The Ergodic Theorems of von Neumann and Birkhoff

Measurability of $T$ makes $f\circ T$ measurable. If $f=g$ off the measurable null set $N$, then $f\circ T=g\circ T$ off $T^{-1}N$, and $\mu (T^{-1}N)=\mu (N)=0$. Thus $U_T$ is representative independent. Pointwise composition distributes over addition and scalar multiplication, so the descended map is linear over either scalar field.

For $1\le p<\mathrm{\infty }$, integral invariance applied to the nonnegative measurable function $\mathrm{\mid }f{\mathrm{\mid }}^p$ gives $\parallel f\circ T{\parallel }_p^p=\int \mathrm{\mid }f{\mathrm{\mid }}^p\circ Td\mu =\int \mathrm{\mid }f{\mathrm{\mid }}^{p}d\mu =\parallel f{\parallel }_p^p\mathrm{.}$ Taking the nonnegative $p$th root proves equality of the norms.

For $p=\mathrm{\infty }$ and every finite $M\ge 0$, the exceptional set for $\mathrm{\mid }f\circ T\mathrm{\mid }\le M$ is $T^{-1}\{\mathrm{\mid }f\mathrm{\mid }>M\}$. Its measure equals that of $\{\mathrm{\mid }f\mathrm{\mid }>M\}$. Hence $M$ is an essential bound for $f\circ T$ exactly when it is one for $f$, and their infima are equal.

Iterating step 1.1 shows that every $U_T^k[f]=[f\circ T^k]$ is well defined and has norm $\parallel f{\parallel }_p$. Finite linear combinations therefore make $S_{n}f$ and $A_{n}f$ well-defined measurable classes.

The real or complex Minkowski inequality and positive homogeneity now give $\parallel A_{n}f{\parallel }_p\le \frac{1}{n}{\sum }_{k=0}^{n-1}\parallel U_T^{k}f{\parallel }_p=\parallel f{\parallel }_p\mathrm{.}$ This includes $n=1$ and $p=\mathrm{\infty }$, and no inverse of $T$ was used.

For $N\ge 1$, set $F_N:=\max (0,S_{1}f,\dots ,S_{N}f),E_N:=\{F_N>0\}\mathrm{.}$ Every $S_{j}f$ is integrable, so $F_N$ is measurable and integrable because a finite maximum of real functions is obtained from addition and absolute value. Also $F_N\ge 0$ and $F_N=0$ on $X\setminus E_N$.

Since $F_N\ge S_{j}f$ for $0\le j\le N-1$, composition and addition give $F_N\circ T+f\ge S_{j+1}f$. Hence $F_N\circ T+f\ge {\max }_{1\le j\le N}{S}_{j}f=F_N\text{on }{E}_N,$ where strict positivity is what permits insertion of the zeroth sum $S_{0}f=0$.

Integrating the preceding inequality over $E_N$, using $F_N=0$ off $E_N$, nonnegativity of $F_N\circ T$, and invariance of its integral, yields ${\int }_{E_N}fd\mu \ge {\int }_X{F}_{N}d\mu -{\int }_{E_N}{F}_N\circ Td\mu \ge {\int }_X{F}_{N}d\mu -{\int }_X{F}_N\circ Td\mu =0.$ All displayed integrals are finite because $F_N$ is integrable.

The sets $E_N$ increase and their union is $E$. Applying monotone convergence separately to $f^{+}{\mathbf{1}}_{E_N}$ and $f^{-}{\mathbf{1}}_{E_N}$ gives ${\int }_{E_N}fd\mu ⟶{\int }_{E}fd\mu \mathrm{.}$ Passing to the limit in the nonnegative inequalities of step 3.1 proves the claim.

The exact identity $A_{n}f(Tx)=\frac{n+1}{n}{A}_{n+1}f(x)-\frac{1}{n}f(x)$ shows, by taking lower and upper limits, that both limiting envelopes have the same value at $Tx$ as at $x$; finite-valuedness of $f(x)$ makes the last term tend to zero. Thus $T^{-1}{E}_{\alpha ,\beta }=E_{\alpha ,\beta }$.

Assume first that $\alpha >0$, and let $C\subseteq E_{\alpha ,\beta }$ be measurable with $\mu (C)<\mathrm{\infty }$. The function $g=f-\alpha {\mathbf{1}}_C$ is integrable. For $x\in E_{\alpha ,\beta }$, some $n$ satisfies $S_{n}f(x)>n\alpha$, while $S_n{\mathbf{1}}_C(x)\le n$; hence $S_{n}g(x)>0$. Therefore $C$ lies in $G:=\{{\sup }_n{S}_{n}g>0\}$.

