Billsingley: Ergodic Theory and Information (1965), pp. 12–14

In the process of proving the ergodic theorem, Billingsley also provides in passing a proof for a weaker theorem related to mixing operations: This states that for a class of processes that satisfy a certain limiting independence condition, visiting frequencies converge to probabilities.

Definitions

More precisely, a time shift $T$ is said to be mixing (p. 12) if$$
P(A \cap T^{-n}B)  \;  \rightarrow  \;  P(A)P(B)
$$for all sets $A$ and $B$. The process is thus mixing if information about the present is almost entirely independent of the far future. (A sample path $\omega$ belongs to the set $T^{-n}B$ if $\omega$ will visit $B$ after $n$ time shifts forward.) Billsingley doesn't comment on the choice of the term "mixing," but I assume it should be read as an ill-chosen synonym for "shuffling" in this context.

We will now be interested in the behavior of the visiting frequencies$$
P_n(A) \; := \;  \frac{1}{n} \sum_{k=0}^{n-1} \mathbb{I}_A(T^k\omega),
$$where $\mathbb{I}_A$ is the indicator function of some bundles of sample paths. This quantity is a random variable with a distribution defined by the distribution on the sample paths $\omega$.

Note that the traditional notion of visiting frequency (i.e., how often is the ace of spades on top of the deck?) is a special case of this concept of visiting frequency, in which the bundle $A$ is defined entirely in terms of the 0th coordinate of the sample path $\omega = (\ldots, \omega_{-2}, \omega_{-2}, \omega_{0}, \omega_{1}, \omega_{2}, \ldots)$. In the general case, the question of whether $\omega\in A$ could involve information stored at arbitrary coordinates of $\omega$ (today, tomorrow, yesterday, or any other day).

Theorem and Proof Strategy

The mixing theorem now says the following: Suppose that $P$ is a probability measure on the set of sample paths, and suppose that the time shift $T$ is mixing and measure-preserving with respect to this measure. Then$$
P_n(A)  \;  \rightarrow  \;  P(A)
$$in probability.

The proof of this claim involves two things:

1. Showing that $E[P_n(A)] = P(A)$;
2. Showing that $Var[P_n(A)] \rightarrow 0$ for $n \rightarrow \infty$.

These two assumptions collectively imply convergence in probability (by Markov's inequality).

Identity in Mean

The time shift $T$ is assumed to preserve the measure $P$. We thus have $$
E[f(\omega)] \; = \; E[f(T\omega)] \; = \; E[f(T^2\omega)] \; = \; \cdots
$$for any measurable function $f$. It follows that$$
E[\mathbb{I}_A(\omega)] \; = \;
E[\mathbb{I}_A(T\omega)] \; = \;
E[\mathbb{I}_A(T^2\omega)] \; = \; \cdots
$$and that these are all equal to $P(A)$. By the linearity of expectations, we therefore get that $E[P_n(A)] = P(A)$.

This proves that $P_n(A)$ at least has the right mean. Convergence to any other value than $P(A)$ is therefore out of the question, and it now only remains to be seen that the variance of $P_n(A)$ also goes to 0 as $n$ goes to infinity.

Vanishing Variance: The Idea

Consider therefore the variance$$
Var[P_n(A)]  \; = \;  E\left[ \left(P_n(A) - P(A)\right)^2 \right].
$$In order to expand the square under this expectation, it will be helpful to break the term $P(A)$ up into $n$ equal chunks and then write the whole thing as a sum:$$
P_n(A) - P(A)  \; = \;  \frac{1}{n} \sum_{k=0}^{n-1} (\mathbb{I}_A(T^k\omega) - P(A)).
$$If we square of this sum of $n$ terms and take the expectation of this expasion, we will get a sum of $n^2$ cross-terms, each of which are expected values of the form$$
Cov(i,j)  \; := \;  E\left[ \left(\mathbb{I}_A(T^i\omega) - P(A)\right) \times \left (\mathbb{I}_A(T^j\omega) - P(A)\right) \right].
$$By expanding this product and using that $E[\mathbb{I}_A(T^k\omega)]=P(A)$, we can find that$$
Cov(i,j)  \; = \;  E\left[\mathbb{I}_A(T^i\omega) \times \mathbb{I}_A(T^j\omega)\right] - P(A)^2.
$$The random variable $\mathbb{I}_A(T^i\omega) \times \mathbb{I}_A(T^j\omega)$ takes the value 1 on $\omega$ if and only if $\omega$ will visit $A$ after both $i$ and $j$ time steps. Hence,$$
Cov(i,j)  \; = \;  P(T^{-i}A \cap T^{-j}A)  -  P(A)^2,
$$and we can write$$
Var[P_n(A)]  \; = \;  \frac{1}{n^2} \sum_{i=0}^{n-1} \sum_{j=0}^{n-1} Cov(i,j).
$$The goal will now be to show the sum of these $n \times n$ converges to $0$ as we increase $n$.

