3

I am reading this Wikipedia article about the Lebesgue measure.

Note that the definition uses a countable number of intervals to cover the set $E\subset \mathbb R$.

Do we get a different measure if we admit uncountable collections of intervals? And if so, what kind of measure do we obtain?

Michael Albanese
  • 99,526
a student
  • 4,365
  • 1
    To expand on the above comment, a reasonable definition of the sum of uncountably many numbers would have that a sum of non-negative numbers was greater than or equal to every finite subsum. As a result if the sum is finite then for every $n$ there can only be finitely many elements in the set of size greater than $1/n$, and taking a limit, only countably many greater than zero. – jxnh Jun 20 '15 at 05:29

1 Answers1

6

Let $I$ be an uncountable set, and $\{a_i\}_{i \in I}$ a collection of non-negative real numbers. If $\sum_{i \in I}a_i < \infty$, then $a_i = 0$ for all but countably many $i$; see this question.

As every nonempty open interval has positive length, an uncountable sum of such lengths would necessarily diverge. Therefore, the sum of lengths of uncountably many open intervals diverges unless all but countably many of the intervals are empty. So, provided you allow empty sets in the cover, the proposed measure is equal to the Lebesgue outer measure. If empty sets are not allowed, you don't even get a measure as every set will be assigned the value $\infty$ (even $\emptyset$).

If you use closed intervals instead of open intervals, you get the trivial measure because any set can be written as the union of singletons which are closed intervals with length zero.

Community
  • 1
Michael Albanese
  • 99,526
  • (+1) Totally agree. And this issue is also related to Zeno's paradox of moving arrow: if each point has zero length in the intuitive sense, and if intervals are consists of points, then how do they "add up" to some nontrivial length for an interval? – Fei Li Aug 25 '16 at 15:46