TITLE: Two-valued measure is a Dirac measure
QUESTION [13 upvotes]: Let $(X,\mathfrak B)$ be a measurable space such that $\{x\}\in \mathfrak B$ for all $x\in X$, and let $\mu$ be a positive measure on this space such that
$$
  \mu(B) \in\{0,1\} \quad\text{for all }B\in \mathfrak B.
$$
What are the mildest conditions on $(X,\mathfrak B)$ that imply that $\mu =\delta_x$ for some $x\in X$?
It is known to hold for $\Bbb R$ with a Borel $\sigma$-algebra, and I believe it fairly easy extends to $\Bbb R^n$. I wonder, though, whether it holds at least for locally compact Polish spaces, or perhaps for more general case. I am also interested in examples of spaces where such statement does not hold.

REPLY [2 votes]: A nice case to consider:  
Let $X$ be a completely regular topological space.  The zero sets in $X$ are sets of the form $\{x \in X : f(x)=0\}$ where $f : X \to \mathbb R$ is continuous.  The sigma-algebra $\mathcal B$ of Baire sets in $X$ is the sigma-algebra generated by the zero sets.  Alternatively, $\mathcal B$ is the smallest sigma-algebra on $X$ such that every continuous real-valued function is $\mathcal B$-measurable.
The condition: 

every (countably-additive) $\{0,1\}$-valued measure on $\mathcal B$ is a Dirac measure  

is equivalent to the condition that $X$ is real-compact.
References:
Gillman & Jerison, Rings of Continuous Funtions.
W. Moran, "Measures and mappings on topological spaces." Proc. London Math. Soc.
19 (1969) 493-508.