Vanishing Variance: The Steps

Since $T$ is assumed to be measure-preserving, the value of $Cov(i,j)$ is completely determined by the distance between $i$ and $j$, not their absolute values. We thus have$$
Cov(i,j)  \; = \;  Cov(n+i, n+j)
$$for all $i$, $j$, and $n$.

By the mixing assumption, $Cov(0,j) \rightarrow 0$ for $j \rightarrow \infty$. Together with the observation about $|i-j|$, this implies that $$Cov(i,j) \;\rightarrow\; 0$$ whenever $|i-j| \rightarrow \infty$. If we imagine the values of  $Cov(i,j)$ tabulated in a huge, square grid, the numbers far away from the diagonal are thus tiny. We will now sum up the value of these $n^2$ numbers $Cov(i,j)$ by breaking them into $n$ classes according to the value of $d = |i-j|$.

To do so, we upper-bound the number of members in each class by $2n$: There are for instance $n$ terms with $|i-j|=0$, and $2(n-1)$ terms with $|i-j|=1$. (This upper-bound corresponds graphically to covering the $n\times n$ square with a tilted $n\sqrt{2} \times n\sqrt{2}$ square, except for some complications surrounding $d=0$.)

We thus have$$
Var[P_n(A)]
\; = \;
\frac{1}{n^2} \sum_{i=0}^{n-1} \sum_{j=0}^{n-1} Cov(i,j)
\;  \leq  \;
\frac{2}{n} \sum_{d=0}^{n-1} Cov(0,d).
$$This last quantity is an average of a list of numbers that converge to $0$; this average itself must therefore also converge to $0$ as $n\rightarrow \infty$. This proves that$$
Var[P_n(A)]  \; \rightarrow \;  0]
$$for $n \rightarrow \infty$, and we can conclude that $P_n(A) \rightarrow P(A)$ in probability.

Applicability

Note that this mixing assumption is very strong. A periodic Markov chain, for instance, will usually not be mixing: Consider for instance a process that alternates between odd and even values; the the joint probability $P(evens \cap T^n odds)$ will alternate between 0 and 1 as $n$ runs through the natural numbers. In many cases of practical interest, we therefore do in fact need the increased generality of the classic ergodic theorem.

Billingsley: Probability and Measure, Ch. 1.4

Prompted by what I found to be quite confusing about Patrick Billingsley's presentation of Kolmogorov's zero-one law, I've been reflecting a bit on the essence of the proof, and I think I've come to a deeper understand of the core of the issue now.

The theorem can be stated as follows: Let $S$ be an infinite collection of events equipped with a probability measure $P$ (which, by Kolmogorov's extension theorem, can be defined exhaustively on the finite subsets of $S$). Suppose further that $A$ is some event in the $\sigma$-extension of $S$ which is independent of any finite selection of events from $S$. Then $A$ either has probability $P(A)=0$ or $P(A)=1$.

The proof is almost evident from this formulation: Since $A$ is independent of any finite selection of events from $S$, the measures $P(\,\cdot\,)$ and $P(\,\cdot\,|\,A)$ coincide on all events that can be defined in terms of a finite number of events from $S$. But by Kolmogorov's extension theorem, this means that the conditional and the unconditional measures extend the same way to the infinite collection. Hence, if P$(A)>0$, this equality also applies to $A$, and thus $P(A\,|\,A)=P(A)$. This implies that $P(A)^2=P(A)$, which is only satisfied by $P(A)=1$.

