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Let $\mu$ be a finite nonatomic measure on a measurable space $(X,\Sigma)$, and for simplicity assume that $\mu(X) = 1$. There is a well-known "intermediate value theorem" of Sierpiński that states that for every $t \in [0,1]$, there exists a set $S \in \Sigma$ with $\mu(S) = t$.

I would like to use the following stronger conclusion for such a measure:

There exists a chain of sets $\{S_t \mid t \in [0,1]\}$ in $\Sigma$, with $S_t \subseteq S_r$ whenever $0 \leq s \leq r \leq 1$, such that $\mu(S_t) = t$ for all $t \in [0,1]$.

(One can view this as the existence a right inverse to the map $\mu \colon \Sigma \to [0,1]$ in the category of partially ordered sets.)

This statement appears (albeit hidden within a proof) on the Wikipedia page for "Atom (measure theory)," and even includes a sketch for the proof! However, I would like to see some mention of this in the literature. I've checked the Wiki references and they both seem to prove the weaker statement. I looked in Fremiln's Measure Theory, vol. 2, and again found the weaker version but not the stronger.

Question: Can anyone provide me with such a reference?


A proof. In case anyone stumbles to this page and wants to see a proof, I'll sketch one that is more constructive than the one that I linked to above. Set $S_0 = \varnothing$ and $S_1 = X$. By Sierpiński, there exists $S_{1/2} \in \Sigma$ of measure $1/2$. For each Dyadic rational $q = m/2^n \in [0,1]$ ($1 \leq m \leq 2^n$), we may proceed by induction on $n$ to construct each $S_q$. Now given $r \in [0,1]$, set $S_r = \bigcup_{q \leq r} S_q$. (This is essentially the same method of proof as the one in the reference provided in Ramiro de la Vega's answer.)

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    $\begingroup$ Btw, it's me who wrote that "sketch of proof". The proof is by myself, but of course I was sure somebody had already made the same statement and proof. $\endgroup$ – Pietro Majer Nov 3 '15 at 22:55
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    $\begingroup$ This post by Dave Renfro contains some references on this fact: mathforum.org/kb/message.jspa?messageID=4109674 . It comes from this MathSE post by Byron Schmuland. $\endgroup$ – Giuseppe Negro Nov 10 '15 at 23:44
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I would say this is folklore (I proved it and used it many years ago on my undergrad thesis), but here is a concrete reference:

Such a family of measurable sets is called a $[0,1]$-family in On the Skorokhod representation theorem by Jean Carlos Cortissoz, PAMS, Vol.135, No. 12, 2007 (see Definition 4.1). A proof that such a family exists in any non-atomic space is given in Lemma 4.1.

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  • $\begingroup$ I was hoping that it could be considered folklore, but thanks for digging up a reference anyway! $\endgroup$ – Manny Reyes Nov 25 '14 at 11:50
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There's a stronger version of that (basic) theorem due to Lyapunov. It is stronger because it concerns vectors of measures, and not only a single measure. It states that given a non-atomic vector measure (a collection of $n$ measures $\mu_1,\ldots, \mu_n$ where each measure is non-atomic) always has an image which is convex (in $\mathbb{R}^n$).

Unfortunately, I could never find a translation of his paper, so I can only link the version in russian. The main statements can be found in French at the end of the paper. There's also a paper of Halmos that proves the result.

Maybe looking at the proof method or subsequent papers you can find the chain statement that you seek.

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    $\begingroup$ Actually, your statement is not strictly stronger than the result the OP is asking for since he wants the sets $S_t$ to be nested. $\endgroup$ – Dirk Nov 25 '14 at 8:24
  • $\begingroup$ @Dirk You're right, I missed that at first. $\endgroup$ – Henrique de Oliveira Nov 25 '14 at 19:36
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It seems to be a special case of Lemma 2.5(chapter 2) of "interpolation of operators"(Bennett, Sharpley).

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It is also a special case of Theorem 15 (p. 43) in:

A. Fryszkowski, Fixed Point Theory for Decomposable Sets, Topological Fixed Point Theory and Its Applications 2, Dordrecht: Kluwer Academic Publishers, 2004.

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