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For a measure-preserving transformation $T$ and $f\in L^1(\mu )$, the ergodic averagesconverge almost everywhere to $\mathbb{E}[f\mid \mathcal{I}]$, the conditional expectation on the invariant sigma-algebra. The limit has the same integral as $f$.

The standard three-cut Chacon transformation is an invertible measure-preserving transformation built by cutting and stacking with spacer vector $(0,1,0)$ at every stage. Cut the current tower into three equal subcolumns, place one spacer above the middle, and stack left to right. With initial width $2/3$ and height one on a unit interval, level counts and widths are $h_j=(3^{j+1}-1)/2$ and $w_j=2/3^{j+1}$. Tower measure tends to one, using total spacer measure $1/3$. Tower words obey $W_{j+1}=W_j{W}_{j}1W_j$.

Cutting and stacking constructs measure-preserving transformations from towers of equal-measure levels. Cut a tower into equal-width subcolumns, add new spacer levels, and stack the resulting columns. Map each level to the one above by a measure-preserving identification; subsequent stages extend the partial map. When uncovered and top-level measures tend to zero, the limiting map is defined modulo null sets.

A probability measure-preserving transformation is strong mixing exactly when $\mu (T^{-n}A\cap A)\to \mu (A{)}^2$ for every measurable $A$. The converse follows by applying decay of autocorrelation implies weak convergence of an observable to $f=1_A$ and then testing against $1_B$. Diagonal polarization alone controls a sum of the two directed cross-correlations and does not suffice as a proof.

A measure-preserving transformation is ergodic when every set in its invariant sigma-algebra has measure zero or full measure. Equivalently, every integrable invariant function is almost everywhere constant.

Ergodic theory studies the long-time statistical behaviour of measure-preserving transformations.

Let $T$ be a measure-preserving transformation of a probability space and $X$ an integrable random variable. If the displayed duality holds for every bounded random variable $Z$, test it on indicator functions to identify $\mathbb{E}[X\mid T^{-1}(\mathcal{F})]=X\circ T$. The measure-preserving transformation gives the same probability distribution to $X$ and $X\circ T$. The principle of conditional expectation preserving a distribution then forces equality almost surely. This does not require an inverse for $T$ or require $X$ to be square-integrable.

The join is the common refinement of two measurable partitions, with atoms given by their nonempty intersections. For a measure-preserving transformation, the block partition is ${\xi }_0^{n-1}={\bigvee }_{j=0}^{n-1}{T}^{-j}\xi$.

For a measure-preserving transformation and real integrable $f$, let $S_{k}f={\sum }_{j<k}f\circ T^j$. For $E_N=\{{\max }_{1\le k\le N}{S}_{k}f>0\}$ the lemma asserts ${\int }_{E_N}fd\mu \ge 0$. Passing to the increasing union gives the corresponding infinite-supremum inequality.

A measure-preserving system is a measure space $(X,\mathcal{B},\mu )$ together with a measure-preserving transformation $T$. In ergodic theory the measure is commonly a probability measure.

Let $(X,\mathcal{B},\mu )$ be a probability space, let $T$ be a measure-preserving transformation, and let $g\in L^1(\mu )$. The Birkhoff ergodic theorem asserts thatfor almost every $x$, where $\mathcal{I}=\{B\in \mathcal{B}:\mu (T^{-1}B△B)=0\}$ is the invariant sigma-algebra. The limit is integrable, satisfies $g^{\ast }\circ T=g^{\ast }$ almost everywhere, and has $\int g^{\ast }d\mu =\int gd\mu$. On a probability space the convergence also holds in $L^1$. No invertibility of $T$ is required.

If $T$ is ergodic, its invariant sigma-algebra is trivial modulo null sets, so its conditional expectation is constant. In that case the concise form is

Outside the countable set of dyadic rational numbers, the binary expansion $x={\sum }_{j\ge 1}{ϵ}_j(x)2^{-j}$ is unique. Multiplication by two modulo one deletes the first binary digit, soLet $g=1_{[1/2,1)}$. By part (b), the doubling map is a measure-preserving transformation and is ergodic for Lebesgue measure. Applying the Birkhoff ergodic theorem to this indicator function givesfor almost every $x$. The dyadic rationals have Lebesgue measure zero, so the choice of their two possible binary expansions does not affect the conclusion. Thus the limiting frequency of the digit one is $1/2$ almost everywhere.

