The Ergodic Theorems of von Neumann and Birkhoff — Examples

Let $C_n$ require only digits $2,4,\dots ,2n$ to be zero. Prescribing the first $2n$ digits leaves $n$ odd-position digits free, so [F1] writes $C_n$ as a disjoint union of $2^n$ half-open cylinders, each of length $2^{-2n}$. Consequently $\lambda (C_n)=2^{-n}$.

For $A\subseteq {\mathbb{N}}_{>0}$, form the binary string whose digit at position $2r-1$ is $1$ exactly when $r\in A$, and whose even-position digits are all $0$. Its series $x_A={\sum }_{r\in A}{2}^{-(2r-1)}$ lies in $[0,1)$. After any place its tail has a forced zero, so the tail value is strictly less than $1$; the greedy recurrence therefore recovers exactly this string. Thus $A\mapsto x_A$ maps $\mathcal{P}({\mathbb{N}}_{>0})$ into $C$.

Every string in $C$ has every even digit zero, so two adjacent digits can never both equal one. The word $11$ has frequency $0$, not the required $1\mathrm{/}4$; hence no member of $C$ is normal in base two.

The sets $C_n$ decrease and $C={\bigcap }_{n\ge 1}{C}_n$. Since $\lambda (C_1)<\mathrm{\infty }$, continuity from above gives $\lambda (C)={\inf }_n{2}^{-n}=0$.

Different subsets have different first differing odd digit, so canonical uniqueness makes the map injective. Conversely, the set of odd positions at which a point of $C$ has digit one recovers it, so the map is a bijection onto $C$.

If $C$ were countable, [F3] would give a surjection $\mathbb{N}\to C$. Composing it with the inverse bijection in step 2.2 and the explicit shift bijection between $\mathbb{N}$ and ${\mathbb{N}}_{>0}$ would give a surjection ${\mathbb{N}}_{>0}\to \mathcal{P}({\mathbb{N}}_{>0})$, contradicting Cantor's theorem. Hence $C$ is uncountable.

Steps 1.3, 2.1, and 3.1 give an uncountable null subset of the exceptional set in [F4]. Countable choice is inherited from the Lebesgue/cylinder suppliers; the family and the coding map are explicit.

The inequalities obtained by squaring positive rational bounds show $1<\sqrt{2}<3\mathrm{/}2$, $4\mathrm{/}3<\sqrt{2}<5\mathrm{/}3$, $5\mathrm{/}4<\sqrt{2}<3\mathrm{/}2$, and $7\mathrm{/}5<\sqrt{2}<8\mathrm{/}5$. Hence the relevant integer parts of $n\sqrt{2}$ for $n=0,\dots ,5$ are $0,1,2,4,5,7$.

Therefore the exact fractional parts are $0,\sqrt{2}-1,2\sqrt{2}-2,3\sqrt{2}-4,4\sqrt{2}-5,5\sqrt{2}-7.$ Direct integer squaring gives ${\left(\frac{14142135}{{10}^7}\right)}^2<2<{\left(\frac{14142136}{{10}^7}\right)}^2,$ so positivity places $\sqrt{2}$ between these two rational numbers; substituting the bounds into the five exact expressions and rounding to four decimal places gives the displayed list.

Since $\sqrt{2}$ is irrational by [F1], [F2] applies and proves that the entire infinite sequence is equidistributed. The six computations in step 2.1 illustrate this orbit; they do not by themselves prove its asymptotic distribution. Countable choice is used only through [F2].

Since $T^2$ is the identity, the summands alternate between $f$ and $f\circ T$. Moreover $f\circ T={\mathbf{1}}_{[1\mathrm{/}2,3\mathrm{/}4)}$.

For $n=2q$, $A_{n}f=(f+f\circ T)\mathrm{/}2=g$. For $n=2q+1$, the counts of the two summands differ by one, so $\mathrm{\mid }{A}_{n}f-g\mathrm{\mid }\le 1\mathrm{/}n$ pointwise.

It follows that $A_{n}f\to g$ at every point. The half-rotation interchanges the two support intervals, so $g\circ T=g$; by [F3] it takes both values $0$ and $1\mathrm{/}2$ on sets of positive measure and is therefore nonconstant.

Finally, [F3] gives $\int gd\lambda =(1\mathrm{/}2)(1\mathrm{/}4+1\mathrm{/}4)=1\mathrm{/}4=\int fd\lambda$. Thus the limit preserves the mean but need not equal the constant mean in this nonergodic example. Countable choice is used only through [F1] and [F3].

Put $N_E(n,x)={\sum }_{k=0}^{n-1}{\mathbf{1}}_E(T^{k}x)$. By [F1], on a conull set $N_E(n,x)\mathrm{/}n\to \mu (E)>0$. Consequently $N_E(n,x)\to \mathrm{\infty }$, so every positive visit number occurs.

For such an $x$ define canonically ${\tau }_m(x)=\min \{r\ge 0:N_E(r+1,x)=m\},m\ge 1.$ Then ${\tau }_m(x)\to \mathrm{\infty }$ and $N_E({\tau }_m(x)+1,x)=m$.

