TITLE: Why is Lebesgue measure theory asymmetric?
QUESTION [20 upvotes]: 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?

REPLY [2 votes]: 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.)