Question

Let $\mathbb{P}$ be a probability measure on some probability space $(\Omega,\mathcal{A})$. Are there conditions on the $\sigma$-algebra $\mathcal{A}$ such that for every real number $c\in [0,1]$ we find a set $A\in\mathcal{A}$ with $\mathbb{P}(A)=c$. It is like the intermediate value theorem for continuous functions.

Answer 1

A measure space $(\mathbb{P},\Omega,\mathcal{A})$ is atomless if for all $A\in\mathcal{A}$ with $\mathbb{P}(A)>0$ there exists $B\subset A, B\in\mathcal{A}$ such that $0<\mathbb{P}(B)<\mathbb{P}(A)$. Now according to a theorem of Sierpinski, the values of an atomless measure space form an interval. In particular, for probability spaces, every value in $[0,1]$ is taken. The original source of the article can be found here (in french). For a proof in english, you can look at on 215D on page 46 in Fremlin's book Measure Theory 2.

Answer 2

A necessary and sufficient condition is that every atom is no larger than the sum of all smaller atoms, plus the non-atomic part.

Answer 3

Here's a concrete example of an atomless measure. Let $f \in L^1$ be an integrable function with total mass 1 (i.e. $\int_0^1 f = 1$). Define $$\mathbb P(A) = \int_A f(x) ~dx$$ for any Borel set $A$. It is a nice exercise to show that $\mathbb P$ is an atomless measure.

Note: $f$ is called the Radon-Nikodym derivative of $\mathbb P$ with respect to Lebesgue measure, and often written $f = \tfrac{d\mathbb P}{dx}$. If a random variable $X$ has distribution $\mathbb P$, then $f$ is called its density function.

Answer 4

This is a property of $\mu$, not that of $\mathcal A$, and it is called being atomless. It is equivalent to not having sets $A \in \mathcal A$ of positive measure such that for all $B \in \mathcal A$, $B \subseteq A$ the measure $\mu(B)$ is either 0 or $\mu(A)$.

edit: Wikipedia article, complete with the proof of the property you describe from atomlessness.

edit: yup, the comments are right and I'm wrong. The precise condition for finite measures composed entirely of atoms to have full range is $a_n \leq \sum_{j>n} a_j$ - it is clearly necessary as $a_n-\varepsilon$ has to be produced somehow, and the greedy algorithm shows sufficiency.