So the core of the proof is that independence of finite selections implies independence in general. What makes Billingsley's discussion of the theorem appear a bit like black magic is that he first goes through a series of steps to define independence in the infinite case before he states the theorem. But this makes things more murky than they are in Kolmogorov's own statement of the theorem, and it hides the crucial limit argument at the heart of the proof.

An example of a "tail event" which is independent of all finite evidence is the occurrence of infinitely many of the events $A_1, A_2, A_3, \ldots$. The reasons that this is independent of an event $B \in \sigma(A_1, A_2, \ldots, A_N)$ is that$$
\forall x\geq 1 \, \exists y\geq x \, : A_y
$$is logically equivalent to$$
\forall x\geq N \, \exists y\geq x \, : A_y.
$$Conditioning on this some initial segment thus does not change the probability of this event.

Note, however, that this is not generally the case for events of the type$$
\forall x\geq 1 \, \exists y\geq 1 \, : A_y.
$$It is only the "$y\geq x$" in the previous case that ensures the equivalence.

A statement like "For all $i$, $X_i$ will be even" is for instance not a tail event, since a finite segment can show a counterexample, e.g., $X_1=17$. Crucially, however, this example fails to be a tail event because the events "$X_i is even$" (the inner quantification) can be written as a disjunctions of finitely many simple events. We can thus give a counterexample to the outer quantification ($\forall i$) by exhibiting a single $i$ for which the negation of "$X_i$ is even" (which is a universal formula) is checkable in finite time.

Reversely, if this were not the case for any of the statements in the inner loop, the event would be a tail event. That is, if the universal quantification were over a list of events which had no upper limit on the potential index of the verifier, then finite data could not falsify the statement. This happens when the existence of a single potential verifier implies the existence of infinitely many (as it does in the case of "infinitely often" statements, since any larger $y$ is an equally valid candidate verifier).

Events of the form $\exists y: A_y$ are also not tail events, since they are not independent of the counterexamples $\neg A_1, \neg A_2, \neg A_3\ldots$. They are, however, independent of any finite selection of positive events (which do not entail the negation of anything on the list).

We thus have a situation in which sets at the two lowest levels of the Borel hierarchy can have probabilities of any value in $[0,1]$, but as soon as we progress beyond that in logical complexity, only the values 0 and 1 are possible.

Illustration by Yoni Rozenshein.

Oddly enough, this means that no probability theory is possible on complex formulas: When a probability measure is defined in terms of a set of simple events, then the $\Delta_2$ events always have probability 0 or 1. This property is conserved at higher orders in the hierarchy, since the quantifications that push us up the hierarchy are countable (and a countable union of null sets is a null set, by the union bound).

Note also that$$
\lim_{N\rightarrow \infty} \cup_{i=1}^{N} \cap_{j=1}^{N} A_{ij}
\;=\;  \cup_{i=1}^{\infty} \cap_{j=1}^{\infty} A_{ij},
$$and since $$
P\left( \cup_{i=1}^{\infty} \cap_{j=1}^{\infty} A_{ij} \right)  \;\in\;  \{0,1\},
$$the probabilities of the finite approximations must converge to these extremes. As we spend more computational power checking the truth of a tail event on a specific sample, we thus get estimates that approach 0 or 1 (although without any general guarantees of convergence speed).

This sheds some light on what the theorem actually means in practice, and how it relates to the zero-one theorem for finite models of first-order logic.

Billingsley: Ergodic Theory and Information (1965)

I'm reading this book (now for the second time) for its proof of Birkhoff's ergodic theorem. I was held up by lots of technical details the first time around, but I've gotten a bit further now.

Definitions

The set-up is the following: A set $X$ is given along with a time shift $T: X\rightarrow X$. We further have a probability distribution on the set of sample paths $\Omega=X^{\infty}$. For each integrable function $f: \Omega\rightarrow \mathbb{R}$, we consider the time-average$$
 A_n f(\omega) \;=\; \frac{1}{n} \sum_{i=0}^{n-1} f(T^i \omega).
$$We are interested in investigating whether such averages converge as $n\rightarrow\infty$. If so, we would also like to know whether the limit is the same on all sample paths.