By [F1], ${\int }_{G}g\ge 0$. Since $C\subseteq G$, $\alpha \mu (C)\le {\int }_{G}fd\mu \le \mid {\int }_{G}fd\mu \mid \le {\int }_G\mathrm{\mid }f\mathrm{\mid }d\mu \le \parallel f{\parallel }_1\mathrm{.}$ Here the middle absolute-value inequality follows because the left side is nonnegative.

From a sigma-finite cover form the increasing finite-measure exhaustion $X_m$ by finite unions, and take $C_m=E_{\alpha ,\beta }\cap X_m$. Step 2.1 gives $\mu (C_m)\le \parallel f{\parallel }_1\mathrm{/}\alpha$, while $C_m\uparrow E_{\alpha ,\beta }$. Continuity from below, which follows from countable additivity of the measure, gives $\mu (E_{\alpha ,\beta })\le \parallel f{\parallel }_1\mathrm{/}\alpha <\mathrm{\infty }$.

If $\alpha \le 0$, then $-\beta >0$. The same set is the oscillation set for $-f$ with upper threshold $-\beta$ and lower threshold $-\alpha$, because $\mathrm{lim\hspace{0.17em}sup}(-A_{n}f)=-\mathrm{lim\hspace{0.17em}inf}{A}_{n}f$ and $\mathrm{lim\hspace{0.17em}inf}(-A_{n}f)=-\mathrm{lim\hspace{0.17em}sup}{A}_{n}f$. Applying steps 1.2–3.1 to $-f$ proves its measure finite.

Suppose first that $f$ is real. Write $u={\mathrm{lim\hspace{0.17em}sup}}_n{A}_{n}f$ and $\mathrm{\ell }={\mathrm{lim\hspace{0.17em}inf}}_n{A}_{n}f$. The exact identity $A_{n}f(Tx)=\frac{n+1}{n}{A}_{n+1}f(x)-\frac{1}{n}f(x)$ shows that $u\circ T=u$ and $\mathrm{\ell }\circ T=\mathrm{\ell }$, with extended values allowed. The divergence set $\{\mathrm{\ell }<u\}$ is the union of the sets $E_{\alpha ,\beta }$ over rational $\beta <\alpha$.

If $f=g$ a.e., let $N=\{f\mathrm{\ne }g\}$. Outside ${\bigcup }_{k\ge 0}{T}^{-k}N$, a null set by preservation and [F4], every summand in $S_{n}f$ equals the corresponding summand in $S_{n}g$. Thus the limits agree a.e.; the construction descends to the $L^1$ class.

Fix such $\beta <\alpha$ and put $E=E_{\alpha ,\beta }$. By [F1], $E$ is strictly invariant and $\mu (E)<\mathrm{\infty }$. For $g=(f-\alpha ){\mathbf{1}}_E$, strict invariance gives $S_{n}g={\mathbf{1}}_E(S_{n}f-n\alpha )\mathrm{.}$ Every $x\in E$ has $S_{n}f(x)>n\alpha$ for some $n$, so $E=\{{\sup }_n{S}_{n}g>0\}$. Applying [F2] gives ${\int }_{E}fd\mu \ge \alpha \mu (E)\mathrm{.}$

Apply the same argument to $-f$ and the thresholds $-\beta >-\alpha$. Its oscillation set is again $E$, so ${\int }_{E}fd\mu \le \beta \mu (E)\mathrm{.}$ Because $f$ is integrable and $E$ has finite measure, both inequalities are finite. Thus $(\alpha -\beta )\mu (E)\le 0$, and $\mu (E)=0$.

There are only countably many rational pairs. Hence [F4] and steps 1.1–3.1 show that $A_{n}f$ has an extended-real limit off a null set. On that set, $\mathrm{\mid }{f}^{\ast }\mathrm{\mid }={\mathrm{lim\hspace{0.17em}inf}}_n\mathrm{\mid }{A}_{n}f\mathrm{\mid },$ and contractivity gives $\int \mathrm{\mid }{A}_{n}f\mathrm{\mid }\le \int \mathrm{\mid }f\mathrm{\mid }$. Fatou therefore yields $\int \mathrm{\mid }{f}^{\ast }\mathrm{\mid }\le \int \mathrm{\mid }f\mathrm{\mid }<\mathrm{\infty }$, which both makes $f^{\ast }$ finite a.e. and proves integrability.

