I am reading from Carothers' Real Analysis. He defines the outer (Lebesgue) measure of a subset $E$ of $\mathbb{R}$ as:

$$m^*(E) = \mathrm{inf}\left\{ \sum_{n=1}^{\infty} l(I_n) L : E \subset \bigcup_{n=1}^{\infty} I_n \right\}$$

where each $I_n$ is a bounded interval and $l(I_n)$ denotes the length of that interval.

I believe there is a technical problem: What if $E$ is an unbounded interval? Then of course we want $m^*(E) = \infty$. Indeed, in this case $\sum_{n=1}^{\infty} l(I_n) = \infty$ for any covering of $E$. But why is it okay to say $\mathrm{inf}\{\infty, \infty, \ldots\} = \infty$? Isn't the infimum supposed to work on sets of real numbers?

  • $\begingroup$ The infimum and supremum are defined on subsets of the extended reals by definition. $\endgroup$ – Gyu Eun Lee Dec 11 '14 at 3:39
  • $\begingroup$ Not usually... hence my question. Usually they are defined on subsets of the reals, and may take values in the extended reals. $\endgroup$ – mb7744 Dec 11 '14 at 3:47
  • $\begingroup$ Basically my question is: is the extension of inf and sup to work on subsets of the extended reals a usual thing, or have I misinterpreted something about the outer measure? $\endgroup$ – mb7744 Dec 11 '14 at 3:48
  • $\begingroup$ The infimum and supremum can be defined on any partially ordered set, which the extended reals certainly are. Your notion of outer measure is fine, and you may work with the extended reals in the natural way. $\endgroup$ – Gyu Eun Lee Dec 11 '14 at 3:49
  • $\begingroup$ Or, if you prefer, the set over which the infimum is being taken could be interpreted as the empty set (if you don't count any of the values $\infty$); and the infimum of the empty set is most naturally defined as $+\infty$. $\endgroup$ – Greg Martin Dec 11 '14 at 4:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.