A measurable set is invariant when ${\theta }^{-1}A=A$, or modulo a null set in the measure-theoretic version. A measurable function is invariant when $f\circ \theta =f$, similarly almost everywhere if appropriate. The map is a measure-preserving transformation when $\mu ({\theta }^{-1}A)=\mu (A)$ for every measurable set. A measure-preserving map is ergodic when every invariant measurable set has measure zero or has null complement. On a finite positive measure space, normalize the measure to obtain the equivalent probability-space definition.

For a probability measure-preserving system, ergodicity means that every measurable set $A$ satisfying $T^{-1}A=A$ modulo null sets has measure zero or one. Equivalently, the only invariant functions are almost everywhere constant. The pointwise ergodic theorem states that, for $f\in L^1(\mu )$,where $\mathcal{I}$ is the invariant sigma-algebra; on a probability space the convergence also holds in $L^1$. For an ergodic transformation, the limit is the constant $\int fd\mu$.

For the Gauss continued-fraction map, the Gauss measure is the probability measureIts density integrates to one and is bounded above and below by positive constants, so it has the same null sets as Lebesgue measure. Apart from countably many endpoints, the inverse branches are $h_a(y)=1/(a+y)$, $a=1,2,\dots$. For a bounded measurable test function $F$, substitution on each branch givesThus $T$ is a measure-preserving transformation. Its definition at zero and the rational branch endpoints does not affect these identities.

For ergodicity, we give the full shrinking-cylinder argument. A digit cylinder with first $n$ partial quotients $a_1,\dots ,a_n$ is an interval $I=h((0,1))$, where $h=h_{a_1}\circ \cdots \circ h_{a_n}$ and $T^n\circ h$ is the identity. The continued fraction convergents giveHere $q_n\ge q_{n-1}>0$ after the initial step, and the recurrence $q_n=a_n{q}_{n-1}+q_{n-2}$ makes $q_n$ tend to infinity at least at the Fibonacci rate. Hence the nested cylinders of every irrational point shrink to that point. The bounded distortion of continued-fraction cylinders is uniform: the ratio of the largest to the smallest value of $|h^{\prime }|$ is at most four. Since $1/(1+h(y))$ ranges between $1/2$ and $1$, the densityhas maximum-to-minimum ratio at most eight. Consequently every measurable $E\subseteq (0,1)$ satisfiesThe last inequality uses $d\nu /dx\le 1/\log 2$.

Now suppose $T^{-1}A=A$ modulo null sets and $\nu (A)>0$. Choose an irrational density point $x\in A$, and let $I_n$ be its successive digit cylinders. The Lebesgue density theorem applies to these intervals even though they need not be centered: an interval of length $\ell$ containing $x$ lies in the centered interval of radius $\ell$, so its relative complement tends to zero. Comparability of the two densities then givesBut invariance gives $1_A(h(y))=1_A(y)$ almost everywhere for each inverse branch of $T^n$. Pulling the exceptional null sets through $h$ is legitimate since $h$ has a positive smooth Jacobian on this fixed branch. Thus the distortion estimate with $E=A^c$ givesLetting $n\to \infty$ forces $\nu (A^c)=0$. Therefore the Gauss continued-fraction map is ergodic for the Gauss measure.

For an irrational $x$, the successive continued fraction digits satisfy $a_j(x)=\lfloor 1/T^{j-1}x\rfloor$. The event that a digit equals two is the interval $(1/3,1/2]$, up to null endpoints. Apply the pointwise ergodic theorem to its indicator:This holds for almost every $x$ with respect to both the Gauss measure and Lebesgue measure, since they have the same null sets.

A measure-preserving transformation is a measurable map $T:E\to E$ satisfying $\mu (T^{-1}A)=\mu (A)$ for every measurable $A$. An invariant set satisfies $T^{-1}A=A$ (or equality modulo null sets, in the measure-theoretic convention).