Define $f^{\ast }=0$ on the exceptional null set. The identity in step 1.1 shows that the convergence set and the limit are invariant wherever the averages converge; after adding the null exceptional orbit set if necessary, $f^{\ast }\circ T=f^{\ast }$ a.e.

For complex $f$, apply the real result to $\mathrm{Re}f$ and $\mathrm{Im}f$ and combine their two conull convergence sets. Their limits give the complex limit, its invariance, and its representative independence. Fatou applied directly to $\mathrm{\mid }{A}_{n}f\mathrm{\mid }$ gives the stated complex $L^1$ bound. Only these two determined components are used, so no choice principle enters.

First let $g$ be real and integrable, and choose its Birkhoff-limit representative $g^{\ast }$ to be the pointwise limit on the convergence set and zero elsewhere. The shifted-average identity and its rearrangement show that the convergence set is strictly invariant and that $g^{\ast }$ is strictly invariant.

Suppose first that $f$ is real. On $(X,\mathcal{I})$ define $\nu (E)={\int }_{E}fd\mu$. This is a finite signed measure by [F3], is absolutely continuous with respect to $\mu {\mathrm{\mid }}_{\mathcal{I}}$, and has finite total variation. By [F4] there is an integrable $\mathcal{I}$-measurable $h$ such that ${\int }_{E}h={\int }_{E}f$ for every $E\in \mathcal{I}$. This is the unique step spending AC.

Fix a positive integer $m$ and, for $k\in \mathbb{Z}$, put $D_{m,k}=\{k\mathrm{/}m\le g^{\ast }<(k+1)\mathrm{/}m\}\mathrm{.}$ These form a measurable, strict-invariant partition of $X$. For $\epsilon >0$, every point of $D_{m,k}$ belongs to the positive-maximal set of $(g-(k\mathrm{/}m-\epsilon )){\mathbf{1}}_{D_{m,k}}$. The maximal theorem and strict invariance therefore give ${\int }_{D_{m,k}}gd\mu \ge (k\mathrm{/}m-\epsilon )\mu (D_{m,k})\mathrm{.}$ Since $\mu (D_{m,k})<\mathrm{\infty }$, letting $\epsilon \downarrow 0$ gives the same inequality with $k\mathrm{/}m$.

On $D_{m,k}$ one has $g^{\ast }<(k+1)\mathrm{/}m$, so step 2.1 gives ${\int }_{D_{m,k}}{g}^{\ast }d\mu \le {\int }_{D_{m,k}}gd\mu +\frac{1}{m}\mu (D_{m,k})\mathrm{.}$ Countable additivity over the partition is legitimate because $g$ and $g^{\ast }$ are integrable. Summing yields ${\int }_X{g}^{\ast }d\mu \le {\int }_{X}gd\mu +\frac{\mu (X)}{m}\mathrm{.}$ Letting $m\to \mathrm{\infty }$ and then applying the same inequality to $-g$, whose limit is $-g^{\ast }$, proves $\int g^{\ast }=\int g$.

Let $E\in \mathcal{I}$ and take $g=f{\mathbf{1}}_E$. Strict invariance gives $A_{n}g=(A_{n}f){\mathbf{1}}_E$ pointwise, so the Birkhoff limit of $g$ is $f^{\ast }{\mathbf{1}}_E$ a.e. In the real case, step 3.1 therefore gives ${\int }_E{f}^{\ast }d\mu ={\int }_{E}fd\mu \mathrm{.}$

Since $h$ is $\mathcal{I}$-measurable, every rational sublevel set of $h$ is strictly invariant; rational separation therefore gives $h\circ T=h$ pointwise. Thus, because $f^{\ast }$ is invariant almost everywhere, $B_r=\{f^{\ast }-h>r\}$ is invariant modulo null sets for every rational $r>0$. Let $E_r\in \mathcal{I}$ be its strict representative from [F2]. Steps 4.1 and 1.2 give ${\int }_{E_r}(f^{\ast }-h)=0$, while $f^{\ast }-h>r$ a.e. on $E_r$. Monotonicity implies $0\ge r\mu (E_r)$, so $E_r$ is null. Applying the same argument to $h-f^{\ast }$ and taking the countable union over positive rational $r$ proves $f^{\ast }=h$ a.e.