Evaluating the limit in step 1.1 along $n={\tau }_m(x)+1$ gives $\frac{m}{{\tau }_m(x)+1}⟶\mu (E)\mathrm{.}$ Positivity permits reciprocals, so $({\tau }_m(x)+1)\mathrm{/}m\to 1\mathrm{/}\mu (E)$; subtracting $1\mathrm{/}m$ proves ${\tau }_m(x)\mathrm{/}m\to 1\mathrm{/}\mu (E)$.

The result is orbitwise and complements [F2], which integrates the first positive return time over $E$. Full choice is used only through the Birkhoff specialization [F1]; the visit times in step 2.1 are least integers, not chosen from an arbitrary family.

The $k$th iterate of the left shift satisfies $({\sigma }^{k}x{)}_0=x_k$. Therefore $f({\sigma }^{k}x)=x_k$.

Summing step 1.1 for $0\le k<n$ and dividing by $n$ gives $A_{n}f(x)=n^{-1}{\sum }_{k<n}{x}_k$.

Applying [F2] to the right side proves $A_{n}f(x)\to 1\mathrm{/}2$ almost surely. The observable is the cylinder indicator ${\mathbf{1}}_{\{x:x_0=1\}}$ from [F1], so this is the usual heads-frequency strong law written dynamically. Countable choice is inherited from [F2].

The digit-one positions are $1,3,5,\dots$, so the represented value is ${\sum }_{j=0}^{\mathrm{\infty }}{2}^{-(2j+1)}=\frac{1\mathrm{/}2}{1-1\mathrm{/}4}=\frac{2}{3}\mathrm{.}$

The string is not eventually one, so [F1] says it is the canonical binary expansion of $2\mathrm{/}3$.

Adjacent pairs alternate between $10$ and $01$. Thus $00$ occurs zero times among every collection of starting positions, and its limiting frequency is $0$.

Since base-two normality would require frequency $2^{-2}=1\mathrm{/}4$ by [F1], $2\mathrm{/}3$ is not normal in base two.

The strict superlevel sets of $f$ are $[0,1)$ for negative levels, $(0,1)$ at level zero, and $(0,\min \{1,1\mathrm{/}a\})$ at positive level $a$ (with the ambient endpoint omitted). They are Borel, so [F1] makes the everywhere finite $f$ measurable.

On $I_k=[1\mathrm{/}(k+1),1\mathrm{/}k)$ one has $f\ge k$. Thus monotonicity and the simple-integral formula give, for every $N$, $\int fd\lambda \ge {\sum }_{k=1}^{N}k\lambda (I_k)={\sum }_{k=1}^N\frac{1}{k+1}\mathrm{.}$

The last sums are unbounded by [F3]. Hence $\int f=+\mathrm{\infty }$, so $f$ is not integrable in the sense of Integrable real and complex functions, and their integrals.

Put $f_m=\min \{f,m\}$ for $m\ge 1$. These are bounded integrable functions, $f_m\uparrow f$, and [F3] yields $c_m:=\int f_{m}d\lambda \uparrow +\mathrm{\infty }$.

For each $m$, [F4] makes $A_n{f}_m$ converge almost everywhere to a constant. Since the averages are bounded by $m$, [F5] identifies that constant as $c_m$.

Outside the countable union of the exceptional null sets in step 4.1, which is null by [F6], all these limits hold simultaneously. Since $f\ge f_m$, there ${\mathrm{lim\hspace{0.17em}inf}}_n{A}_{n}f\ge {\lim }_n{A}_n{f}_m=c_m$ for every $m$. Letting $m\to \mathrm{\infty }$ gives $A_{n}f\to +\mathrm{\infty }$.

Thus this finite measurable but nonintegrable $f$ is the promised counterexample. Countable choice is inherited from the Lebesgue/doubling suppliers; the truncations are explicit.

Since $T^2$ is the identity, $f\circ T={\mathbf{1}}_{[1\mathrm{/}2,3\mathrm{/}4)}$ and the summands in $A_{n}f$ alternate between $f$ and $f\circ T$.

Put $g=(f+f\circ T)\mathrm{/}2$. For $n=2q$, $A_{n}f=g$ exactly; for $n=2q+1$, $\mathrm{\mid }{A}_{n}f-g\mathrm{\mid }\le 1\mathrm{/}n$ everywhere. Therefore $A_{n}f\to g$ both pointwise and in $L^2$, since the circle has measure one.

The function $g$ is $\frac{1}{2}{\mathbf{1}}_{[0,1\mathrm{/}4)\cup [1\mathrm{/}2,3\mathrm{/}4)}$. The half-rotation interchanges its two support intervals, so $g\circ T=g$; it is nonconstant because [F5] gives positive measure to both its support and complement.

By [F2], the same sequence $A_{n}f$ has $L^2$ limit $P_{M}f$. Uniqueness of limits in the $L^2$ norm and step 2.1 therefore give $P_{M}f=g$. Step 2.1 also strengthens the pointwise almost-everywhere conclusion of [F3] to convergence at every point in this example.

Full choice is used only through the projection theorem [F2], as recorded by [F4]; the period-two computation itself is explicit.