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Measurable fields of weighted shifts over the 2-adic integers

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

This lesson constructs the operator \(S\) of OpenAI's theorem [OAI]. It acts on square-integrable fields of vectors in \(K=\ell^2(\mathbb Z)\) over the \(2\)-adic integers \(X\). At every point \(x\in X\) it is a bilateral weighted shift \(S_x\) whose closed invariant subspaces are exactly the tails, by Corollary 5.2 of the first lesson. The weights are chosen so that a later operator can link neighbouring fibres. The lesson also proves the estimate \(\|S^n\|\le\exp(-10\lfloor(n+1)^2/4\rfloor)\), so \(S\) is nonzero and \(\|S^n\|^{1/n}\to0\).

Section 1 describes the probability space and the odometer \(x\mapsto x+1\). Section 2 treats square-integrable fields and proves that every operator commuting with the scalar multipliers is given by a measurable field of operators [OAI-C, Lemma 3.1]. Section 3 defines the weights and checks Corollary 5.2 of the first lesson in every fibre, and Section 4 proves the power estimate.

We use: the product of probability spaces and the uniqueness of measures that agree on a family closed under finite intersections, from Section 9 of the double commutant lesson (Theorem "Arbitrary products of probability spaces" and Lemma "Uniqueness from a generating family"); monotone convergence and Fatou's lemma, Theorems 2.1 and 2.2 of the measure tools lesson, and completeness of \(L^2\), Theorem 3.2 there; from the Hilbert space lesson, Theorem 9.1 (an operator on \(L^2\) of a finite measure space that commutes with all multiplication operators is a multiplication operator), the projection theorem (Theorem 2.2) and closedness of the compact operators (Theorem 5.1).

1. The 2-adic integers and the odometer

Let \(X=\{0,1\}^{\{1,2,3,\ldots\}}\) be the set of digit sequences \(x=(x_1,x_2,\ldots)\), with the sigma-algebra \(\Sigma\) generated by the coordinates, and let \(\mu\) be the product of the probability measures on \(\{0,1\}\) giving mass \(\frac12\) to each point; it exists and is unique by the product theorem of the double commutant lesson. We read \(x\) as the \(2\)-adic integer \(\sum_ix_i2^{i-1}\), and \(\mu\) as its Haar probability measure, but use only the concrete description. For \(m\ge1\) put \[ \begin{aligned} &N_m=2^m,\\ &T_m(x)\\ &\quad =\sum_{i=1}^mx_i2^{i-1}\in\{0,1,\ldots,N_m-1\}. \end{aligned} \] So \(T_m(x)\) is "\(x\bmod 2^m\)", and the residue class \(\{T_m=r\}\) is the set of sequences whose first \(m\) digits are the binary digits of \(r\). Its measure is \(2^{-m}\), and \[ \begin{aligned} &T_{m+1}(x)-T_m(x)\\ &\quad =x_{m+1}N_m\in\{0,N_m\}. \end{aligned}\tag{1.1} \]

The odometer \(\theta:X\to X\) adds \(1\) with carry. If the first \(0\) among the digits of \(x\) is at position \(k\), then \(\theta x\) has digits \(0\) at positions \(1,\ldots,k-1\), the digit \(1\) at position \(k\), and the digits of \(x\) after position \(k\); and \(\theta(1,1,1,\ldots)=(0,0,0,\ldots)\). Subtraction of \(1\) with borrow is its inverse, so \(\theta\) is a bijection. For every \(m\ge1\) the rule (1.2) holds: \[ \begin{aligned} &T_m(\theta x)\\ &\quad=\begin{cases}T_m(x)+1,&T_m(x)<N_m-1,\\0,&T_m(x)=N_m-1.\end{cases} \end{aligned} \] Indeed, if the first \(m\) digits are all \(1\), they all become \(0\). Otherwise the first \(0\) is at a position \(k\le m\), and \(T_m\) changes by \(2^{k-1}-\sum_{i<k}2^{i-1}=1\).

Lemma 1.1.

  1. The residue classes generate \(\Sigma\). Together with \(\varnothing\) and \(X\) they form a family closed under finite intersections.
  2. \(\theta\) and \(\theta^{-1}\) are measurable, and \(\mu(\theta^{-1}E)=\mu(\theta E)=\mu(E)\) for \(E\in\Sigma\).
  3. Every countable subset of \(X\) is null.
  4. (Ergodicity.) If \(E\in\Sigma\) and \(\mu(E\,\triangle\,\theta^{-1}E)=0\), then \(\mu(E)\in\{0,1\}\).
  5. The indicators of the residue classes span a dense subspace of \(L^2(X,\mu)\). In particular \(L^2(X,\mu)\) is separable, and it is infinite-dimensional.

