2

As I understand it, the outer measure $\mu^{*}(A)$ is used to find the length of the smallest cover that covers $A$.

However, in another definition, the outer measure is defined as the largest lower bound.

My question is, Shouldn't the outer measure be defined as the least upper bound since we are trying to find the smallest cover that covers $A$.

Sorry, if this sounds like a stupid question, but I just started learning about outer measure and I am a little bit confused about the definition.

Thanks.

user3879021
  • 375
  • 1
    The "another definition" is called *inner measure*. – Zhanxiong May 19 '16 at 17:38
  • does someone has a good and intuitive example of a set $E \subset \mathbb{R}$ such that $m(E) \ne m^*(E)$ ? I only found http://math.stackexchange.com/a/148557/276986 and maybe https://en.wikipedia.org/wiki/Regular_measure#Inner_regular_measures_that_are_not_outer_regular – reuns May 19 '16 at 17:46
  • @user1952009 Standard examples all require the axiom of choice. In fact if you abandon the axiom of choice, there are models of set theory in which all sets are measurable and no such sets E exist. See http://www.math.wisc.edu/~miller/old/m873-03/solovay.pdf. – btilly May 19 '16 at 18:01

1 Answers1

0

It should not be the least upper bound, because there generally is no upper at all bound on how big a covering set can be. Also there may not exist any cover that is the exact size of the outer measure.

For example if you construct the outer measure out of the union of open intervals of rational length, an interval of irrational length will not have any cover that is exactly its size. And it will have covers of arbitrary size.

btilly
  • 1,859