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arXiv:1105.0663v1 [math.LO] 3 May 2011

UNCOMPUTABLY NOISY ERGODIC LIMITS

JEREMY AVIGAD

Abstract. V’yugin [2, 3] has shown that there are a computable shift-invariant

measure on 2Nand a simple function f such that there is no computable bound

on the rate of convergence of the ergodic averages Anf. Here it is shown that

in fact one can construct an example with the property that there is no com-

putable bound on the complexity of the limit; that is, there is no computable

bound on how complex a simple function needs to be to approximate the limit

to within a given ε.

Let 2Ndenote Cantor space, the space of functions from N to the discrete space

{0,1} under the product topology. Viewing elements of this space as infinite se-

quences, for any finite sequence σ of 0’s and 1’s let [σ] denote the set of elements

of 2Nthat extend σ. The collection B of Borel sets in the standard topology are

generated by the set of such [σ]. For each k, let Bk denote the finite σ-algebra

generated by the partition {[σ] | length(σ) = k}. If a function f from 2Nto Q is

measurable with respect to Bk, I will call it a simple function with complexity at

most k.

Let µ be any probability measure on (2N,B), and let f be any element of L1(µ).

Say that a function k from Q+to N is a bound on the complexity of f if, for every

ε > 0, there is a simple function f′of complexity at most k(ε) such that ?f−f′? < ε.

If (fn) is any convergent sequence of elements of L1(µ) with limit f, say that r(ε)

is a bound on the rate of convergence of (fn) if for every n ≥ r(ε), ?fn− f? < ε.

(One can also consider rates of convergence in any of the Lpnorms for 1 < p < ∞,

or in measure. Since all the sequences considered below are uniformly bounded,

this does not affect the results below.)

Now suppose that µ is a computable measure on 2Nin the sense of computable

measure theory [1, 4]. Then if f is any computable element of L1(µ), there is a

computable sequence (fn) of simple functions that approaches f with a computable

rate of convergence r(ε); this is essentially what it means to be a computable

element of L1(µ). In particular, setting k(ε) equal to the complexity of fr(ε)provides

a computable bound on the complexity of f. But the converse need not hold: if

r is any noncomputable real number and f is the constant function with value r,

then f is not computable even though there is a trivial bound on its complexity.

It is not hard to compute a sequence of simple functions (fn) that converges to

a function f even in the L∞norm with the property that there is no computable

bound on the complexity of the limit, with respect to the standard coin-flipping

measure on 2N. Notice that this is stronger than saying that there is no computable

bound on the rate of convergence of (fn) to f; it says that there is no way of

computing bounds on the complexity of any sequence of good approximations to

f.

2010 Mathematics Subject Classification. 03F60, 37A25.

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2 JEREMY AVIGAD

To describe such a sequence, let (qi) be the sequence of rational numbers where

qi is equal to 4−eif i codes a pair (e,s) and Turing machine e halts in exactly

s steps, and 0 otherwise. For each k, let hk be the Bk-measurable Rademacher

function, defined by

hk=

?

{σ | length(σ)=k}

where σk−1denotes the last bit of σ and 1[σ]denotes the characteristic function of

the cylinder set [σ]. Intuitively, hkis a “noisy” function of complexity k. Finally,

let fn=?

that any Turing machine with index at most m that halts in fact halts before stage

n, then |fi(x)−fn(x)| ≤?

not hard to verify that if f is the L1limit of this sequence and g is a simple function

of complexity at at most n such that µ({x | |g(x) − f(x)| > 4−(m+1)}) < 1/2, then

any Turing machine with index at most m that halts necessarily halts before stage

n. Thus any bound on the complexity of f solves the halting problem.

The sequence (fn) just constructed is contrived, and one can ask whether similar

sequences arise “in nature.” Letting Anf denote the ergodic average1

the mean ergodic theorem implies that for every measure µ on 2Nand f in L1(µ), the

sequence (Anf) converges in the L1norm. However, V’yugin [2, 3] has shown that

there is a computable shift-invariant measure µ on Cantor space such that there is no

computable bound on the rate of convergenceof (An1[1]). In V’yugin’s construction,

the limit doesn’t have the property described in the last paragraph; in fact, it is very

easy to bound the complexity of the limit in question, which places a noncomputable

mass on the string of 0’s and the string of 1’s, and is otherwise homogeneous. The

next theorem shows, however, that there are computable measures µ such that the

limit does have this stronger property.

(−1)σk−11[σ],

i≤nqihi. The sequence (fn) converges in the L∞norm, since if n is such

j≥m4−j< 1/(3·4m) for every x. At the same time, it is

n

?

i<nf ◦Tn,

Theorem 1. There is a computable shift-invariant measure µ on 2Nsuch that

if f = limnAn1[1], the halting problem can be computed from any bound on the

complexity of f.