Proof. (1) A residue class is a coordinate cylinder, and \(\{x_i=1\}\) is the union of the classes \(\{T_i=r\}\) over the \(r\) whose \(i\)-th binary digit is \(1\). By (1.1), two classes \(\{T_m=r\}\) and \(\{T_{m'}=r'\}\) with \(m\le m'\) are disjoint unless \(r'\equiv r\pmod{2^m}\), in which case their intersection is \(\{T_{m'}=r'\}\).

(2) By (1.2), \(\theta^{-1}\{T_m=r\}=\{T_m=r-1\bmod N_m\}\) and \(\theta\{T_m=r\}=\{T_m=r+1\bmod N_m\}\). The sets whose preimage under \(\theta\) (respectively \(\theta^{-1}\)) is measurable form a sigma-algebra containing the residue classes, hence containing \(\Sigma\). The probability measures \(E\mapsto\mu(\theta^{-1}E)\) and \(\mu\) agree on the family of (1), so they agree on \(\Sigma\) by the uniqueness lemma. Applying this to \(\theta E\) gives \(\mu(E)=\mu(\theta E)\).

(3) \(\{x\}\subseteq\{T_m=T_m(x)\}\) for every \(m\), so \(\mu\{x\}\le2^{-m}\) for all \(m\).

(4) For a residue class \(\{T_m=r\}\), using (2), \(\theta^{-1}\{T_m=r+1\bmod N_m\}=\{T_m=r\}\), and \(\mu(E\,\triangle\,\theta^{-1}E)=0\), \[ \begin{aligned} &\mu(E\cap\{T_m=r+1\bmod N_m\})\\ &\quad =\mu(\theta^{-1}E\cap\{T_m=r\})\\ &\quad =\mu(E\cap\{T_m=r\}). \end{aligned} \] So \(\mu(E\cap\{T_m=r\})\) does not depend on \(r\); summing over \(r\) shows that it equals \(\mu(E)/N_m=\mu(E)\mu\{T_m=r\}\). The finite measures \(A\mapsto\mu(E\cap A)\) and \(A\mapsto\mu(E)\mu(A)\) therefore agree on the family of (1), hence on \(\Sigma\). With \(A=E\) we get \(\mu(E)=\mu(E)^2\).

(5) Let \(f\in L^2(X,\mu)\) be orthogonal to all residue-class indicators. Since these indicators are real, the real and imaginary parts of \(f\) are orthogonal to them too, so we may take \(f\) real. Then \(\int_Xf\,d\mu=\int_{\{T_1=0\}}f+\int_{\{T_1=1\}}f=0\). The finite measures \(A\mapsto\int_Af^+d\mu\) and \(A\mapsto\int_Af^-d\mu\) therefore agree on the family of (1), hence on \(\Sigma\). With \(A=\{f>0\}\) this gives \(\int f^+d\mu=0\), and similarly \(\int f^-d\mu=0\), so \(f=0\). By the projection theorem the indicators span a dense subspace; they are countably many. For each \(m\), the \(N_m\) indicators of level \(m\) are nonzero and mutually orthogonal, so the dimension is infinite. \(\square\)

2. Square-integrable fields and decomposable operators

Let \(K=\ell^2(\mathbb Z)\) with basis \((e_j)\). On \(X\times\mathbb Z\) consider the sets \(E\) whose sections \(E_j=\{x:(x,j)\in E\}\) lie in \(\Sigma\), and the measure \(\mu\otimes\#(E)=\sum_j\mu(E_j)\); countable additivity follows from that of \(\mu\) and the rearrangement of nonnegative double series. Put \[ H_0=L^2(X\times\mathbb Z,\mu\otimes\#), \] a Hilbert space by Theorem 3.2 of the measure tools lesson. For \(f\in H_0\) and \(x\in X\) write \(f(x)=(f(x,j))_{j\in\mathbb Z}\). By monotone convergence, \[ \|f\|^2=\int_X\|f(x)\|_K^2\,d\mu(x), \] so \(f(x)\in K\) for almost every \(x\). Conversely, a map \(x\mapsto f(x)\in K\) whose coordinates are measurable and with \(\int_X\|f(x)\|^2d\mu<\infty\) is an element of \(H_0\). For \(v\in K\) the constant field \(1\otimes v\) is \(x\mapsto v\). For \(g\in L^\infty(X,\mu)\) the scalar multiplier is \((M_gf)(x)=g(x)f(x)\). A field is finitely supported if only finitely many coordinates \(f(\cdot,j)\) are nonzero; truncation shows that finitely supported fields are dense in \(H_0\).

