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A classical Banach limit is very usefull concept for me, but there is a problem with the integration and even with the measurability, this means for a sequence $(f_n)_{n\in \mathbb{N}}$ of measurable (eg borel) and uniformly bounded functions, the function $x \mapsto L((f_n(x))_{n\in\mathbb{N}})$, where $L$ is Banach limit, is not measurable in general.

I heard about a stronger concept of Banach limit of Mokobodzki called "medial limit", that exists, assuming the continuum hypothesis. This functional apparently has the same proporties like Banach limit and additionaly, which is important for me, commutes with integration in appropriate settings, this means it would preserve the measurebility and satisfied condition $M((\int f_n(x)\mu(dx))_{n\in\mathbb{N}}))=\int M((f_n(x))_{n\in\mathbb{N}})\mu(dx)$, with appropriate assumptions.

Unfortunately, I found only a brief mention of this notion in the available literature. Could somebody explain to me this idea more precisely or point me to available literature on this subject?

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Mokobodzki, right? Do you read French? –  t.b. Jul 30 '11 at 4:07
    
sorry for the typo, so i dont know french, but if there are no other options i will try to cope.. I will be grateful for any materials –  dawid Jul 30 '11 at 4:14
    
dawid: it would be very nice if you started accepting some of the answers you received. You can do so by clicking on the grey checkmark sign on the left of the answers that were most helpful to you (you can only accept one answer per question). –  t.b. Jul 30 '11 at 5:14
    
Thank you for accepting some answers! I added a small bit below. By the way, if you find some good references, please let me know. –  t.b. Jul 30 '11 at 6:03
    
@Asaf: I added (set-theory) and (cardinals), as set-theoretic measure theory and questions on (cardinals) are involved in the answer here. –  t.b. Jul 30 '11 at 17:07

1 Answer 1

up vote 25 down vote accepted

You are talking about what is often called a measure-linear mean or limite médiale in the sense of Mokobodzki. $\newcommand{mypart}[1]{\unicode{x2014;}\text{ #1 }\unicode{x2014;}}$

Edit: Given my recent revisiting the literature on this topic, it seems like medial limit (when thought of as an ultrafilter on $\mathbb{N}$) or medial mean (when thought of as linear functional on $\ell^{\infty}(\mathbb{N})$) are the most wide-spread terms nowadays.


Edit: Since this answer grew longer than I ever would have imagined, I thought a short overview might be in order. I wanted to keep the chronology visible. The text roughly grew in the order it is displayed below:

  1. Definition
  2. Simple comments on the definition
  3. Some pointers to the literature
  4. A Birkhoff-type result with a one-line proof
  5. Reference to Fremlin's book, weakening and elaborating on (seemingly necessary) hypotheses
  6. Larson's result on the necessity of assuming more than ZFC

[Meta: A big thank you to Vadim Kaĭmanovich, Freddy Delbaen and Andreas Blass.]


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A medial mean is a linear functional $\mathfrak{m}: \ell^{\infty}(\mathbb{N}) \to \mathbb{R}$ satisfying the following properties:

  1. It is positive, i.e., $\mathfrak{m}{(f)} \geq 0$ if $f \geq 0$.

  2. It is normalized, i.e., $\mathfrak{m}{(1_{\mathbb{N}})} = 1$.

  3. It is shift-invariant: if $Tf(x) = f(x+1)$ then $\mathfrak{m}{(Tf)} = \mathfrak{m}{(f)}$.

  4. It is universally measurable on the unit ball $B = [0,1]^{\mathbb{N}}$ of $\ell^{\infty}$: More precisely, if $\mu$ is a Borel probability measure with respect to the weak$^{\ast}$-topology (= product topology) on $B$ then $\mathfrak{m}$ is $\mu$-measurable.

  5. It is measure-linear: If $f_n : [0,1] \to [0,1]$ is a sequence of Borel measurable functions then the function $f = \mathfrak{m}((f_n))$ is measurable (!) with respect to any probability measure $\mu$ on $B$, and we have the identity: $$\int f \,d\mu = \int \mathfrak{m}(f_n)\,d\mu = \mathfrak{m}\left[\int f_n\,d\mu\right].$$ Equivalently, this holds for a sequence of functions defined on any probability space (this is quite non-trivial).


