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?

Solution 1:

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.