A measurable field of operators is a map \(x\mapsto A_x\in B(K)\) such that \(x\mapsto\langle A_xe_i,e_j\rangle\) is measurable for all \(i,j\). Then \(x\mapsto\langle A_xv,e_j\rangle=\sum_iv_i\langle A_xe_i,e_j\rangle\) is measurable for every \(v\in K\), as a pointwise limit of measurable partial sums.

Lemma 2.1 (decomposable operators). Let \(x\mapsto A_x\) be a measurable field of operators with \(\|A_x\|\le C\) for almost every \(x\).

  1. \((Af)(x)=A_xf(x)\) defines an operator \(A\in B(H_0)\) with \(\|A\|\le C\), which commutes with every scalar multiplier.
  2. If \(x\mapsto B_x\) is a second such field defining \(B\), then \(x\mapsto A_xB_x\) and \(x\mapsto A_x^*\) are such fields, and they define \(AB\) and \(A^*\).

Proof. (1) The coordinates \(x\mapsto\langle A_xf(x),e_j\rangle=\sum_if(x,i)\langle A_xe_i,e_j\rangle\) are measurable, being limits of measurable partial sums (for almost every \(x\), \(f(x)\in K\) and the series converges). Moreover \(\int\|A_xf(x)\|^2d\mu\le C^2\|f\|^2\). Scalars commute with \(A_x\).

(2) \(\langle A_xB_xe_i,e_j\rangle=\sum_k\langle B_xe_i,e_k\rangle\langle A_xe_k,e_j\rangle\) and \(\langle A_x^*e_i,e_j\rangle=\overline{\langle A_xe_j,e_i\rangle}\) are measurable, and \(\|A_xB_x\|,\|A_x^*\|\) are bounded almost everywhere. Clearly \((ABf)(x)=A_xB_xf(x)\). Finally \(\langle Af,g\rangle=\int\langle A_xf(x),g(x)\rangle d\mu\), which is \(\int\langle f(x),A_x^*g(x)\rangle d\mu\), so the field \(A_x^*\) defines \(A^*\). \(\square\)

Operators of this form are called decomposable. The next theorem says that every operator commuting with the scalar multipliers is decomposable. It is the only place where the structure of \(H_0\) as a space of fields enters the proof of transitivity.

Theorem 2.2 (fields from scalar multipliers). Let \(A\in B(H_0)\) commute with every scalar multiplier \(M_g\), \(g\in L^\infty(X,\mu)\). There is a measurable field of operators \(x\mapsto A_x\) with \(\|A_x\|\le\|A\|\) for every \(x\), such that \((Af)(x)=A_xf(x)\) for almost every \(x\), for each \(f\in H_0\). Two such fields agree almost everywhere. If \(A\) is an orthogonal projection, then \(A_x\) is an orthogonal projection for almost every \(x\).

Proof. Matrix entries. Let \(\iota_j:L^2(X,\mu)\to H_0\) place a function in the \(j\)-th coordinate, and let \(\pi_i:H_0\to L^2(X,\mu)\) take the \(i\)-th coordinate. Both intertwine the multiplication operators on \(L^2(X,\mu)\) with the scalar multipliers on \(H_0\). So \(\pi_iA\iota_j\) commutes with every multiplication operator on \(L^2(X,\mu)\), and by Theorem 9.1 of the Hilbert space lesson it is multiplication by some \(a_{ij}\in L^\infty(X,\mu)\). Fix measurable representatives. For a finitely supported \(f\), \[ \begin{aligned} &(Af)(\cdot,i)\\ &\quad =\sum_j\pi_iA\iota_jf(\cdot,j)\\ &\quad =\sum_ja_{ij}f(\cdot,j)\\ &(i\in\mathbb Z). \end{aligned}\tag{2.1} \]

A pointwise bound. Let \(\mathcal V\) be the countable set of finitely supported \(v\in K\) with coordinates in \(\mathbb Q+i\mathbb Q\). For \(v\in\mathcal V\) and \(E\in\Sigma\), apply (2.1) to \(f=1_E\otimes v\) and use \(\|Af\|\le\|A\|\|f\|\) and monotone convergence: \[ \begin{aligned} &\int_E\sum_i\Big|\sum_ja_{ij}(x)v_j\Big|^2d\mu(x)\\ &\quad \le\|A\|^2\|v\|^2\mu(E). \end{aligned} \] For \(E\) the set where the integrand exceeds \(\|A\|^2\|v\|^2\), this forces \(\mu(E)=0\). Let \(N\) be the union of these countably many null sets.