Notice that $f$ is a function of the whole sample path, not just of a single coordinate. The function $f$ could for instance measure the difference between two specific coordinates, or it could return a limiting frequency associated with the sample path.

If $f$ has the same value on the entire orbit $$\omega,\, T\omega,\, T^2\omega,\, T^3\omega,\, \ldots$$we say that the function $f$ is time-invariant. Global properties like quantifications or tail events are time-invariant.

A set $A$ is similarly time-invariant if $TA=A$. (I like to call such invariants as "trapping sets.") This is a purely set-theoretic notion, and doesn't involve the probability measure on the sample paths. However, we say that an invariant set $A$ is a non-trivial if $0 < P(A) < 1$.

Statement

One of the things that are confusing about the so-called ergodic theorem is that it actually doesn't involve ergodicity very centrally. Instead it makes the following statement:

1. Suppose that the time shift $T$ is measure-preserving in the sense that $P(A)=P(TA)$ for all measurable sets $A$. Then the time-averages of any integrable function converges almost everywhere:$$P\left\{\omega : \lim_{n\rightarrow \infty} A_n f(\omega) \textrm{ exists} \right\} \;=\; 1.$$
2. Suppose in addition that the time shift is ergodic in the sense that all its trapping are trivial with respect to $P$. Then the time averages are all the same on a set with $P$-probability 1:$$P \left\{(\omega_1, \omega_2) : \lim_{n\rightarrow\infty} A_n f(\omega_1) \,=\, \lim_{n\rightarrow\infty} A_n f(\omega_2) \right\} \;=\; 1.$$

Almost all the mathematical energy is spent proving the first part of this theorem. The second part is just a small addendum: Suppose a function $f$ has time averages which are not almost everywhere constant. Then the threshold set$$
\{A_n f \leq \tau\}
$$must have a probability strictly between 0 and 1 for some threshold $\tau$. (This is a general feature of random variables that are not a.e. constant.)

But since the limiting time-average is a time-invariant property (stating something about the right-hand tail of the sample path), this thresholding set is an invariant set. Thus, if the limiting time-averages are not a.e. unique, the time shift has a non-trivial trapping set.

So again, the heavy lifting lies in proving that the time-averages actually converge on all sample paths when the time shift is measure-preserving. Billingsley gives two proofs of this, one following idea by von Neumann, and one following Birkhoff.

The Von Neumann Proof

Let's say that a function is "flat" if it has converging time-averages. We can then summarize von Neumann's proof along roughly these lines:

1. Time-invariant functions are always flat.
2. Increment functions of the form $$s(\omega) \;=\; g(\omega) - g(T\omega),$$ where $g$ is any function, are also flat.
3. Flatness is preserved by taking linear combinations.
4. Any square-integrable function can be obtained as the $L^2$ limit of (linear combinations of) invariant functions and increment functions.
5. Moreover, the $L^2$ limit of a sequence of flat $L^2$ functions is flat; in other words, flatness is preserved by limits.
6. Hence, all $L^2$ functions are flat.

This argument seems somewhat magical at first, but I think more familiarity with the concept of orthogonality and total sets in the $L^p$ Hilbert spaces could make it seem more natural.

Just as an example of how this might work, consider the coordinate function $f(\omega) = X_0(\omega)$. In order to obtain this function as a limit of our basis functions, fix a decay rate $r < 1$ and consider the function$$
g(\omega) \;=\; X_0(\omega) + r X_1(\omega) + r^2 X_2(\omega) + r^3 X_3(\omega) + \cdots
$$Then for $r \rightarrow 1$, we get a converging series of increment functions, $g(\omega) - g(T\omega) \;\rightarrow\; X_0(\omega)$. If the coordinates are bounded (as in a Bernoulli process), this convergence is uniform across all sample paths. Billingsley uses an abstract and non-constructive orthogonality argument to show that such decompositions are always possible.

In order to show that the limit of flat functions is also flat, suppose that $f_1, f_2, f_3, \ldots$ are all flat, and that $f_k \rightarrow f^*$. We can then prove that the averages $A_1f^*, A_2f^*, A_3f^*, \ldots$ form a Cauchy sequence by using the approximation $f^* \approx f_k$ to show that $$| A_n f^* - A_m f^* | \;\approx\; | A_n f_k - A_m f_k | \;\rightarrow\; 0.$$ Of course, the exact argument needs a bit more detail, but this is the idea.

