9
$\begingroup$

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.

$\endgroup$

4 Answers 4

17
$\begingroup$

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.

$\endgroup$
1
  • 3
    $\begingroup$ 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. $\endgroup$ Feb 19, 2010 at 22:45
6
$\begingroup$

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

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

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.

$\endgroup$
3
  • $\begingroup$ @TG: Instead of "Exercise: Prove...", maybe "It is a nice exercise to show..." or something like that? $\endgroup$ Feb 19, 2010 at 17:01
  • $\begingroup$ Pete, I like your rephrasing. I'll change my post. $\endgroup$ Feb 19, 2010 at 18:06
  • 5
    $\begingroup$ Only a mathematician would start a concrete example with "Let $f \in L^1$..." $\endgroup$ Feb 19, 2010 at 20:03
4
$\begingroup$

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.

$\endgroup$
2
  • 10
    $\begingroup$ In fact, certain atomic measures also have range [0,1], so "atomless" is sufficient but not necessary. $\endgroup$ Feb 19, 2010 at 14:48
  • 3
    $\begingroup$ 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 $\endgroup$ Feb 19, 2010 at 20:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.