The field. For \(x\notin N\) and finitely supported \(v\in K\), put \((A_xv)_i=\sum_ja_{ij}(x)v_j\). This is linear in \(v\), and \(\|A_xv\|\le\|A\|\|v\|\) for \(v\in\mathcal V\). A finitely supported \(v\) is a coordinatewise limit of vectors \(v^{(n)}\in\mathcal V\) with the same support; each coordinate of \(A_xv^{(n)}\) converges to that of \(A_xv\), so Fatou's lemma for sums gives \(\|A_xv\|\le\|A\|\|v\|\). Hence \(A_x\) extends to an operator on \(K\) with \(\|A_x\|\le\|A\|\) and \(\langle A_xe_j,e_i\rangle=a_{ij}(x)\). For \(x\in N\) put \(A_x=0\). The field is measurable.

Representation. Let \(\tilde A\) be the decomposable operator of the field (Lemma 2.1). By (2.1), \(\tilde Af=Af\) for finitely supported \(f\). Both operators are bounded and the finitely supported fields are dense, so \(\tilde A=A\).

Uniqueness. If two fields represent \(A\), then for each \(j\) both send \(e_j\) to \((A(1\otimes e_j))(x)\) for almost every \(x\). Outside a countable union of null sets they agree on the basis, hence everywhere on \(K\).

Projections. If \(A=A^*=A^2\), then by Lemma 2.1(2) the fields \(A_x^*\) and \(A_x^2\) also represent \(A\), and uniqueness gives \(A_x=A_x^*=A_x^2\) almost everywhere. \(\square\)

3. The weights

Fix the constants \[ \begin{gathered} c=10,\qquad D=100,\\ \ell_0=1,\qquad \ell_m=8^m\ (m\ge1). \end{gathered} \] The blocks \(I_m=\{k\in\mathbb Z:\ell_{m-1}\le k<\ell_m\}\), \(m\ge1\), partition the positive integers. Put \(J_m=D(N_m-2)\), so \(J_1=0\) and \(J_{m+1}=J_m+DN_m\). Define \(a^+_0(x)=a^-_0(x)=c\) and, for \(k\in I_m\), \[ \begin{aligned} &a^+_k(x)=c(k+1)+J_m\\ &\quad +DT_m(x),\\ &a^-_k(x)=c(k+1)+J_m\\ &\quad +D\big(N_m-1-T_m(x)\big). \end{aligned}\tag{3.1} \] On the whole integer line put \[ \begin{gathered} a_j(x)=\begin{cases}a^+_j(x),&j\ge0,\\a^-_{-j-1}(x),&j\le-1,\end{cases}\\ \beta_j(x)=e^{-a_j(x)}, \end{gathered} \] and let \(S_x\) be the bilateral weighted shift \(S_xe_j=\beta_j(x)e_{j+1}\).

Each \(a^\pm_k\) is the linear term \(c(k+1)\) plus a correction that depends on the residue of \(x\) at the level of the block of \(k\). The linear terms make the logarithms of the products of weights grow quadratically, as Corollary 5.2 of the first lesson requires. Under \(x\mapsto\theta x\) the corrections at a block change by \(+D\) in \(a^+\) and by \(-D\) in \(a^-\), unless the residue wraps around; the third lesson builds a commuting operator from these opposite changes. The jumps \(J_{m+1}-J_m\) keep the weights monotone across block boundaries.

