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A set $E\subseteq \mathbb{R}^d$ is said to be Jordan measurable if its inner measure $m_{*}(E)$ and outer measure $m^{*}(E)$ are equal.However, Lebesgue mesure theory is developed with only outer measure.

A function is Riemann integrable iff its upper integral and lower integral are equal.However, in Lebesgue integration theory, we rarely use upper Lebesgue integral.

Why are outer measure and lower integral more important than inner measure and upper integral?

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    $\begingroup$ Lebesgue measurability can also be defined via inner and outer measure, and I believe that was how it was originally defined, the "outer measure only" definition by Caratheodory coming later. That said, I can't speak to why the Caratheodory definition seems to be presented more frequently. $\endgroup$
    – Noah Schweber
    Commented Aug 22, 2018 at 2:41
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    $\begingroup$ I do not think it is asymetric.. It may be true that for a set $E\subseteq \mathbb{R}$, lebesgue outer measure is defined by covering it from out side by open intervals... But, you call a set $E$ Lebesgue measurable only if for any $A\subseteq \mathbb{R}$ you have $\mu(A)=\mu(A\cap E)+\mu(A\cap E^c)$.. This can be seen as taking care of outer approximation (in case when you consider $A\cap E$) and inner approximation (when you consider $A\cap E^C$).. Does it make some sense.. ?? It is vague but I can not make it any better :D $\endgroup$
    – Praphulla Koushik
    Commented Aug 22, 2018 at 2:55
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    $\begingroup$ I agree with Noah and Praphulla — it’s rather an expository choice. As further evidence, see e.g. Bourbaki, Chap. IV, §4, exercises 6b) and 7b) or Godement (1948, p. 4, fourth displayed equation). $\endgroup$
    – Francois Ziegler
    Commented Aug 22, 2018 at 8:52
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    $\begingroup$ Also “Historical note” in Chap. V, pp. 126–127: In Lebesgue’s thesis, “imitating the Peano–Jordan method, the ‘outer measure’ of a bounded set $A\subset\mathbf R$ is defined... then, if $I$ is an interval containing $A$, the ‘inner measure’ of $A$ is the difference of the outer measures of $I$ and $I - A$; one thus obtains a notion of ‘measurable set’...” $\endgroup$
    – Francois Ziegler
    Commented Aug 22, 2018 at 9:25

6 Answers 6

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I have a (possibly idiosyncratic) view that the natural form of measure theory is for finite measure spaces and bounded functions. Other cases are obviously very important, but we have to work harder to get them. You can see this is many of the proofs, where the finite case is easier, and we have to work a bit more to generalize it. For example, the usual proof of the Radon-Nikodym theorem works that way.

In the finite measure space case, with bounded functions, everything can be made symmetric. The symmetry is broken in the general case, because allowing infinity breaks it. In integration, this asymmetry shows up in the way we have to have separate theorems for the non-negative measurable functions and the integrable functions. For bounded functions on finite measure spaces you don't need to impose any extra conditions.

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    $\begingroup$ One very explicit way in which the symmetry is broken is by declaring $0 \cdot \infty = \infty \cdot 0$ to equal $0$ rather than $\infty$. On the non-negative extended reals $[0,+\infty]$, this makes multiplication continuous from below, but not from above: if $a_n,b_n \in [0,+\infty]$ increase to $a,b$ respectively, then $a_n b_n$ increases to $ab$, but the same is not true for decreasing sequences. I think this already explains much of the asymmetry in Lebesgue measure theory. $\endgroup$
    – Terry Tao
    Commented Aug 22, 2018 at 14:56
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    $\begingroup$ If one adopted the opposite convention $0 \cdot \infty = \infty \cdot 0 = \infty$, and reversed the roles of upper and lower integrals, I think one also gets a self-consistent theory, but one that is completely degenerate as soon as one allows $\infty$ as a possible value of a function or as a possible value of a measure: all integrals would be infinite or divergent, because all functions are infinite on the empty set. $\endgroup$
    – Terry Tao
    Commented Aug 22, 2018 at 15:03
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    $\begingroup$ I'm sorry if I'm being obtuse, but I don't understand how everything can be made symmetric in this case either: a fat Cantor set has outer measure equal to its measure, but if I try to define inner measure in "Jordan way but with countable collections of intervals" I will still get zero (there are no intervals inside). I'm forced to use Lebesgue's definition of inner measure instead, which does not seem completely symmetric in the way Jordan case is (in the finite case, complement of a union of intervals is a union of intervals, which explains the symmetry, and fails for the countable case). $\endgroup$
    – Max M
    Commented Apr 11, 2019 at 18:21
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I think your statement about Jordan is actually wrong. If $m_*(E) = \infty$ and $m^*(E) = \infty$, then $E$ need not be Jordan measurable. If you talk only about bounded sets $E$, then your characterization is correct. But it is also correct for Lebesgue measure (using Lebesgue inner and outer measure).