Preimages preserve complements and countable unions. Therefore the whole space is invariant, the complement of an invariant set is invariant, and a countable union of invariant sets is invariant. Thus the invariant sets form a $\sigma$-algebra. The same proof works modulo null sets because a countable union of null exceptional sets is null.

Work with a probability measure-preserving system, so $\mu (X)=1$. PutThe maximal ergodic theorem, also called the maximal ergodic lemma, says that for real $f\in L^1(\mu )$,No ergodic transformation or invertibility assumption is needed for this maximal ergodic lemma.

Here is a proof of the maximal ergodic lemma. Set $M_0=0$ and $M_N=\max (0,S_{1}f,\dots ,S_{N}f)$. These are integrable: $0\le M_N\le {\sum }_{j=0}^{N-1}|f|\circ T^j$. Splitting off the first summand givesOn $E_N=\{M_N>0\}$, this implies $f=M_N-M_{N-1}\circ T\ge M_N-M_N\circ T$. Outside $E_N$ the right side is $-M_N\circ T\le 0$. ConsequentlyIntegrating and using the measure-preserving transformation property gives ${\int }_{E_N}fd\mu \ge 0$. Since $E_N$ increases to $E$, the dominated convergence theorem, with dominating function $|f|$, proves the infinite-supremum assertion of the maximal ergodic lemma.

For example, applying the maximal ergodic lemma to $|f|-\epsilon$ yields the useful maximal boundIndeed on the latter set $G$, the maximal ergodic lemma gives $\epsilon \mu (G)\le {\int }_G|f|d\mu$.

The pointwise ergodic theorem states that for a probability measure-preserving system and $f\in L^1(\mu )$,where $\mathcal{I}$ is the invariant sigma-algebra. The conditional expectation is invariant and has integral $\int fd\mu$. If $T$ is an ergodic transformation, this conditional expectation is the constant $\int fd\mu$. The assertion applies to real or complex integrable functions; it is the Birkhoff ergodic theorem.

For a nonnegative measurable $f$, if $\int fd\mu <\infty$, the pointwise ergodic theorem already proves the required conclusion. For the remaining case, let $f_K=\min (f,K)$, $K=1,2,\dots$. Each $f_K$ is integrable on the probability measure-preserving system. The pointwise ergodic theorem and ergodic transformation property give, on a single set of full measure obtained by a countable intersection,Since $f\ge f_K$, every such $x$ satisfies ${\mathrm{liminf}}_n{A}_{n}f(x)\ge \int f_{K}d\mu$ for all $K$. The monotone convergence theorem gives $\int f_{K}d\mu \uparrow \int fd\mu =+\infty$. Thus the nonnegative ergodic averages with infinite integral tend to $+\infty$, rather than merely having an unbounded subsequence. Combining the two cases,Nonnegativity is what makes truncation a lower bound. The probability normalization is also essential to the displayed value: for a finite measure of total mass $c>0$, the corresponding constant is $c^{-1}\int fd\mu$.

For a probability measure-preserving system, strong mixing means that for every pair of measurable sets $A,B$,A sequence $a_n$ has convergence in density of a sequence to $a$ when for each $\epsilon >0$ the exceptional set $\{n:|a_n-a|\ge \epsilon \}$ has natural density zero. A weakly mixing measure-preserving transformation has convergence in density of $\mu (T^{-n}A\cap B)$ to $\mu (A)\mu (B)$ for every $A,B$. Equivalently, because these correlations are bounded,Indeed the mean of the absolute discrepancy is at least $\epsilon$ times the exceptional frequency, while it is at most $\epsilon$ plus a uniform bound times that frequency. A signed Cesaro convergence of a sequence without absolute values is insufficient to define weak mixing.