After going through this $L^2$ proof, Billingsley explains how to extend the proof to all of $L^1$. Again, this is a limit argument, showing that the flatness is preserved as we obtain the $L^1$ functions as limits of $L^2$ functions.

The Birkhoff Proof

The second proof follows Birkhoff's 1931 paper. The idea is here to derive the the ergodic theorem from the so-called maximal ergodic theorem (which Birkhoff just calls "the lemma"). This theorem states that if $T$ preserves the measure $P$, then$$
P\left\{ \exists n: A_n f > \varepsilon \right\}
\;\leq\;
\frac{1}{\varepsilon} \int_{\{f>\varepsilon\}} f \;dP.
$$I am not going to go through the proof, but here a few observations that might give a hint about how to interpret this theorem:

1. For any arbitrary function, we have the inequality $$P(f>\varepsilon) \;\leq\; \frac{1}{\varepsilon} \int_{\{f>\varepsilon\}}f\;dP.$$This is the generalization of Markov's inequality to random variables that are not necessarily positive.
2. If $f$ is time-invariant, then $A_n f = f$, because the average $A_n f$ then consists of $n$ copies of the number $$f(\omega) \;=\; f(T\omega) \;=\; f(T^2\omega) \;=\; \cdots \;=\; f(T^{n-1}\omega).$$ In this case, the theorem thus follows from the generalized Markov bound above.
3. Even though we do not assume that $f(\omega)=f(T\omega)$ for a fixed sample path $\omega$, all distributional statements we can make about $f(\omega)$ also apply to $f(T\omega)$ because $T$ is assumed to be measure-preserving; for instance, the two functions have the same mean when $\omega \sim P$. This fact of course plays a crucial part in the proof.
4. By putting $(f,\varepsilon) := (-f,-\varepsilon)$, we see that the maximal ergodic theorem is equivalent to$$P\left\{ \exists n: A_n f < \varepsilon \right\}
   \;\leq\;
   \frac{1}{\varepsilon} \int_{\{f<\varepsilon\}} f \;dP.
   $$

The maximal ergodic theorem can be used to prove the general (or "pointwise") ergodic theorem. The trick is to find both an upper and a lower bound on the non-convergence probability
$$
q \;=\; P\{ \omega :\, \liminf A_n f(\omega) < a < b < \limsup A_n f(\omega) \},
$$with the goal of showing that these bounds can only be satisfied for all $a<b$ when $q=0$.

Comments

One of the things I found confusing about the ergodic theorem is that it is often cited as a kind of uniqueness result: It states, in a certain sense, that all sample paths have the same time-averages.

However, the problem is that when the time shift has several trapping sets, there are multiple probability distributions that make the time shift ergodic. Suppose that $T$ is non-ergodic with respect to some measure $P$; and suppose further that $A$ is a minimal non-trivial trapping set in the sense that it does not itself contain any trapping sets $B$ with $0 < P(B) < P(A) < 1$. Then we can construct a measure with respect to which $T$ will be ergodic, namely, the conditional measure $P(\,\cdot\,|\,A)$.

This observation reflects the fact that if we sample $\omega \sim P$, the sample path we draw may lie in any of the minimal non-trivial trapping sets for $T$. Although it might not be clear based on a finite segment of $\omega$ (because such a segment might not reveal which trapping set we're in), such a sample path will stay inside this minimal trap set forever, and thus converge to the time-average characteristic of that set.

A small complication here is that even a minimal trapping set might in fact contain other trapping sets that are smaller: For instance, the set of coin flipping sequences whose averages converge to 0 are a trapping set, but so is the sequence of all 0s. However, this singleton set will either have positive probability (in which case the limit set is not minimal) or probability 0 (in which case it can never come up as a sample from $P$).

Lastly, suppose the sample path $\omega$ is not sampled from $P$, but from some other measure $Q$ which is not preserved by $T$. Can we say that $A_nQ\rightarrow P$?

Not always, but a sufficient condition for this two be the case is that $Q$ is absolutely continuous with respect to $P$ ($Q \ll P$). If that is the case, then the $Q$-probability of observing a non-converging average will be 0, since the $P$-probability of this event is 0, and $P$ dominates $Q$.

