheodory came up with the idea of testing all sets not just $X$ itself (since it clearly doesn't work).

  • $\begingroup$ Thompson How can you define the outer measure like that when $\mu^\ast(\{1\}) + \mu^\ast(\{2\}) + \mu^\ast(\{3\}) = 1 + 1 + 1 = 3 \neq 2 = \mu^\ast(X)$? $\endgroup$ – eurocoder Mar 11 '17 at 9:43
  • $\begingroup$ Check the definition of an outer measure. The requirement is that it be subadditive, not (necessarily) additive. Thus your computation might be written as $$2 = \mu^*(X) \leq \mu^*(\{1\}) + \mu^*(\{2\}) + \mu^*(\{3\}) =3.$$ This is what we want. Indeed the fact that there is strict subadditivity shows that these three sets cannot be measurable. $\endgroup$ – B. S. Thomson Mar 12 '17 at 16:01

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