The standard three-cut Chacon map is obtained by cutting and stacking. Start on $[0,1)$ with normalized Lebesgue measure, an initial tower consisting of $[0,2/3)$, and a reservoir $[2/3,1)$ for spacers. At stage $j$, cut every level of the current tower into three equal subintervals, producing three subcolumns. Add one new interval of the same width above the middle subcolumn. Stack the first subcolumn at the bottom, then the middle subcolumn with its spacer, then the third at the top. Define the partial transformation by translation from each level to the next, leaving the current top unmapped. These assignments extend the earlier partial transformation. Repeat indefinitely.

If $h_j$ is the number of levels and $w_j$ their width, thenHence $h_j=(3^{j+1}-1)/2$, $w_j=2/3^{j+1}$, and the stage-$j$ tower has measure $1-3^{-j-1}$. The spacers consume total measure ${\sum }_{j\ge 0}{w}_{j+1}=1/3$, precisely the reservoir. The increasing partial maps give the Chacon map modulo null sets. The tower tops and unused reservoir have measures tending to zero, and the construction yields an invertible measure-preserving transformation. With $0$ marking an original level and $1$ a spacer, the tower words satisfy $W_0=0$ and $W_{j+1}=W_j{W}_{j}1W_j$, so the first new word is $0010$. This describes the spacer placement without needing a proof of well-definedness. The three-cut convention agrees with the classical constant spacer vector $(0,1,0)$ described in Ryzhikov's construction.

For the correlation assertion, take the L2 inner product to be $⟨u,v⟩=\int u\overline{v}d\mu$, linear in its first argument, and put $U=U_T$. The Koopman operator is an isometry on $L^2$, even when $T$ is not invertible. For $n\ge k$ the hint givesTo extend rigorously to all test functions, centre the observable: $v=f-a1$. Invariance of the integral gives $⟨U^{n}v,v⟩=⟨U^{n}f,f⟩-|a{|}^2\to 0$. LetFor each fixed $k$, $⟨U^{n}v,U^{k}v⟩\to 0$ by the same isometry identity. Therefore the limit is zero for every finite linear combination of these orbit vectors. By the Cauchy-Schwarz inequality and $\Vert U^{n}v{\Vert }_2=\Vert v{\Vert }_2$, approximation extends this conclusion to every $g\in M$: the approximation error in the correlation is bounded by $\Vert v{\Vert }_2\Vert g-g_0{\Vert }_2$ uniformly in $n$. For $g\in M^{\perp }$, the correlation is identically zero because $U^{n}v\in M$. The orthogonal decomposition by a closed subspace now gives the result for all $g\in L^2$. ThusThis is decay of autocorrelation implies weak convergence of an observable; no assumption that $T$ is an ergodic transformation, and no invertibility hypothesis was used, and $a=0$ is included.

Finally, strong mixing immediately implies the stated diagonal limit by taking $B=A$. Conversely, suppose that limit holds for every $A$. Apply the correlation result with $f=1_A$ and $g=1_B$. Its hypothesis is exactly $⟨U^n{1}_A,1_A⟩\to \mu (A{)}^2$, and its conclusion isHence strong mixing is equivalent to diagonal set-correlation criterion for mixing. Using the orbit-span proof avoids an invalid polarization argument that would recover only the sum of the two directed cross-correlations from diagonal correlations.

The Birkhoff ergodic theorem states: for a measure-preserving transformation $\theta$ of a finite measure space and $f\in L^1$, the averages $A_{n}f=n^{-1}{\sum }_{j=0}^{n-1}f\circ {\theta }^j$ converge almost everywhere and in $L^1$ to an invariant function $f^{\ast }$. On a probability space,where $\mathcal{I}$ is the invariant sigma-algebra modulo null sets. In particular the integral is preserved, and if $\theta$ is ergodic then $f^{\ast }=\int fd\mu$ almost everywhere; for a finite space of nonunit mass, this constant is $\mu (E{)}^{-1}\int fd\mu$ when $\mu (E)>0$.

The maximal ergodic lemma says that for real integrable $f$, with $S_{k}f={\sum }_{j=0}^{k-1}f\circ {\theta }^j$ and $E_N=\{{\max }_{1\le k\le N}{S}_{k}f>0\}$,The same inequality holds on $E_{\infty }=\{{\sup }_{k\ge 1}{S}_{k}f>0\}$, by taking the increasing limit and dominated convergence. These statements do not require the transformation to be invertible or the space to be ergodic.