I am not sure whether this condition is also necessary. A degenerate measure $Q$ which places all mass on a single sample path, or even on countably many, can often be detected by some integrable function $f$ such as the indicator function of everything except the points on the countably many countable sample paths.

Billingsley: Probability and Measure (1979), ch. 2

In the chapter on probability measures, Patrick Billingsley points out a potential fallacy that I can see that I have been making (pp. 24–25).

I have been thinking about the $\sigma$-field generated by a system $S$ as the limit of a series of $\sigma$-operations such as complements and countable unions. Such a recursive procedure for generating $\sigma(S)$ fails to generate all the required sets when, as I tacitly imagined, it is applied only countably many times.

This can be seen by means of a clever diagonal argument which involves constructing a bijection between the sets thus generated and the real unit interval. The gist of the proof is that when we only use countably many recursive steps, we can enumerate the resulting sets by means of sequences of integers, and this leaves the construction open to a diagonal counterexample.

The Playing Field

Let's first agree on a canonical defition of the Borel field $\mathcal{B}$ over $(0,1]$: For the present purposes, we will identify it as the $\sigma$-field generated by the intervals with rational endpoints, including the empty set.

This seed set $S$ is countable and can therefore be enumerated, for instance as

$\emptyset$, $(0,1]$, $(0,\frac{1}{2}]$, $(\frac{1}{2},1]$, $(0,\frac{1}{3}]$, $(\frac{1}{3},\frac{1}{2}]$, $(\frac{1}{3},1]$, $(0,\frac{1}{4}]$, $(\frac{1}{4},\frac{1}{3}]$, $(\frac{1}{4},\frac{1}{2}]$, …

For the purposes of this argument, let's further restrict the permitted $\sigma$-operations to

1. complementation;
2. countable unions.

These two operations can compactly be described as an application of the function
$$
\Psi(A_1,A_2,A_3,A_4,\ldots)\;=\; A_1^c \cup A_2 \cup A_3 \cup A_4 \cup \ldots
$$to a countable sequence of sets. We can then translate
\begin{eqnarray}
A^c &=& A^c \cup \emptyset \cup \emptyset \cup \emptyset \cup \emptyset \ldots \\
\cup_{i=1}^{\infty}A_i &=& \emptyset \cup A_1 \cup A_2 \cup A_3 \cup \ldots
\end{eqnarray}Any $\sigma$-operation is then a matter of applying $\Psi$ to some countable sequence of sets.

The Elements of Construction

This allows us to describe our inductive procedure in terms of a recursive function $R$ that uses $\Psi$ as a subroutine. We define the base case as a projection:
$$
R_0(A_1,A_2,A_3,\ldots) \;=\; A_1.
$$In order to define the recursion step, we first need to recall that $\mathbb{N}$ and $\mathbb{N}\times \mathbb{N}$ are equipotent, as can be seen from the following array:
$$
\begin{array}{ccccc}
1 & 2 & 4 & 7 & \ldots \\
3 & 5 & 8 & 12 & \ldots \\
6 & 9 & 13 & 18 & \ldots \\
10 & 14 & 19 & 25 & \ldots \\
\vdots & \vdots & \vdots & \vdots &
\end{array}

$$This array can be used to map a countable sequences of sets onto a countable sequence of countable sequences. We use this unraveling trick for the recursive step:
$$
R_n(A_1,A_2,\ldots) \;=\; \Psi( R_{n-1}(A_1,A_2,A_4,\ldots), R_{n-1}(A_3,A_5,A_8,\ldots), \ldots )
$$This way, we can, by a careful selection of rational intervals $A_1,A_2,A_3,\ldots$, select any countable sequence of $R_{n-1}$-sets we like and feed that sequence into a new $\Psi$-construction.

This trick, then, transforms a sequence of rational intervals into an $R_n$-set. In other words, any level $n$ set can be obtained in a (nominally) single step by feeding an appropriate sequence of level 0 sets into the recursive function $R_n$.

