# Is there a probability measure on $[0,1]$ with no subsets with measure $\frac{1}{2}$?

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?

• Show function $\phi(t) := \mu([0,t])$ is continuous. – GEdgar Jul 27 '16 at 0:05
• Yes, it works (to show the non-existence of such a measure). Thanks! – John Samples Jul 27 '16 at 0:10
• 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 $\mu^*$? – John Samples Jul 27 '16 at 0:12
• 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 – GEdgar Jul 27 '16 at 0:21
• 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 – GEdgar Jul 27 '16 at 0:59

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