Let $(X,\mu)$ be a probability measure space and $A$ be a measurable subset of $X$ such that $0 \le \mu(A) < p < 1$.


When is it true that there exists a measurable $B \subseteq X$ such that $A \subseteq B$ and $\mu(B)=p$ ?

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    $\begingroup$ It is true if there are no atoms (that means: any subset of positive measure has a subset of positive but smaller measure.) $\endgroup$ – Fedor Petrov Nov 29 at 14:20
  • $\begingroup$ Ah, great thanks. I've just come across essentially this same fact at the moment, the so-called Sierpiński's Theorem, a kind of "Intermediate-Value Theorem" for measures. $\endgroup$ – dohmatob Nov 29 at 14:26
  • $\begingroup$ (now I remember I wrote a short section in a wiki article on this: en.wikipedia.org/wiki/Atom_(measure_theory)#Non-atomic_measures) $\endgroup$ – Pietro Majer Nov 29 at 14:43
  • $\begingroup$ Yes, that's I was referring to in my comment. $\endgroup$ – dohmatob Nov 29 at 14:44

An equivalent condition is that the probability space $(X, \mathcal{M}, \mu)$ is non-atomic, meaning that for every $A\in \mathcal{M}$ with $\mu(A)>0$ there exists $B \in \mathcal{M}$ such that $B\subset A$ and $0<\mu(B)<\mu(A)$.

If $(X, \mathcal{M}, \mu)$ is non-atomic then $\mu: \mathcal{M}\to [0,1]$ has a monotone section: that is, a map $E:[0,1]\to \mathcal{M}$ such that for all $0\le t\le s\le1$ one has $\emptyset=E_0\subset E_t\subset E_s\subset E_1=X$ and $\mu(E_t)=t$.

In fact, any "partial monotone section of $\mu$" (meaning a family $E:S\to \mathcal{M}$ as above, but with possibly smaller domain $S\subset [0,1]$) can be extended to a monotone section of $\mu$ (which in particular solves your problem).

I think this result is essentially due to Sierpiński; the proof uses the Zorn lemma on the set of graphs of the partial monotone sections of $\mu$, ordered by "extension" i.e. inclusion of graphs.

By the way, the maximality argument via Zorn lemma generalizes to a proof of the Lyapunov convexity theorem: if a $\mathbb{R}^n$ valued bounded vector measure $\mu=(\mu_1,\dots,\mu_n)$ is non atomic (meaning, all $\mu_i$ are non-atomic bounded measures), then the range of $\mu$ is a compact convex set.

  • $\begingroup$ Thanks for the input. The question has been answered in the comments. Someone also downvoted it for the fun of it... $\endgroup$ – dohmatob Nov 29 at 14:41
  • $\begingroup$ Thanks for mentioning the Chebychev convexity theorem. New good stuff. $\endgroup$ – dohmatob Nov 29 at 14:55
  • $\begingroup$ Which minimal restrictions on my problem (i.e on a non-atomic measure) would remove the need for the (somewhat controversial) Zorn's lemma in the proof ? $\endgroup$ – dohmatob Nov 29 at 14:57
  • $\begingroup$ What you want is to prove that a non-atomic probability measure is divisible, meaning the range is [0,1]. I guess the axiom of Dependent Choice is sufficient $\endgroup$ – Pietro Majer Nov 29 at 16:10
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    $\begingroup$ What exactly was proved by Chebyshev? $\endgroup$ – Fedor Petrov Nov 29 at 16:39

An argument for Sierpiński's Intermediate Value Theorem for non-atomic measures without using Zorn's lemma (but possibly implicitly using countable axiom of choice, that I am not qualified to judge), it was told to me by Alexander Kuznetsov.

Denote for a measurable subset $C\subset X$ $$f(C,q):=\sup_{D\subset C,\mu(D)\leqslant q} \mu(D).$$ We want to prove that $f(C,q)=q$ if $\mu(C)\geqslant q$. Choose consecutively disjoint subsets $C_1,C_2,\ldots$ in $C$ so that $$ q-\sum_{j<i}\mu(C_j)\geqslant \mu(C_i)\geqslant f\left(C\setminus \cup_{j<i} C_j,q-\sum_{j<i}\mu(C_j)\right)-\frac 1i$$ for all $i=1,2,\ldots$. Then for $C_0=\cup C_i$ we have $\mu(C_0)=\sum \mu(C_i)\leqslant q$. Assume that $\mu(C_0)=q-a$ for $a>0$. The set $C\setminus C_0$ has positive measure, thus it contains a subset $E$ for which $0<\mu(E)<a$ (since our measure is non-atomic, the set $C\setminus C_0$ may be partitioned onto two subsets of positive measure, take the part with smaller measure and partition it again and so on). But it means that for large $i$ the choice of $C_i$ was not $1/i$-optimal in above sense: we could use $E$ instead. A contradiction.

  • $\begingroup$ Great constructive answer. Thanks! $\endgroup$ – dohmatob Nov 29 at 17:17

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