4

I have been told that a measure's atoms are at most a countable set. This has not been proved to me, and my book leaves it as an exercise. The only possible way I can think of is to try by contradiction, so suppose there is a measure whose atoms are more than countably many. So what? How do I go on? How do I prove this?

MickG
  • 8,645
  • Are you assuming the measure to be $\sigma$-finite? This isn't true for the counting measure. –  Jun 28 '14 at 17:38
  • My teacher didn't make any assumption, whereas the exercise in the book is about finite measures, but the statement is used in a proof related to the Lebesgue measure. So what assumption should I make? – MickG Jun 28 '14 at 17:42
  • Well, it's false for general measures but true for finite and $\sigma$-finite measures. Lebesgue measure has no atoms. –  Jun 28 '14 at 17:43
  • Yes, the statement is used precisely to prove that the Lebesgue measure has no atoms, combining the countability of the set of atoms with translation invariance. – MickG Jun 28 '14 at 17:46

1 Answers1

3

Here's how to prove your claim, with the appropriate assumption. Let $S\subset X$ be the set of atoms for some measure $\mu$ on $X$. Let $\{U_i\}$ be a countable measurable partition of $X$. Then if $S$ is uncountable, some $U_i$ contains an uncountable subset $S'$ of $S$, and $\mu(U_i)\geq \sum_{x\in S'}\mu(x)=\infty$ since any uncountable sum of positive numbers diverges. Thus $\mu$ is not $\sigma$-finite.

Kevin Carlson
  • 52,457
  • 4
  • 59
  • 113
  • How do you prove that every uncountable sum of positives converges? – MickG Jun 29 '14 at 12:42
  • We may assume the sum is of numbers less than or equal to 1 by normalizing, since of course an unbounded sum diverges. Decompose such a sum into subsums of numbers between $1/n$ and $1/n+1$. One of these subsums must be infinite, indeed uncountable, since otherwise the entire sum decomposes as a countable collection of countably many numbers. – Kevin Carlson Jun 29 '14 at 17:17
  • So the statement holds for all $\sigma$ finite measure, right? – John Nov 23 '14 at 07:41
  • Yes, what I showed is that $S$ uncountable implies $\mu$ not $\sigma$-finite, so if $\mu$ is $\sigma$-finite, then $S$ is countable. – Kevin Carlson Nov 23 '14 at 20:15