Proof. If σ is any finite binary sequence, let σ∗denote the element σσσ ... of Cantor

space. For each e, define a measure µeas follows: if Turing machine e halts in s

steps, let µeput mass uniformly on these 8s elements:

• all 4s shifts of (1s03s)∗

• all 4s shifts of (13s0s)∗

Otherwise, let µedivide mass uniformly between 0∗and 1∗. Each measure µe is

shift invariant, by construction. I will show, first, that µeis computable uniformly

in e, which is to say, there is a single algorithm that, given e, σ, and ε > 0, computes

µe([σ]) to with ε. I will then show that information as to the complexity needed to

approximate f in (2ω,B,µe) allows one to determine whether or not Turing machine

e halts. The desired conclusion is then obtained by defining µ =?

If Turing machine e does not halt, µe([σ]) = 1/2 if σ is a string of 0’s or a string

of 1’s, and µe([σ]) = 0 otherwise. Suppose, on the other hand, that Turing machine

e halts in s steps, and suppose k < s. Then there are 2(k − 1) additional strings σ

with length k such that µe([σ]) > 0, each consisting of a string of 1’s followed by a

string of 0’s or vice versa. For each of these σ, µe([σ]) = 1/4s, and if σ is a string

of 0’s or a string of 1’s of length k, µe([σ]) = 1/2 − (k − 1)/4s. Thus when s is

large compared to k, the non-halting case provides a good approximation to µe([σ])

e

1

2e+1µe.

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UNCOMPUTABLY NOISY ERGODIC LIMITS3

when length(σ) ≤ k, even though e eventually halts. Thus, to compute µe([σ]) to

within ε, it suffices to simulate the eth Turing machine O(k/ε) steps. If it halts

before then, that determines µeexactly; otherwise, the non-halting approximation

is close enough.

Now consider f = limnAn1[1]in (2ω,B,µe). Note that (An1[1])(ω) counts the

density of 1’s among the first n bits of ω. If Turing machine e does not halt,

f(ω) = 1 if ω is the sequence of 1’s and f(ω) = 0 if ω or the sequence of 0’s. Up

to a.e. equivalence, these are all that matters, since the mass concentrates on these

two elements of Cantor space. If Turing machine e halts in s steps, then f(ω) = 1/4

on the shifts of (1s03s)∗, and f(ω) = 3/4 on the shifts of (13s0s)∗.

Suppose g is Bk-measurable. If Turing machine e halts in s steps and k is much

less than s, then roughly 3/4 of the shifts of (1s03s)∗lie in [0k] and roughly 1/4 lie

in [1k]; and roughly 3/4 of the shifts of (13s0s)∗lie in [1k] and roughly 1/4 lie in

0k. But f(ω) only takes on the values 1/4 and 3/4, and g is constant on [0k] and

[1k]. So if k is much less than s, µe({ω | |f(ω)−g(ω)| > 1/8}) > 1/4. Turning this

around, given the information that µe({ω | |f(ω)−g(ω)| > 1/8}) ≤ 1/4 for some g

of complexity at most k enables one to determine whether or not Turing machine

e halts; namely, one simulates the Turing machine for O(k) steps, and if it hasn’t

halted by then, it never will.

Set µ =?

e

µe({ω | |f(ω) − g(ω)| > 1/8}) ≤ µ({ω | |f(ω) − g(ω)| > 1/(8 · 2e+1)}),

1

2e+1µe. Since, for any g,

knowing a kefor each e with the property that µ({ω | |f(ω)−g(ω)| > 1/(8·2e+1)} <

1/4 for some g of complexity at most keenables one to solve the halting problem.

But such a ke can be obtained from a bound on the complexity of f. Thus µ

satisfies the statement of the lemma.

?

The proof above relativizes, so for any set X there is a measure µ on 2N, com-

putable from X, such that no bound on the rate of complexity of f can be computed

from X. As the following corollary shows, this implies that limnAn1[1]can have

arbitrarily high complexity.

Corollary 2. For any v : Q+→ N there is a measure µ on 2Nsuch that if f =

limnAn1[1]and k(ε) is a bound on the complexity of f, then limsupε→0k(ε)/v(ε) =

∞.

Proof. Let µ be such that no bound on the complexity of f that can be computed

from v. If the conclusion failed for some k, then there would be a rational ε′> 0

and N such that for every ε < ε′, k(ε) < N ·v(ε). But then k′(ε) = N ·v(min(ε,ε′))

would be a bound on the complexity of f that is computable from v, contrary to

our choice of µ.

?

References

[1] Mathieu Hoyrup and Crist´ obal Rojas. Computability of probability measures and Martin-L¨ of

randomness over metric spaces. Inform. and Comput., 207(7):830–847, 2009.

[2] V. V. V’yugin. Ergodic convergence in probability, and an ergodic theorem for individual

random sequences. Teor. Veroyatnost. i Primenen., 42(1):35–50, 1997.

[3] V. V. V’yugin. Ergodic theorems for individual random sequences. Theoret. Comput. Sci.,

207(2):343–361, 1998.

[4] Klaus Weihrauch. Computability on the probability measures on the Borel sets of the unit

interval. Theoret. Comput. Sci., 219(1-2):421–437, 1999.

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4 JEREMY AVIGAD

Departments of Philosophy and Mathematical Sciences, Carnegie Mellon University,

Pittsburgh, PA 15213