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  1. The first three properties are saying that $\mathfrak{m}$ is a Banach limit. In particular, $\mathfrak{m}$ has norm one and gives the usual limit for convergent sequences.
  2. The measurability property is essential and subtle. One can show that a weak$^{\ast}$-measurable linear functional is automatically weak$^{\ast}$-continuous on $\ell^{\infty}$, and this would mean that $\mathfrak{m}$ is representable by an $\ell^1$ function, which it obviously can't be in view of 1,2,3. Thus weak$^{\ast}$-measurability must be weakened. If you take just any Hahn-Banach extension from the convergent sequences to the bounded sequences of the limit functional, there is no reason for this extension to be measurable.
  3. Property 5. can be seen as a version of Fubini involving one finitely additive measure. Another way of looking at it would be to see it as a very strong version of the bounded convergence theorem. Indeed, write $\mathrm{LIM}$ instead of $\mathfrak{m}$.

    Edit: Note that 5. does not follow immediately from 4. as I carelessly asserted earlier (we're composing measurable maps "the wrong way around").
  4. The existence of a measurable mean is (as far as I know) not provable in ZFC. The usual assumption people make is the continuum hypothesis, but Martin's Axiom MA is sufficient. MA is in some very naïve sense "half the continuum hypothesis": Gödel, in What is Cantor's Continuum Problem famously listed 6 reasons why the continuum hypothesis should be considered implausible, and MA "only" implies 3 of them.

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These surprising gadgets were discovered by Gabriel Mokobodzki, Ultrafiltres rapides sur N. Construction d'une densité relative de deux potentiels comparables., Séminaire de Théorie du Potentiel, dirigé par M. Brelot, G. Choquet et J. Deny: 1967/68, Exp. 12, 22 pp. Secrétariat mathématique, Paris 1969. The construction was simplified and clarified by Pierre-André Meyer Limites médiales, d'après Mokobodzki, Séminaire de Probabilités, VII (Univ. Strasbourg, année universitaire 1971–1972), pp. 198–204. Lecture Notes in Math., Vol. 321, Springer, Berlin, 1973.

These medial limits were studied rather in-depth by J.P.R. Christensen, but I don't remember in which papers exactly. One nice paper that makes use of them is Kaĭmanovich-Fisher, A Poisson formula for harmonic projections, Ann. Inst. H. Poincaré Probab. Statist. 34 (1998), no. 2, 209–216. See also A. Fisher, Convex-invariant means and a pathwise central limit theorem, Adv. in Math. 63 (1987), no. 3, 213–246.


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Added: Here's one classic sample application of medial limits (taken fom Kaĭmanovich-Fisher):

Let $T: X \to X$ be an ergodic and measure-preserving transformation on a probability space $(X,\mu)$. If $f$ is a bounded and measurable function then we can consider its time translates $(T^{\ast n}f)(x) = f(T^nx)$. If we fix a medial limit $\mathfrak{m}$ then the time average $\mathcal{T}\,f = \mathfrak{m}((T^{\ast n}f))$ is $T$-invariant and measurable, hence constant by ergodicity of $T$. By measure-linearity and our assumptions on $T$ we have $$\mathcal{T}\,f = \int \mathcal{T}\,f\,d\mu = \int \mathfrak{m}{(T^{\ast n}f)}\,d\mu = \mathfrak{m}\left[\int T^{\ast n}f\,d\mu\right] = \mathfrak{m}\left[\int f\,d\mu\right] = \int f\,d\mu = \mathcal{S}_\mu f$$ where $\mathcal{S}_\mu$ denotes the space average with respect to $\mu$. This is at least morally the content of the classical Birkhoff ergodic theorem. See also this related MO-thread.


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Added Later: An English reference giving a careful proof of the existence of a medial limits is volume 5 Part I of Fremlin's Measure Theory compendium. In fact, he proves the existence in Theorem 538S on page 277 under the quite a bit weaker assumption than Martin's axiom that $\mathbb{R}$ can't be covered by strictly less than $\mathfrak{c}$ meager sets. A rather detailed discussion of medial limits is contained from 538P (page 272) on where you'll find variants of Fubini, dominated convergence, Vitali and similar things (in case you'll find yourself needing them).

