12
$\begingroup$

I have a decidedly weird question.

Does there exist a probability measure $(\mu, \mathcal{F})$ on $[0,1]$ such that

1) $\mu(x) = 0$ for every $x \in [0,1]$

2) For every $r \in [0,1] \setminus \lbrace \frac{1}{2} \rbrace$, there exists $A \in \mathcal{F}$ with $\mu(A) = r$, and

3) There is no $A \in \mathcal{F}$ with $\mu(A) = \frac{1}{2}$?

The question comes about from a prelim problem in which the existence of a measure-$\frac{1}{2}$ set was assumed, and it made me wonder whether the assumption was for convenience or necessary. Pigeonhole principle? Ultrafilters?

$\endgroup$
  • 4
    $\begingroup$ Show function $\phi(t) := \mu([0,t])$ is continuous. $\endgroup$ – GEdgar Jul 27 '16 at 0:05
  • $\begingroup$ Yes, it works (to show the non-existence of such a measure). Thanks! $\endgroup$ – John Samples Jul 27 '16 at 0:10
  • $\begingroup$ Oh sorry, your solution is actually incorrect. The measurability of those sets isn't assumed. Will $\phi$ still be continuous if we replace $\mu$ with $\mu^*$? $\endgroup$ – John Samples Jul 27 '16 at 0:12
  • $\begingroup$ OK, if this is not assumed a Borel measure, then you will have to show it is a nonatomic measure then see en.wikipedia.org/wiki/Atom_%28measure_theory%29 $\endgroup$ – GEdgar Jul 27 '16 at 0:21
  • 4
    $\begingroup$ My second hint is also no good ... 1) and 2) do not imply the measure is nonatomic. So we have to go to... "Halmos showed that the range of a non-negative, finite measure is a closed subset of real numbers." see mathoverflow.net/questions/160338 $\endgroup$ – GEdgar Jul 27 '16 at 0:59
12
$\begingroup$

Since $\mathcal F$ is any $\sigma$-algebra on $[0,1]$, this is essentially the general case of:

Halmos [1] showed that the range of a non-negative, finite measure is a closed subset of real numbers.


[1] Halmos, Paul R. On the set of values of a finite measure. Bull. Amer. Math. Soc. 53 (1947), no. 2, 138--141. http://projecteuclid.org/euclid.bams/1183510408.

$\endgroup$

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.