Proposition 3.1. For every \(x\in X\):

  1. \(a^\pm_{k+1}(x)-a^\pm_k(x)\ge c\) for all \(k\ge0\).
  2. \(0\le a^\pm_k(x)-c(k+1)\le4Dk^{1/3}\) for \(k\ge1\).
  3. With \(d(j)=j+1\) for \(j\ge0\) and \(d(j)=|j|\) for \(j\le-1\), we have \(a_j(x)\ge c\,d(j)\). Moreover \(a_{-1}(x)=a_0(x)=c\), and \(a_j(x)\ge2c\) for \(j\notin\{-1,0\}\).
  4. The closed invariant subspaces of \(S_x\) are exactly the tails \(K_{\ge h}\), \(h\in\mathbb Z\cup\{\pm\infty\}\), and \(\|S_x\|=e^{-c}\).
  5. Each \(\beta_j\) is measurable on \(X\); \(x\mapsto S_x\) is a measurable field of operators.

Proof. (1) Inside a block, both \(a^\pm_k\) increase by exactly \(c\). From \(k=0\) to \(k=1\in I_1\): \(a^\pm_1\ge2c=a^\pm_0+c\). At a boundary, \(k=\ell_m-1\in I_m\) and \(k+1=\ell_m\in I_{m+1}\), put \(\delta=T_{m+1}(x)-T_m(x)\in\{0,N_m\}\) (by (1.1)). The increments are \[ \begin{aligned} &a^+_{k+1}-a^+_k=c+DN_m+D\delta\ge c,\\ &a^-_{k+1}-a^-_k=c+DN_m\\ &\quad +D(N_{m+1}-N_m)-D\delta=c\\ &\quad +D(N_{m+1}-\delta)\ge c . \end{aligned} \]

(2) For \(k\in I_m\), \(k\ge8^{m-1}\) gives \(k^{1/3}\ge2^{m-1}=N_m/2\). The correction is at most \(J_m+D(N_m-1)<2DN_m\le4Dk^{1/3}\), and it is nonnegative because \(0\le T_m\le N_m-1\) and \(J_m\ge0\).

(3) By (2), \(a^\pm_k\ge c(k+1)\) for all \(k\ge0\). For \(j\le-1\), \(a_j=a^-_{-j-1}\ge c(-j)=c|j|\). The values at \(-1\) and \(0\) are \(a^-_0=a^+_0=c\), and \(d(j)\ge2\) for the other \(j\).

(4) By (1), \(a_0\le a_1\le\cdots\) and \(a_{-1}\le a_{-2}\le\cdots\), and by (3) the lower bounds of Corollary 5.2 of the first lesson hold with \(\kappa=c=10>2\log3\). That corollary gives the invariant subspaces. The norm of a weighted shift is the supremum of its weights, here \(\beta_0(x)=e^{-c}\).

(5) \(T_m\) is measurable, so each \(a_j\) is. The matrix entries of \(S_x\) are \(\beta_j(x)\) or \(0\). \(\square\)

By Lemma 2.1, the field \(x\mapsto S_x\) defines an operator \[ \begin{gathered} S\in B(H_0),\\ (Sf)(x)=S_xf(x),\qquad \|S\|\le e^{-c}, \end{gathered} \] which commutes with every scalar multiplier.

4. The power estimate

Proposition 4.1. For every \(n\ge1\), \[ \|S^n\|\le\exp\Big(-c\Big\lfloor\frac{(n+1)^2}4\Big\rfloor\Big). \] Moreover \(\|S\|=e^{-c}\), so \(S\neq0\); \(\lim_n\|S^n\|^{1/n}=0\); and \(S\) is not a scalar multiple of the identity.

Proof. By Lemma 2.1(2), \(S^n\) is the decomposable operator of the field \(S_x^n\), and by Exercise 6.2 of the first lesson, \[ \begin{aligned} &\|S_x^n\|=\sup_{j\in\mathbb Z}\exp\Big(-\sum_{r=0}^{n-1}a_{j+r}(x)\Big)\\ &\quad \le\sup_{j\in\mathbb Z}\exp\Big(-c\sum_{r=0}^{n-1}d(j+r)\Big). \end{aligned} \] The function \(d\) takes each value \(v\ge1\) exactly twice, at \(v-1\) and at \(-v\). So the \(n\) values \(d(j),\ldots,d(j+n-1)\) at distinct integers add up to at least the sum of the \(n\) smallest entries of the list \(1,1,2,2,3,3,\ldots\). For \(n=2m\) that sum is \(m(m+1)\), and for \(n=2m+1\) it is \((m+1)^2\); in both cases it is \(\lfloor(n+1)^2/4\rfloor\). This bounds \(\|S_x^n\|\) for every \(x\), and Lemma 2.1(1) transfers the bound to \(S^n\).