For complex $f$, apply steps 1.2–5.1 to its real and imaginary parts and set $h=h_1+i{h}_2$. This $h$ is integrable and $\mathcal{I}$-measurable, equals $f^{\ast }$ a.e., and has all asserted event-integral identities. Uniqueness follows componentwise from Radon–Nikodym uniqueness. If $\mu (X)=0$, the same proof gives the zero density and all assertions are vacuous off a null set.

Assume $p=1$. Put $E_{n,M}=\{\mathrm{\mid }{A}_{n}f\mathrm{\mid }>M\}$. Contractivity and Markov give $\mu (E_{n,M})\le \parallel f{\parallel }_1\mathrm{/}M$. For $K>0$, split $f=f_K+r_K$ by radial clipping, so $\mathrm{\mid }{f}_K\mathrm{\mid }\le K$ and $\mathrm{\mid }{r}_K\mathrm{\mid }=(\mathrm{\mid }f\mathrm{\mid }-K{)}_{+}$. Pointwise, $\mathrm{\mid }{A}_{n}f\mathrm{\mid }\le A_n\mathrm{\mid }f\mathrm{\mid }\le K+A_n\mathrm{\mid }{r}_K\mathrm{\mid }\mathrm{.}$ Consequently ${\int }_{E_{n,M}}\mathrm{\mid }{A}_{n}f\mathrm{\mid }d\mu \le \frac{K\parallel f{\parallel }_1}{M}+\parallel r_K{\parallel }_1,$ where invariance gives $\int A_n\mathrm{\mid }{r}_K\mathrm{\mid }=\parallel r_K{\parallel }_1$.

Now assume $1<p<\mathrm{\infty }$. Let $f_m=f{\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }\le m\}}$. Then $f_m$ is bounded and belongs to $L^p$ by [F4], while dominated convergence applied to $\mathrm{\mid }f-f_m{\mathrm{\mid }}^p$ gives $\parallel f-f_m{\parallel }_p\to 0$.

As $K\to \mathrm{\infty }$, $\mathrm{\mid }{r}_K\mathrm{\mid }\downarrow 0$ and is dominated by $\mathrm{\mid }f\mathrm{\mid }$, so its integral tends to zero. Choose $K$ and then $M$ in step 1.1; the bound is uniform in $n$ and proves uniform integrability of $(A_{n}f)$. For complex $f$, it also proves uniform integrability of the real and imaginary parts because each component modulus is bounded by $\mathrm{\mid }{A}_{n}f\mathrm{\mid }$.

Let $f_m^{\ast }$ be the Birkhoff limit of $f_m$. For fixed $m$, both $\mathrm{\mid }{A}_n{f}_m\mathrm{\mid }$ and $\mathrm{\mid }{f}_m^{\ast }\mathrm{\mid }$ are bounded by $m$. Their pointwise difference tends to zero a.e.; dominated convergence on the finite measure space therefore gives $\parallel A_n{f}_m-f_m^{\ast }{\parallel }_p\to 0$.

By [F1] the averages converge a.e., hence by [F2] in measure. Vitali applied to the real and imaginary parts gives $L^1$ convergence to their corresponding components of $f^{\ast }$. The complex triangle inequality combines the two component conclusions.

Since $A_n(f-f_m)\to f^{\ast }-f_m^{\ast }$ a.e., Fatou and contractivity imply $\parallel f^{\ast }-f_m^{\ast }{\parallel }_p\le {\mathrm{lim\hspace{0.17em}inf}}_n\parallel A_n(f-f_m){\parallel }_p\le \parallel f-f_m{\parallel }_p\mathrm{.}$ Thus $f^{\ast }\in L^p$, and ${\mathrm{lim\hspace{0.17em}sup}}_n\parallel A_{n}f-f^{\ast }{\parallel }_p\le 2\parallel f-f_m{\parallel }_p\mathrm{.}$ Letting $m\to \mathrm{\infty }$ proves the claim. The two cases exhaust $1\le p<\mathrm{\infty }$; the proof never supplies uniform-norm convergence.

Normalize the measure to $\widehat{\mu }=\mu \mathrm{/}\mu (X)$. This does not change measurable sets, null sets, invariance, or ergodicity, and $T$ preserves $\widehat{\mu }$. By [F1], $A_{n}f\to f^{\ast }$ a.e. and $f^{\ast }\circ T=f^{\ast }$ a.e.