In fact, the consistency needed for set-theoretic measure theory (about which I don't know much) depends on certain cardinal numbers to coincide. These are subsumed in what is called Cichoń's diagram amalgamated with Martin's diagram by Fremlin (most of the entries in the picture below are not of interest here):

Cichoń-Martin diagram

The entries appearing here are cardinal numbers which increase from bottom up and from left to right. Assuming the continuum hypothesis, which is equivalent to $\omega_1 = \mathfrak{c}$ (the continuum is equal to the first uncountable ordinal), all these cardinalities collapse to one. Martin's axiom can be expressed as $\mathfrak{m} = \mathfrak{c}$, where $\mathfrak{m}$ is Martin's number (this is of course a different $\mathfrak{m}$ than the one in the main body of this answer). Now the assertion that $\mathbb{R}$ can't be covered by less than continuum many meager sets is equivalent to $\operatorname{cov}{\mathcal{M}} = \mathfrak{c}$. This should give some impression on how much weaker this assumption is. In volume 5 Fremlin collects facts that are (usually) undecidable in ZFC and proves them with reasonably weak assumptions beyond.

Even Later: The question of necessity of going beyond ZFC seems to be open if you're willing to take Fremlin's word for it: Apparently, it is unknown whether it is relatively consistent with ZFC that there are no medial limits at all. In fact, this is asked as problem in 538Z. I'm quoting the very last paragraph of 538 (page 281):

For most of the classes of filter [sic] here, there is a question concerning their existence. Subject to the continuum hypothesis, there are many Ramsey ultrafilters, and refining the argument we find that the same is true if $\mathfrak{p = c}$ (538Yb). There are many ways of forcing non-existence of Ramsey ultrafilters, of which one of the simplest is in 553H below. With more difficulty, we can eliminate $p$-point ultrafilters (Wimmers 82) or rapid filters (Miller 80) or nowhere dense filters and therefore measure-centering ultrafilters (538Hd, Shelah 98). It is not known for sure that we can eliminate medial limits or measure-converging filters (538Z).

The relevant statement mentioned here is 538Z on page 280:

538Z Problem Show that it is relatively consistent with ZFC to suppose that there are no measure-converging filters on $\mathbb{N}$. (Note that in this case there are no medial limits, by 538Rd.)


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Update: Andreas Blass kindly pointed out to me on meta.MO that the problem was settled in 2009 by Paul Larson in The filter dichotomy and medial limits, J. Math. Logic, vol. 9, no. 2 (2009) pp. 159-165 (see also the ESI preprint 2173). Indeed, based on results by Blass and Laflamme concerning models constructed by Miller and Blass–Shelah, Paul Larson establishes

Corollary 3.3 If ZFC is consistent, then so is ZFC + “there exist no medial limits.”

I'd like to point out that Larson mentions—reference [10] is Fremlin's book I mentioned above:

As far as we know, the weakest hypothesis known to be sufficient [for the existence of medial limits] (see 538S of [10]) is the statement that the reals are not a union of fewer than continuum many meager sets (i.e., that the covering number for the meager ideal is the continuum).

As a final remark: Larson's paper gives quite a few references on results concerning medial limits (I already mentioned quite a few of them above), he lists in particular numerous papers by Christensen and attributes the weakening to Martin's Axiom to Dag Norman.

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1  
While they are convenient due to their strong formal properties, I think it's usually better to try and avoid them. Indeed, a proof using them may usually be much quicker than a proof in ZFC (using amenability of the semigroup $\mathbb{N}$ and the ensuing averaging concepts, for example), but it usually doesn't yield much insight. –  t.b. Jul 30 '11 at 4:57
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I think I've convinced myself at some point that there exists such a measure-linear mean on any countable (or even locally compact second countable) amenable semigroup, but I haven't written this up and don't know if it can be found in the literature. –  t.b. Jul 30 '11 at 5:09
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I've removed the wiki-mode from this answer. –  Zev Chonoles Jul 31 '11 at 15:46
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I'm fascinated by the evolution of this answer! –  Andres Caicedo Jul 31 '11 at 21:39
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@Andres: That's nice to hear, thanks a lot! Maybe I should have added a tag (reality-soap): "watch Theo do his homework and give live updates" :) It is something I worked on in summer '03 (!) and never found the motivation to go back. Now I jumped on the opportunity... –  t.b. Jul 31 '11 at 22:32

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