Since \(\beta_0(x)=e^{-c}\) for all \(x\), \(S(1\otimes e_0)=e^{-c}(1\otimes e_1)\), so \(\|S\|\ge e^{-c}\). Next, \(\|S^n\|^{1/n}\le\exp(-c\lfloor(n+1)^2/4\rfloor/n)\to0\). Finally, if \(S=\lambda1\) then \(\|S^n\|^{1/n}=|\lambda|\), which forces \(\lambda=0\), contradicting \(S\neq0\). \(\square\)

An operator with \(\|S^n\|^{1/n}\to0\) is called quasinilpotent; the spectral radius formula identifies this with \(\sigma(S)=\{0\}\), but we do not need it.

5. Exercises

Exercise 5.1 (medium; compact fibres, a noncompact operator). Show that every \(S_x\) is compact, but \(S\) is not.

Solution. By Proposition 3.1(3), \(\beta_j(x)\to0\) as \(|j|\to\infty\). Let \(F_N\) be the weighted shift with the weights \(\beta_j(x)\) for \(|j|\le N\) and \(0\) otherwise; it has finite rank, and \(\|S_x-F_N\|=\sup_{|j|>N}\beta_j(x)\to0\). The compact operators are norm closed (Theorem 5.1 of the Hilbert space lesson), so \(S_x\) is compact. For \(S\), the sets \(E_n\) of the \(x\) with \(x_1=\cdots=x_{n-1}=0\) and \(x_n=1\) are disjoint with \(\mu(E_n)=2^{-n}\), so \(\varphi_n=2^{n/2}1_{E_n}\) is an orthonormal sequence in \(L^2(X,\mu)\). The fields \(\varphi_n\otimes e_0\) are orthonormal in \(H_0\), and \(S(\varphi_n\otimes e_0)=e^{-c}\varphi_n\otimes e_1\). These images are orthogonal with the same norm \(e^{-c}\), so no subsequence converges. A compact operator maps bounded sequences to sequences with convergent subsequences, so \(S\) is not compact. \(\square\)

Exercise 5.2 (easy; the multipliers are not maximal abelian). Let \(Q_0\) be the orthogonal projection of \(H_0\) onto the fields supported in the coordinate \(0\). Show that \(Q_0\) commutes with every scalar multiplier but is not a scalar multiplier, and that \(Q_0\notin\{S\}'\).

Solution. \(Q_0\) is the decomposable operator of the constant field \(x\mapsto\) (projection onto \(\mathbb Ce_0\)), so it commutes with all \(M_g\) by Lemma 2.1(1). If \(Q_0=M_g\), then \(Q_0(1\otimes e_0)=1\otimes e_0\) gives \(g=1\) almost everywhere, while \(Q_0(1\otimes e_1)=0\) gives \(g=0\) almost everywhere. Finally \(Q_0S(1\otimes e_0)=e^{-c}Q_0(1\otimes e_1)=0\), whereas \(SQ_0(1\otimes e_0)=e^{-c}(1\otimes e_1)\neq0\). \(\square\)

So the abelian algebra of scalar multipliers, which lies in \(\{S\}'\), is not maximal abelian in \(B(H_0)\): it has infinite multiplicity.

Exercise 5.3 (easy; the exceptional orbit). Show that \(T_m(x)=N_m-1\) holds for infinitely many \(m\) only for \(x=(1,1,1,\ldots)\). Let \(X_*\) be the set of sequences that are not eventually constant. Show that \(X_*\) is conull, that \(\theta X_*=X_*\), and that \((1,1,1,\ldots)\notin X_*\).

Solution. \(T_m(x)=N_m-1\) means that the first \(m\) digits are \(1\); if this holds for infinitely many \(m\), all digits are \(1\). The complement of \(X_*\) is countable, hence null by Lemma 1.1(3). The odometer and its inverse change only finitely many digits of a sequence, so they preserve being eventually constant, and \(\theta X_*=X_*\). The constant sequence \((1,1,1,\ldots)\) is eventually constant. \(\square\)

Where this leads

The next lesson constructs a contraction \(V\) commuting with \(S\) that sends the fibre over \(x\) to the fibre over \(\theta x\) and moves every tail one step backward. With Theorem 2.2 and the ergodicity of \(\theta\), it proves that \(S\), \(V\) and the scalar multipliers have no common invariant closed subspace except \(\{0\}\) and \(H_0\).

References