# If $0 \le \mu(A) < p < 1$, when is it true that there exists a measurable $B \supseteq A$ such that $\mu(B)=p$?

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

# Question

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

• It is true if there are no atoms (that means: any subset of positive measure has a subset of positive but smaller measure.) – Fedor Petrov Nov 29 at 14:20
• 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. – dohmatob Nov 29 at 14:26
• (now I remember I wrote a short section in a wiki article on this: en.wikipedia.org/wiki/Atom_(measure_theory)#Non-atomic_measures) – Pietro Majer Nov 29 at 14:43
• Yes, that's I was referring to in my comment. – 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.

• Thanks for the input. The question has been answered in the comments. Someone also downvoted it for the fun of it... – dohmatob Nov 29 at 14:41
• Thanks for mentioning the Chebychev convexity theorem. New good stuff. – dohmatob Nov 29 at 14:55
• 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 ? – dohmatob Nov 29 at 14:57
• 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 – Pietro Majer Nov 29 at 16:10
• What exactly was proved by Chebyshev? – 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 $$q-\sum_{j for all $$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) (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.

• Great constructive answer. Thanks! – dohmatob Nov 29 at 17:17