When does a probability measure take all values in the unit interval? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T02:40:07Z http://mathoverflow.net/feeds/question/15804 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15804/when-does-a-probability-measure-take-all-values-in-the-unit-interval When does a probability measure take all values in the unit interval? vitp 2010-02-19T10:20:34Z 2010-02-25T21:31:51Z <p>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.</p> http://mathoverflow.net/questions/15804/when-does-a-probability-measure-take-all-values-in-the-unit-interval/15807#15807 Answer by Thorny for When does a probability measure take all values in the unit interval? Thorny 2010-02-19T11:04:20Z 2010-02-22T08:17:31Z <p>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 $\mu(A)$.</p> <p>edit: <a href="http://en.wikipedia.org/wiki/Atom%5F%28measure%5Ftheory%29" rel="nofollow">Wikipedia article</a>, complete with the proof of the property you describe from atomlessness.</p> <p>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.</p> http://mathoverflow.net/questions/15804/when-does-a-probability-measure-take-all-values-in-the-unit-interval/15808#15808 Answer by Michael Greinecker for When does a probability measure take all values in the unit interval? Michael Greinecker 2010-02-19T11:07:22Z 2010-02-19T11:07:22Z <p>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 $0&lt;\mathbb{P}(B)&lt;\mathbb{P}(A)$. 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 <a href="http://matwbn.icm.edu.pl/ksiazki/fm/fm3/fm3125.pdf" rel="nofollow">here</a> (in french). For a proof in english, you can look at on 215D on page 46 in Fremlin's book <a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.116.7816&amp;rep=rep1&amp;type=pdf" rel="nofollow">Measure Theory 2</a>. </p> http://mathoverflow.net/questions/15804/when-does-a-probability-measure-take-all-values-in-the-unit-interval/15820#15820 Answer by Tom LaGatta for When does a probability measure take all values in the unit interval? Tom LaGatta 2010-02-19T16:10:17Z 2010-02-19T20:02:02Z <p>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. </p> <p>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.</p> http://mathoverflow.net/questions/15804/when-does-a-probability-measure-take-all-values-in-the-unit-interval/16453#16453 Answer by angel for When does a probability measure take all values in the unit interval? angel 2010-02-25T21:31:51Z 2010-02-25T21:31:51Z <p>A necessary and sufficient condition is that every atom is no larger than the sum of all smaller atoms, plus the non-atomic part.</p>