From Sequences to Sets

We now finally put
$$
R(A_1,A_2,\ldots) \;=\; R_1(A_1,A_2,\ldots) \cup R_2(A_3,A_5,\ldots) \cup R_3(A_6,A_9,\ldots) \cup \ldots
$$This format covers all sets that can be obtained by any finite level of recursion.

For instance, if a set can be produced by level 17 recursion, we can choose the sequence $A_1,A_2,A_3,\ldots$ so that all $R_i$ except $R_{17}$ return the empty set, while $R_{17}$ returns the set we are interested in. This is achieved by constructing the sequence $A_1,A_2,A_3,\ldots$ so that it splits into just the right subsequences.

In other words, we now have a method for picking out any set that requires only finite recursion by specifying an appropriate sequence of rational intervals. Since there are only countably many rational intervals, the selection of a specific finite-level recursive set is therefore a matter of selecting an appropriate sequence of natural numbers.

Stone to Stone

The set of countably infinite sequences of positive integers is equipotent with the real unit interval.

Specifically, Billingsley suggests that we use the bijection $b$ which for its $k$th integer takes the distance between the $(k-1)$st and the $k$th occurrence of the digit 1 in the binary expansion of the number $\omega\in(0,1]$.

Under this bijection,
$$
b\left(\frac{1}{3}\right) \;=\; b\left(.010101\ldots\right) \;=\; (2,2,2,2,\ldots).
$$For this definition to work, there has to be infinitely many 1s in the binary expansion. This can be guaranteed by always choosing the non-terminating binary expansion of $\omega$, so that, for instance,
$$
b\left(\frac{1}{2}\right) \;=\; b\left(.111111\ldots\right) \;=\; (1,1,1,1,\ldots).
$$Incidentally, this choice also fixes the ambiguity problems we might otherwise have, making $b$ a true bijection.

Because of the construction above, the output of this function can be fed into another function that translates the infinite sequences of integers into an infinite sequence of rational intervals, which is then translated into a recursive set. In other words, we can map any real number $\omega\in(0,1]$ onto a finite-recursion set $s(\omega)$. This establishes a surjection of the real numbers into the system of sets that can be obtained by finitely many $\sigma$-operations on the rational intervals.

The Diagonal

The set we now need for our diagonal counterargument is
$$
B = \{\omega\in (0,1]: \omega \not\in s(\omega)\}.
$$This set cannot be obtained by finite application of $\sigma$-operations to the rational intervals.

If it could, there would be at least one $\omega_0\in (0,1]$ such that $B=s(\omega_0)$. But by construction, $\omega_0\in s(\omega_0)$ holds if and only if $B\not \in s(\omega_0)$; so $B$ and $s(\omega_0)$ must disagree at least about the element $\omega_0$ and possibly about others. This contradicts the assertion that $B=s(\omega_0)$. Hence, $B$ cannot be obtainable by a finite number of recursive operations.

On the other hand, $B$ is a Borel set. Billingsley gives a complicated argument for this which is a notational nightmare, but I think he is just expressing a proof by induction.

Proof Sketch

The first step of the argument introduces a building block called $D_k$. This is the set of real numbers that are members of $A_k(\omega)$, the $k$th interval on their own list. For instance, $\omega=1/3$ is mapped onto the constant interval list
$$
(0,1], (0,1], (0,1], (0,1], \ldots
$$since $b(1/3)=(2,2,2,\ldots)$, and $(0,1]$ is the second interval on the list. It thus follows that $1/3\in A_k(1/3)$ for all $k$, and thus $1/3 \in D_k$ for all $k$.

The sets $D_k$ are Borel sets. This follows from the fact that the set of binary numbers with a specific $k$th waiting time is a Borel set, and that any rational interval is a Borel set.

The second part of the argument uses a translation between the set
$$
R_n(A_{m_1}(\omega), A_{m_2}(\omega), A_{m_3}(\omega), \ldots)
$$and the set
$$
R_n(D_{m_1}, D_{m_2}, D_{m_3}, \ldots)
$$for an arbitrary integer sequence $m_1, m_2, m_3, \ldots$. This equality is proven by induction.

Using this translation, the set $B^c = \{\omega\in s(\omega)\}$ can then be rewritten as a countable union of sets of this form. This shows that $B$ is indeed a Borel set, since the $D_k$ are.