Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $(X,d)$ be a metric space equipped with a probability measure $\mu$ (defined on the Borel $\sigma$-algebra on the topology induced by the metric $d$). I am interested in the different values that the following can take

$\sup\lbrace\mu(A):A\subset X\text{ is measurable and }\mu(A)\leq1/2\rbrace~~~~~~~~~~(\ast)$.

If there exists a measurable set $A\subset X$ such that $\mu(A)=1/2$, then it clearly is the case that $(\ast)$ is $1/2$.

If there is no set of measure $1/2$, then it is possible for $(\ast)$ to be different than $1/2$ (Dirac probability measure comes to mind). What I am wondering is if it could still be $1/2$.

In other words, I'm wondering if we could prove the existence (ideally with an example) or the nonexistence of a probability measure on a metric space $(X,d)$ such that for every measurable $A\subset X$, $\mu(A)\neq1/2$ and such that there is a sequence $\lbrace A_n\rbrace$ where $\mu(A_n)\to1/2$.

A few obvious remarks, if we were to find a probability measure as described in the above paragraph, the sequence $A_n$ cannot be nested, otherwise continuity from above or continuity from below would imply that the union or the intersection of the $A_n$ has a measure of $1/2$. Furthermore, the $A_n$ cannot be pairwise disjoint, otherwise $\mu$ cannot be a probability measure: for all $\epsilon>0$, infinitely many of the $A_n$ are contained in $(1/2-\epsilon,1/2+\epsilon)$, which implies by countable additivity that $\mu(X)\geq\infty$.

share|improve this question

2 Answers 2

up vote 6 down vote accepted

Short version: the set $\{\mu(B):B\in\mathcal{B}\}$ is a closed set for any probability space $(X,\mathcal{B},\mu)$. For atomic spaces this follows from an elementary topological argument, and for non-atomic spaces it is a closed interval by a classical (and easy) result of Sierpinski.

Longer version with details:

Let $A_n=A'_n\cup A''_n$ where $A'_n$ is the atomic part of $A_n$ and $A''_n$ is the rest. By passing to a subsequence we may assume that $\mu(A'_n)$ converges to some $\alpha\in [0,1/2]$.

I claim that there is an atomic set $B'$ such that $\mu(B')=\alpha$. Indeed, let $\{ x_k\}$ be the collection of the $\mu$-measures of all atoms of all $A'_n$.

The set $X=\{ \sum_{k\in I} x_k: I\subset \mathbb{N}\}$ can be checked to be closed, since it is the continuous image of the $\{0,1\}^\mathbb{N}$ under the map $I\to \sum_{k\in I} x_k$, where we identify sequences $\omega$ in $\{0,1\}^\mathbb{N}$ with the set $I=\{ j: \omega_j=1\}$. Since $\alpha$ is in the closure of $X$ (as $\mu(A'_n)\to\alpha$), $\alpha\in X$, as claimed.

Now the non-atomic part of the space has mass at least $1/2-\alpha$ (since it has mass at least $\mu(A''_n)$ for each $n$). But it is well known that if $\nu$ is a non-atomic measure of mass $c$, then for any $c'\in [0,c]$ there is a measurable set of $\nu$-mass $c'$ (see for example wikipedia). Applying this to $\nu$=restriction of $\mu$ to non-atomic part and $c'=1/2-\alpha$, we get a measurable non-atomic set $B''$ of $\mu$-measure $1/2-\alpha$.

Hence $B=B'\cup B''$ satifies $\mu(B)=1/2$.

share|improve this answer

Presuming your space is separable, the measure space $(X,\mu)$ is always Lebesgue, i.e., isomorphic to a disjoint union of an interval (with the Lebesgue measure on it) and an at most countable family of atoms.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.