I understand that a measure on a set is basically a way to assign a number to each suitable subset of that set and this number may be interpreted as "size".

Intuitively, what is the outer measure?


2 Answers 2


Sometimes the intuition is aided by a narrative rather than just doing the mathematics. Here is the story which I don't usually mind repeating.

For a long time the Riemann integral was considered the correct integral to handle bounded functions. Research continued on how best to do improper integrals but there wasn't much doubt that the Riemann integral did its job correctly on bounded functions. Until Volterra found a differentiable function $F$ with a bounded derivative $F'$ that was not Riemann integrable. Now a derivative $F'$ is not some weird function--it should be integrable. That doomed the Riemann integral! (Well it is having a long slow death but some of us would like to put it out of its misery.)

Lebesgue proposed to generalize the Riemann integral to handle the Volterra example. He started by imagining that he could produce an integral that handled all bounded functions and that possessed the properties that one would require. He argued that if he could do that then, at the same time (for free) he would have also produced a nice measure theory for how to estimate the length $\ell(E)$ of all sets $E\subset [a,b]$.

The integration theory predicted that the length function would have this property:$$\ell\left(\bigcup_{n=1}^\infty E_n \right) = \sum_{n=1}^\infty \ell( E_n )\ \ \ \ \ (\sharp) $$ whenever the sets $\{E_n\}$ are pairwise disjoint. This says the length of the set is equal to the sum of the lengths of it's pieces---reasonable? A very nice feature was that the integration theory produced the measure theory (for free) and it also works backwards: if you have the measure theory then (almost free with a bit of work) you can produce the integration theory.

So Lebesgue set out to develop the measure theory (using ideas that Borel had already published). As it turns out you cannot quite do this: not for all sets, only for nice sets.

To find out which sets were the nice sets he turned to an ancient method that the Greeks used (the method of exhaustion): if you want a correct measurement of an area or a volume, take your best shot at an overestimate and call it the upper area or upper volume. Then take your best shot at an underestimate and call it the lower area or lower volume. If the two agree then you have succeeded. If they disagree then either your methods are bad or else there really is no area or volume that makes sense.

So Lebesgue used this idea to produce an upper length $\ell^*(E)$ and also a lower length $\ell_*(E)$. If (just as the Greeks would have done) $\ell^*(E)=\ell_*(E)$ then he called the set measurable (i.e., nice). He was able to prove the identity $(\sharp)$ for measurable sets.

This meant, however, a slight change in the integration theory. Now he couldn't integrate all bounded functions, only ones that were closely related to the nice measurable sets. He called these functions measurable (using the same name as for sets). Then instead of producing an integral that integrates all bounded functions, it integrates all bounded measurable functions. But, restricted to these functions, the integral has all the strong integration properties that he required.

So the outer measure of the problem really traces back to this story with a lot of credit to the Greeks for the upper/lower or outer/inner idea, and a lot of credit to Lebesgue who managed to connect integration theory with measure theory.

Does that help the intuition? For a more technical but still heuristic discussion of Lebesgue measure see Section 2.1 of our Real Analysis textbook. Free PDFs available here.

  • $\begingroup$ Where does Caratheodory's construction fit in? $\endgroup$
    – Al jabra
    Commented Nov 5, 2015 at 10:44
  • $\begingroup$ Lebesgue, for his outer measure, started with a measure for open sets and then approximated for arbitrary sets from above by open sets containing them. Caratheodory realized that in any space you could construct an outer measure from a "premeasure" by a covering argument that is vaguely similar to Lebesgue's idea and reduces to it if you are on the real line and use open intervals for your cover. $\endgroup$ Commented Nov 5, 2015 at 22:47

Ideally any measure you define would have 4 desirable properties:

1.) The measure is defined for every subset of $\mathbb{R}$.

2.) The measure of any interval is equal to its length.

3.) The measure of a countable union of disjoint sets is equal to the sum of measures of each of the disjoint sets.

4.) The measure should be translation invariant.

It turns out that no matter how you attempt to define a measure, you can never satisfy these 4 ideas simultaneously!

The outer measure of A is defined to be the smallest sum of lengths of intervals that cover A and it turns out that the outer measure does not have property 3 for example!

  • $\begingroup$ What does the "outer" have to do in outer measure? $\endgroup$
    – Al jabra
    Commented Nov 4, 2015 at 13:53
  • $\begingroup$ The outer terminology in outer measure comes from the fact that we are using intervals that cover the set. The outer measure is the smallest measurement over all interval combinations that have some part outside of our set. As a contrast, an inner measure would be using the largest intervals that our set can cover. The inner measure is the largest measurement over all interval combinations that lie completely inside our set. $\endgroup$ Commented Nov 4, 2015 at 16:31
  • $\begingroup$ You will see the term "outer measure" applied to the situation in which $\mu\left(\bigcup_{n=1}^\infty E_n \right) \leq \sum_{n=1}^\infty \mu( E_n )$ and there is no restriction on what the sets are. The term measure (without the "outer") is reserved for situations where, when the sets $\{E_n\}$ are special pairwise disjoint sets we have equality and not just inequality: $\mu\left(\bigcup_{n=1}^\infty E_n \right) = \sum_{n=1}^\infty \mu( E_n )$. But some textbooks and research papers that use outer measures drop the "outer" terminology and just refer to everything as "measures." $\endgroup$ Commented Nov 4, 2015 at 17:50

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