Typicality and the Role of the Lebesgue Measure in Statistical Mechanics

Abstract

Consider a finite collection of marbles. The statement “half the marbles are white” is about counting and not about the probability of drawing a white marble from the collection. The question is whether non-probabilistic counting notions such as half, or vast majority can make sense, and preserve their meaning when extended to the realm of the continuum. In this paper we argue that the Lebesgue measure provides the proper non-probabilistic extension, which is in a sense uniquely forced, and is as natural as the extension of the concept of cardinal number to infinite sets by Cantor. To accomplish this a different way of constructing the Lebesgue measure is applied. One important example of a non-probabilistic counting concept is typicality, introduced into statistical physics to explain the approach to equilibrium. A typical property is shared by a vast majority of cases. Typicality is not probabilistic, at least in the sense that it is robust and not dependent on any precise assumptions about the probability distribution. A few dynamical assumptions together with the extended counting concepts do explain the approach to equilibrium. The explanation though is a weak one, and in itself allows for no specific predictions about the behavior of a system within a reasonably bounded time interval. It is also argued that typicality is too weak a concept and one should stick with the fully fledged Lebesgue measure. We show that typicality is not a logically closed concept. For example, knowing that two ideally infinite data sequences are typical does not guarantee that they make a typical pair of sequences whose correlation is well defined. Thus, to explain basic statistical regularities we need an independent concept of typical pair, which cannot be defined without going back to a construction of the Lebesgue measure on the set of pairs. To prevent this and other problems we should hold on to the Lebesgue measure itself as the basic construction.

This chapter was written by Itamar Pitowsky for this volume shortly before his death. As we did not have Itamar‘s LaTex file, it had to be retyped, and proofread by us. Remaining typos are, therefore, our fault. We thank William Demopoulos for reading the chapter and streamlining some of its formulations.

Appendices

Appendix 1: Proof of Theorem 1

Theorem 1

Let $E \in \mathcal{L}$ be any Lebesgue measurable set and let $\varepsilon > 0$; then there is ${F_\varepsilon } \in \mathcal{F}$ such that $\mu [(E\backslash {F_\varepsilon }) \cup ({F_\varepsilon }\backslash E)] < \varepsilon .$

Proof

Consider first a Lebesgue measurable subset $E \subseteq [0,1].$ By the regularity of the Lebesgue measure (see [23], page 230) given any $\varepsilon > 0$ there is an open set U, with $E \subseteq U$ and $\mu (U\backslash E) < \frac{\varepsilon }{2}.$ The family of open intervals with dyadic endpoints forms a basis for the usual topology on [0, 1] (recall that a dyadic number is a rational whose denominator is a power of 2). Thus, we can represent U as a disjoint countable union $U = \cup_{j = 1}^\infty ({c_j},{d_j}),$ where c _j and d _j are dyadic, and $\mu (U) = \sum\nolimits_{j = 1}^\infty {({d_j} - {c_j}).}$ By choosing a sufficiently large natural number N we can make sure that $U\prime = \cup_{j = 1}^N({c_j},{d_j}) \subseteq \,U$ satisfies $\mu (U\prime) > \mu (U) - \frac{\varepsilon }{2}.$ Now define $U\prime\prime$ to be the set obtained from $U\prime$ by adding the endpoints of each interval: $U\prime\prime = \cup_{j = 1}^N({c_j},{d_j})$. Since we have added just finitely many points the measure of $U\prime\prime$ is the same as that of $U\prime$, and therefore, $\mu [(E\backslash U\prime\prime) \cup (U\prime\prime\backslash E)] < \varepsilon .$

Now apply the map$\sum\nolimits_{j = 1}^\infty {{a_j}2^{ - j} \to ({a_1},\,} {a_2},{a_3}, \ldots )$ which takes real numbers in [0, 1] to their sequence of binary coefficients in ${\{ 0,1\}^\omega }$. Dyadic rationals have two binary developments, one ending with an infinite sequence of zeroes, and the other ending with an infinite sequence of ones. Adopt the convention that in case of a dyadic rational d, the map takes d to its two binary sequences. Since the set of dyadic numbers has measure zero the map is measure preserving. The set E is then mapped to a subset of $\{ 0,1\}^\omega$which we shall also denote by E. The set $U\prime\prime$ is mapped to a finite subset of $\{ 0,1\}^\omega$ which we will denote by ${F_\varepsilon } \in \,\mathcal{F}$. The reason is that every closed interval with dyadic endpoints is mapped to a finite set, for example,$\left[\frac{1}{4},\frac{5}{8}\right] \to \{ (0,1,0),(0,1,1),(1,0,0)\} \times \{ 0,1\} \times \{ 0,1\} \times \ldots \subseteq \{ 0,1\}^\omega,.$ and $U\prime\prime$ is a finite union of such intervals. This completes the proof.

Appendix 2: Proof of Theorem 2

This theorem and proof appeared first in [24] as part of a criticism of the frequency interpretation of probability.

Theorem 2

Let A ⊂ {0, 1}^ω be any measurable set with $\mu (A) > \frac{1}{2}$, then there are a, b ∈ A such that a⋅b has a divergent sequence of averages.

Proof

Denote by a ⊕ b the XOR of the elements a and b, in other words (a ⊕ b) _i  = a _i  + b _i (mod 2). We first show that $\mu (A) > \frac{1}{2}$ implies that A ⊕ A = {a ⊕ b; a, b ∈ A} = {0, 1}^ω. Indeed if c ∉ A ⊕ A, then (c ⊕ A) ∩A = φ, where c ⊕ A = {c ⊕ a; a ∈ A}. Otherwise, if d ∈ (c ⊕ A) ∩ A then d ∈ A and d = (c ⊕ a) for some a ∈ A. Hence c = (d ⊕ a) ∈ A ⊕ A, contradiction. Therefore, (c ⊕ A) ∩ A = φ, but this also leads to a contradiction since $\mu ({\mathbf{c}}\, \oplus \,A) = \mu (A) > \frac{1}{2}$, hence A ⊕ A = {0, 1}^ω.

We can assume without loss of generality that all elements of A have a convergent sequence of averages. This is the case because the set of elements of {0, 1}^ω whose averages diverge has measure zero. Let c ∈ {0, 1}^ω be some sequence with a divergent sequence of averages. Then by the above argument there are a, b ∈ A such that c = (a ⊕ b), that is c _i  = a _i  + b _i (mod 2) = a _i  + b _i − 2a _i b _i and therefore

$$ \frac{1}{n}\sum\limits_{i = 1}^n {{c_i} = \;\frac{1}{n}\sum\limits_{i = 1}^n {{a_i}} } + \frac{1}{n}\sum\limits_{i = 1}^n {{b_i}} - \frac{2}{n}\sum\limits_{i = 1}^n {{a_i}{b_i}} . $$

The sequence on the left diverges, and the first two sequences on the right converge. Hence, ${n^{ - 1}}\sum\nolimits_{i = 1}^n {{a_i}{b_i}}$ diverges. This completes the proof.