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?



Why is Lebesgue measure theory asymmetric?
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.