Applying [F3] to the normalized probability system makes $f^{\ast }=c$ a.e. for some scalar $c$. The event identity of [F2] at $E=X$ gives $c\mu (X)={\int }_X{f}^{\ast }d\mu ={\int }_{X}fd\mu ,$ so $c=\mu (X{)}^{-1}{\int }_{X}fd\mu$. This is where the AC-dependent Radon–Nikodym identification is used.

The case $p=1$ of [F4] gives $\parallel A_{n}f-f^{\ast }{\parallel }_1\to 0$. Combining with step 2.1 proves both modes of convergence. For complex $f$, [F3] and the integral identity apply to its two components, producing the same complex constant formula.

By [F1], $U_T$ is a linear isometry of complex $L^2$. Its fixed vectors are precisely the classes $g$ with $g\circ T=g$ a.e., namely $\mathcal{M}$ by [F3].

The operator averages in [F2] satisfy $\frac{1}{n}{\sum }_{k=0}^{n-1}{U}_T^{k}f=\frac{1}{n}{\sum }_{k=0}^{n-1}f\circ T^k=A_{n}f\mathrm{.}$ Applying [F2] therefore gives $A_{n}f\to P_{\mathcal{M}}f$ in $L^2$. AC is spent exactly in the published projection/Cesaro supplier. That supplier treats a nonsurjective isometry through its closed range, so no inverse of $T$ or of $U_T$ on all of $L^2$ has been assumed.

This argument used neither the pointwise Birkhoff theorem nor any finite-measure or ergodicity hypothesis. In particular its norm conclusion alone makes no assertion about pointwise convergence of the full sequence.

Let $F\subseteq K$ be closed and nonempty. By [F3], its distance function is continuous, vanishes on $F$, and is strictly positive off $F$. For $m\ge 1$, set ${\varphi }_m(x)=\max \{0,1-md(x,F)\}\mathrm{.}$ Then $0\le {\varphi }_m\le 1$, every ${\varphi }_m$ is continuous, and ${\varphi }_m\downarrow {\mathbf{1}}_F$ pointwise.

The assumed continuous-test identity applies to each ${\varphi }_m$. Dominated convergence for each probability measure gives $\mu (F)={\lim }_m\int {\varphi }_{m}d\mu ={\lim }_m\int {\varphi }_{m}d\nu =\nu (F)\mathrm{.}$ For $F=\mathrm{\varnothing }$ the same equality is immediate.

Closed subsets of $K$ form a pi-system containing $K$, and their complements are the open sets, so they generate the Borel sigma-algebra. The two probabilities have equal total mass one and agree on every closed set by step 2.1. Applying [F2] proves $\mu =\nu$.

Assume unique ergodicity with invariant probability $\mu$. If uniform convergence to $\int fd\mu$ failed for some continuous $f$, countable choice would give $\epsilon >0$, strictly increasing $n_j$, and $x_j\in K$ such that $\mid A_{n_j}f(x_j)-\int fd\mu \mid \ge \epsilon \mathrm{.}$ Define the empirical Borel probabilities ${\eta }_j=n_j^{-1}{\sum }_{k<n_j}{\delta }_{T^k{x}_j}$; finite additivity and countable additivity of each point mass make this a probability.

By [F1], pass to a subsequence, not relabelled, and a Borel probability $\eta$ such that $\int gd{\eta }_j\to \int gd\eta$ for every $g\in C(K,\mathbb{R})$. For such $g$, $\int g\circ Td{\eta }_j-\int gd{\eta }_j=\frac{g(T^{n_j}{x}_j)-g(x_j)}{n_j}⟶0,$ because $g$ is bounded by [F4]. Taking limits shows that $\eta$ and its pullback probability $E\mapsto \eta (T^{-1}E)$ have identical continuous test integrals; [F2] makes them equal. Hence $\eta$ is invariant.

Unique ergodicity gives $\eta =\mu$, but $\int fd{\eta }_j=A_{n_j}f(x_j)$ and the closed inequality in step 1.1 passes to the limit, contradicting $\int fd\eta =\int fd\mu$. Thus $A_{n}f\to \int fd\mu$ uniformly for every $f$.

