Why is Lebesgue measure theory asymmetric? 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?
 A: 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 $\mathbb{R}$.)
A: 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.
A: 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.)
A: 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.
A: 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 )
A: 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.