Use the probability-system convention $\mu (X)=1$. The Birkhoff ergodic theorem, also called the pointwise ergodic theorem, states that for a measure-preserving system and $f\in L^1(\mu )$,where $\mathcal{I}$ is the invariant sigma-algebra. The limit is integrable and has the same integral as $f$. On a probability space the convergence also holds in $L^1$, as in the allowed mean ergodic theorem. If $T$ is an ergodic transformation, $\mathcal{I}$ is trivial modulo null sets, giving

The integer multiplication map on the circle preserves Lebesgue measure: for any integrable $g$ on $[0,1)$,To prove the ergodicity of integer multiplication on the circle, suppose $u\in L^2(m)$ satisfies $u\circ T_K=u$. Let $c_j$ be its Fourier coefficients in the Fourier basis $e_j(x)=e^{2\pi ijx}$. Since $e_j\circ T_K=e_{Kj}$, the Fourier coefficients of $u\circ T_K$ at index $j$ are zero if $K$ does not divide $j$, and are $c_{j/K}$ otherwise. This identity holds for all $L^2$ functions by approximation with trigonometric polynomials and the isometry $u\mapsto u\circ T_K$. Invariance givesEvery nonzero integer $j$ can be divided by $K$ only finitely often. Thus $c_j=0$ for every $j\ne 0$, and completeness of the Fourier basis makes $u$ constant almost everywhere. Applying this to the indicator function of an invariant set gives measure zero or one, so

A normal number in base $K$ has every word $w=(w_1,\dots ,w_r)$ of $r$ base-$K$ digits occurring with limiting overlapping frequency $K^{-r}$. Use the expansion that is not eventually equal to $K-1$ when there are two expansions. The word $w$ corresponds to the half-open intervalA word starting at position $n+1$ occurs exactly when $T_K^{n}x\in I_w$. The Birkhoff ergodic theorem, applied to $1_{I_w}$, gives frequency $m(I_w)=K^{-r}$ almost everywhere. There are countably many pairs $(K,w)$, so their full-measure sets have a full-measure intersection. In particular,These are absolutely normal numbers, so existence follows as well. This interval description also proves normality and equidistribution under integer multiplication: the base-$K$ intervals form arbitrarily fine grids, so their frequencies imply the correct frequency for every interval by approximation from inside and outside.

For the growth assertion, put $g=|f|$. For every $\epsilon >0$, the Tonelli theorem gives the useful summability boundSince $T$ is a measure-preserving transformation, $\mu (g\circ T^n>\epsilon n)=\mu (g>\epsilon n)$. The first Borel-Cantelli lemma shows that $g(T^{n}x)>\epsilon n$ occurs only finitely often almost everywhere. Intersecting the resulting full-measure sets for $\epsilon =1,1/2,1/3,\dots$ proves the linear growth bound for integrable observables, $f(T^{n}x)/n\to 0$. Multiplication by $n^{1-a}\le 1$ then gives

The threshold is sharp. For $0<a<1$, choose $p$ with $1<p<1/a$, and take the Bernoulli shift on $X=(0,1{)}^{\mathbb{Z}}$ with the product measure of independent uniform coordinates. The left shift preserves that measure because it preserves the probability of every finite-coordinate event. Define $f(x)=x_0^{-1/p}$; it is integrable becauseThe variables $f(T^{n}x)=x_n^{-1/p}$ are independent. For any fixed $M\ge 1$,The probability sum diverges, so the second Borel-Cantelli lemma makes these events occur infinitely often almost surely. Intersecting over positive integer $M$ even yields ${\mathrm{limsup}}_{n}f(T^{n}x)/n^a=\infty$. For $a\le 0$, the constant observable $f=1$ already fails to give limit zero. Thus the sharpness of the linear growth bound for integrable observables gives

A probability measure $\mu$ is $T$-invariant whenEquivalently, the pushforward measure satisfies $T_{\ast }\mu =\mu$, so $T$ is a measure-preserving transformation.