Conversely, assume all continuous averages converge uniformly to constants $c(f)$. Fix $x_0\in K$ and form ${\eta }_n=n^{-1}{\sum }_{k<n}{\delta }_{T^k{x}_0}$. By [F1], a subsequence has a continuous-test limit probability $\eta$. The telescoping calculation of step 2.1 makes $\eta$ invariant, and uniform convergence gives $\int fd\eta ={\lim }_r{A}_{n_r}f(x_0)=c(f)\mathrm{.}$

If $\rho$ is any invariant Borel probability, then [F3] gives $\int A_{n}fd\rho =\int fd\rho$. Moreover [F5] gives $\mid \int A_{n}fd\rho -c(f)\mid \le \parallel A_{n}f-c(f){\parallel }_{\mathrm{\infty }}⟶0.$ Hence $\int fd\rho =c(f)=\int fd\eta$ for every continuous real $f$. By [F2], $\rho =\eta$. Thus an invariant probability exists and is unique, and its integral is the asserted constant.

Steps 1.1–3.1 prove the forward implication and steps 3.2–4.1 prove the converse. Countable choice is used precisely through [F1] and to select the failure witnesses in step 1.1; no stronger choice principle is invoked.

Define $q:[0,1]\to \mathbb{T}$ by $q(t)=t$ for $t<1$ and $q(1)=0$. Equivalently $q(t)=\{t\}$. For $s,t\in [0,1]$, the circle-distance formula gives $d(q(s),q(t))\le \mathrm{\mid }s-t\mathrm{\mid },$ including when one endpoint is $1$. Thus $q$ is continuous. It is surjective because every $x\in [0,1)$ equals $q(x)$.

By [F1], $[0,1]$ is compact. Its continuous image under $q$ is all of $\mathbb{T}$, so [F2] makes the circle compact; [F3] reads this in the metric sense. Finally $0\in \mathbb{T}$, so it is nonempty.

Fix $f\in C(\mathbb{T},\mathbb{R})$. Compactness makes $f$ bounded, hence $f\in L^1(\lambda )$. By [F2]–[F3], $A_{n}f$ converges on a conull set $Y_f$ to a constant $c_f$. Since $\mathrm{\mid }{A}_{n}f\mathrm{\mid }\le \parallel f{\parallel }_{\mathrm{\infty }}$, dominated convergence and invariance give $c_f=\int c_{f}d\lambda ={\lim }_n\int A_{n}fd\lambda =\int fd\lambda \mathrm{.}$

Given $\epsilon >0$, uniform continuity of $f$ gives $\delta >0$ such that $d(x,y)<\delta$ implies $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }<\epsilon \mathrm{/}3$. Rotations are isometries, so the same $\delta$ gives $\mathrm{\mid }{A}_{n}f(x)-A_{n}f(y)\mathrm{\mid }<\epsilon \mathrm{/}3$ for every $n$ whenever $d(x,y)<\delta$.

The set $Y_f$ is dense. Indeed, every nonempty circle-open set contains a nondegenerate ordinary interval, possibly on one side of the cut, and such an interval has positive Lebesgue measure by the interval-measure formula. A conull set must meet it.

Compactness supplies a finite $\delta \mathrm{/}3$-net $x_1,\dots ,x_m$. By density choose, using only finite choice, $y_i\in Y_f$ with $d(x_i,y_i)<\delta \mathrm{/}3$. For all sufficiently large $n$, every $\mathrm{\mid }{A}_{n}f(y_i)-c_f\mathrm{\mid }<\epsilon \mathrm{/}3$. Given $x$, choose $i$ with $d(x,x_i)<\delta \mathrm{/}3$; then $d(x,y_i)<2\delta \mathrm{/}3$, and step 1.2 gives $\mathrm{\mid }{A}_{n}f(x)-c_f\mathrm{\mid }<2\epsilon \mathrm{/}3$. Thus $A_{n}f\to c_f$ uniformly.

The argument applies to every real continuous $f$, with $c_f=\int fd\lambda$. By [F5], $R_{\alpha }$ is uniquely ergodic and its unique invariant probability is the already invariant $\lambda$. Countable choice is inherited from [F2] and [F5]; the finite net and finite choices in step 3.1 require no stronger principle. No Fourier series or Weyl criterion was used.

If $a=c$, use the zero function; if $(a,c)=(0,1)$, use the constant-one function. Otherwise, for every sufficiently small $\delta >0$, circular distance gives continuous functions $0\le {\mathrm{\ell }}_{\delta }\le {\mathbf{1}}_I\le u_{\delta }\le 1$ as follows: ${\mathrm{\ell }}_{\delta }$ is zero on the closed complementary arc and rises linearly to one within distance $\delta$ inside $I$, while $u_{\delta }$ is one on the closed arc $[a,c]$ and falls linearly to zero within distance $\delta$ outside it. Thus they differ from ${\mathbf{1}}_I$ only in the two boundary arcs of total length at most $4\delta$.