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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.

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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.

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    $\begingroup$ So, the condition holds on spaces where the total mass of the atoms is at most 1/2, and on spaces where the sums of the atoms have gaps of width at most the mass of the space excluding atoms. $\endgroup$
    – Douglas Zare
    Commented Feb 19, 2010 at 22:45
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A necessary and sufficient condition is that every atom is no larger than the sum of all smaller atoms, plus the non-atomic part.

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    $\begingroup$ In particular a measure consisting only of atoms of mass $1/2, 1/4, 1/8, \ldots$ just barely suffices. $\endgroup$
    – Michael Lugo
    Commented Feb 25, 2010 at 21:45
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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.

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  • $\begingroup$ @TG: Instead of "Exercise: Prove...", maybe "It is a nice exercise to show..." or something like that? $\endgroup$
    – Pete L. Clark
    Commented Feb 19, 2010 at 17:01
  • $\begingroup$ Pete, I like your rephrasing. I'll change my post. $\endgroup$
    – Tom LaGatta
    Commented Feb 19, 2010 at 18:06
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    $\begingroup$ Only a mathematician would start a concrete example with "Let $f \in L^1$..." $\endgroup$
    – Tom LaGatta
    Commented Feb 19, 2010 at 20:03
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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.

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    $\begingroup$ In fact, certain atomic measures also have range [0,1], so "atomless" is sufficient but not necessary. $\endgroup$
    – Gerald Edgar
    Commented Feb 19, 2010 at 14:48
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    $\begingroup$ If you take a measure made up purely of atoms (the atoms having measures (a_1, a_2, a_3...) in decreasing order, what are the conditions guaranteeing full range? For example, a_i <= 2a_{i+1} suffices, but may not be necessary $\endgroup$
    – Kevin P. Costello
    Commented Feb 19, 2010 at 20:11

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