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For independent and identically distributed random variables $X_j$ that are integrable random variables, with $S_n={\sum }_{j=1}^n{X}_j$, symmetry gives $\mathbb{E}[X_1\mid S_n,S_{n+1},\dots ]=S_n/n$. The decreasing sigma-algebras generated by these future sums permit application of the reverse martingale convergence theorem.

For an infinite sequence of real exchangeable random variables, conditional on its invariant sigma-algebra of an exchangeable sequence, the coordinates are independent and identically distributed. In particular, for Borel sets $A_0,\dots ,A_p$,Here is the core proof. Let ${\mathcal{S}}_m$ denote invariance under permutations of coordinates $0,\dots ,m$. Symmetrization as conditional expectation identifies the empirical average of each indicator with its conditional expectation on ${\mathcal{S}}_m$. The reverse martingale convergence theorem gives its limit conditional on $\mathcal{S}$. Multiplying the averages sums over all ordered coordinate tuples, including repetitions. Symmetrization of the product itself sums over the distinct tuples. The fraction of repeated tuples is at most $\left(\genfrac{}{}{0ex}{}{p+1}{2}\right)/(m+1)$, so these bounded averages have the same limit. This proves the displayed factorization. Permutation invariance also makes the individual conditional laws identical. A regular conditional distribution on the real line packages that common law as a random probability measure, giving the equivalent mixture-of-iid formulation.

For a real sequence of exchangeable random variables, every event $A$ in its invariant sigma-algebra of an exchangeable sequence agrees almost surely with an event of its tail sigma-algebra. Put $F=1_A$. Approximate $F$ in $L^1$ by its conditional expectation on the first $n+1$ coordinates. Swap that block with the next $n+1$ coordinates. Invariance of $A$ and the law shows that the conditional expectation on the second block is an equally good approximation. It is measurable with respect to the tail after index $n$. The L1 contraction of conditional expectation then bounds $\Vert F-\mathbb{E}[F\mid {\mathcal{G}}_n]{\Vert }_1$ by twice the approximation error. The reverse martingale convergence theorem gives $F=\mathbb{E}[F\mid {\bigcap }_n{\mathcal{G}}_n]$ almost surely. Thresholding the latter at $1/2$ supplies the required tail event. Together with the reverse inclusion, the two sigma-algebras have identical probability completions.

Here $F=1_A$ is bounded and hence lies in $L^2$ as well as $L^1$. Conditional expectation on a sub-sigma-algebra is the orthogonal projection in $L^2$ onto its measurable subspace. We can use these projections to justify both limits without any symmetry assumption on the probability measure.

The increasing coordinate sigma-algebras ${\mathcal{E}}_n$ generate the whole product sigma-algebra $\mathcal{F}$. Functions depending on finitely many coordinates are dense in $L^2(\mathcal{F})$: their cylinder indicators generate $\mathcal{F}$, and the usual monotone-class approximation proves density of their linear span. If a cylinder function $V$ is ${\mathcal{E}}_N$-measurable, then for $n\ge N$ the $L^2$ projection contraction givesApproximate $F$ arbitrarily well by such $V$ to prove the first $L^2$ limit.

For the decreasing tail sigma-algebras ${\mathcal{G}}_n$, put $Y_n=\mathbb{E}[F\mid {\mathcal{G}}_n]$. If $m\ge n$, the tower property gives $\mathbb{E}[Y_n\mid {\mathcal{G}}_m]=Y_m$. Projection orthogonality therefore yieldsThe norms on the right decrease and are nonnegative, so $(Y_n)$ is Cauchy in $L^2$ and has a limit $Y$. Choose an almost-surely convergent subsequence and take its pointwise upper limit. For every fixed $n$ its tail consists of ${\mathcal{G}}_n$-measurable functions, so this supplies a version of $Y$ measurable with respect to every ${\mathcal{G}}_n$, hence to $\mathcal{G}={\bigcap }_n{\mathcal{G}}_n$. For any $C\in \mathcal{G}$, $\mathbb{E}[Y_n{1}_C]=\mathbb{E}[F{1}_C]$; passing to the limit identifies $Y=\mathbb{E}[F\mid \mathcal{G}]$.

