According with the book Lages Lima - Análise real, a subset $X \subset \mathbb{R}$ has zero measure whenever for every $\epsilon>0$ there is a countable cover made up of open intervals, such that the sum of the lenght of all those intervals is less than $\epsilon$.

My question is : why do we require the cover to be countable?

  • $\begingroup$ In addition to the A from Zachary Selk: If $X\subset \cup F$ where $F$ is a family of intervals of $\mathbb R,$ there exists countable $G\subset F$ with $X\subset \cup G.$ $\endgroup$ – DanielWainfleet Jan 6 '17 at 14:27
  • $\begingroup$ The point is rather that you need to define what is the sum of uncountably many numbers. Whereas the sum of countably many positive number is pretty standard. $\endgroup$ – user251257 Jan 6 '17 at 14:44

Because the sum of uncountably many positive numbers is necessarily infinite. See:

Can we add an uncountable number of positive elements, and can this sum be finite?

| cite | improve this answer | |
  • $\begingroup$ But then it is not necessary to impose the condition of countability since if the cover is made up of uncountably many open intervals then that cover will not satisfy the condition of having a total lenght of less than $\epsilon$, right? $\endgroup$ – la flaca Jan 6 '17 at 14:17
  • $\begingroup$ @laflaca I guess technically you're right but it's just emphasizing the point. $\endgroup$ – user223391 Jan 6 '17 at 14:21

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.