The reason for Caratheodory's criterion is to define measurability when even bounded sets could have infinite measure, so that restricting to bounded sets no longer helps. One of Caratheorory's examples was an "arc length" measure for sets in $\mathbb R^n$. In that case, there is no obvious way to define inner measure. But we still can define outer measure. And then we need a criterion for measurability that uses only outer measure.

More recent mathematicians have developed a way to start only with an "inner measure" and go from there.

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I think, the reason is that if the ground space has infinite measure, you can not define the measurable sets as those for which inner measure equals the outer measure: it may happen that both are infinite, while the set is still not measurable.

Note also (this may be related) that outer and inner regularity behave differently in general. For example, a sigma-finite Borel measure on a Polish space is inner regular, but not always outer regular (example: counting measure of rational numbers as a measure on $\mathbb{R}$.)

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My two cents. Outer measures are sub-additive on countable coverings: $$A\subset \cup _{j\in\mathbb{N}} A_j\quad \Rightarrow \quad \mu( A)\le \sum_{{j\in\mathbb{N}}}\mu(A_j)$$ which is somehow a nicer and more practical property than the analogous dual property of super-additivity for inner measures (even in the case of finite measures). It gives a bound on the set $A$ in terms of the supposedly simpler sets $A_j$. Also, we like sub-additivity more than super-additivity, because it recalls norms.

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The asymmetry has only historical reasons. It is possible to develop Lebesgue theory (moreover, all extension theorems and thus the theory of product measures) from the “inner approach”. This was done by Heinz König in a couple of papers and monographs.

Although this “inner approach” is for the Lebesgue measure equivalent (more general, in the $\sigma$-finite case, IIRC), it has huge advantages for the non-$\sigma$-finite case, since the "outer approach" loses a lot of information - intuitively, there are “bubbles of measure $\infty$” which lost all information. For example, the product of Haar measures with the “inner” approach is compatible with the Fubini-Tonelli theorem one obtains from the approach by Haar integrals for functions with compact support. (For the “outer” approach this holds only for the $\sigma$-finite case, in general.)

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Concerning Tao comment that the symmetry is broken by declaring 0⋅∞ = ∞⋅0 = 0, I would like to add that this is the reason why Lebesgue integral does not satisfy Newton-Leibniz formula. Namely, for Cantor-Lebesgue function f, f(1)–f(0) = 1 but ∫01f’ = 0 because f’ = ∞ on Cantor set C which has measure 0 (and f’ = 0 on its complement). But if we realize that the measure of C is 0 = (1, 2/3, 4/9, ... ) and f’ = ∞ = (1, 3/2, 9/4, ...) then we see that this particular 0⋅∞ is not 0 but exactly 1, as it should be by Newton-Leibniz formula. We have the similar problem with countable additivity of limiting frequencies, which is usually contradicted by an infinite lottery with tokens 1,2,3,4,…. This contradiction also depends on ∞⋅0 = 0 and disappears if we really calculate the relevant ∞⋅0 ( https://www.researchgate.net/publication/290606552_A_Note_on_Probability_Frequency_and_Countable_Additivity )

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