Finally $\Vert W{\Vert }_1\le \Vert W{\Vert }_2$ on a probability space, by the Cauchy-Schwarz inequality. ThusThese are the increasing-filtration and reverse martingale convergence theorem limits; the argument above establishes the requested norm convergence directly.

Let $T_{\pi }$ permute coordinates $0,\dots ,m$, fixing the later coordinates. For any integrable random variable $V$, its group averageis invariant under this group, hence ${\mathcal{S}}_m$-measurable. If $C\in {\mathcal{S}}_m$, invariance of $C$ and measure preservation of each $T_{\pi }$ give $\mathbb{E}[1_{C}V\circ T_{\pi }]=\mathbb{E}[1_{C}V]$. Thus $A_{m}V$ satisfies the integral definition of $\mathbb{E}[V\mid {\mathcal{S}}_m]$. This proves symmetrization as conditional expectation.

Apply it to $V=1_{A_k}(X_k)$ for $m\ge k$. Under a uniform permutation of $m+1$ coordinates, its selected coordinate is uniform among all of them, soThe reverse martingale convergence theorem says that, for decreasing sigma-algebras ${\mathcal{H}}_m$ and an integrable $V$, these conditional expectations converge almost surely and in $L^1$ to $\mathbb{E}[V\mid {\bigcap }_m{\mathcal{H}}_m]$. Here ${\mathcal{S}}_m\downarrow \mathcal{S}$, because every finitely supported permutation belongs to one of the finite groups. ThereforeIntersect these probability-one events for the finitely many $k=0,\dots ,p$ and multiply the limits. Expanding the product of finite sums givesThe sum here includes repeated indices. No independence of the original coordinates was assumed; exchangeability and reverse conditional-expectation convergence are precisely what replaced it.

Let $U_m$ be the average over all $d=p+1$ coordinate tuples in part (a), and $V_m$ the distinct-tuple average in part (b). Every product summand lies in $[0,1]$. A proportion ${\alpha }_m=(N{)}_d/N^d$, where $N=m+1$, of all tuples have distinct coordinates. Thuswhere $W_m\in [0,1]$ is the average over the remaining tuples. If there are no remaining tuples, the error is zero. Hence $|U_m-V_m|\le 1-{\alpha }_m$.

For $d$ independent uniform choices from $N$ indices, any particular pair coincides with probability $1/N$. A union bound over the $\left(\genfrac{}{}{0ex}{}{d}{2}\right)$ pairs givesThis deterministic sampling with and without replacement comparison for bounded products is the needed link between the two averages.

Part (a) identifies the almost-sure limit of $U_m$ as ${\prod }_{k=0}^p\mathbb{E}[1_{A_k}(X_k)\mid \mathcal{S}]$. By part (b) and the reverse martingale convergence theorem, $V_m$ converges almost surely to $\mathbb{E}[{\prod }_{k=0}^p{1}_{A_k}(X_k)\mid \mathcal{S}]$. Their difference vanishes, soThis is precisely conditional independence. Exchangeability also makes the single-coordinate conditional laws identical, by applying a coordinate transposition and testing against invariant events. The result is the De Finetti theorem, established here by finite symmetrization and the vanishing collision fraction.

The sigma-algebras ${\mathcal{G}}_n=\sigma (S_n,S_{n+1},\dots )$ decrease with $n$, and (b) gives $S_n/n=\mathbb{E}[X_1\mid {\mathcal{G}}_n]$. The permitted reverse martingale convergence theorem therefore gives an integrable random variable $L$ such that $S_n/n\to L$ almost surely and with convergence in L1. In particular,It remains to prove that $L$ is constant. We must not assume that ${\bigcap }_n{\mathcal{G}}_n$ is the tail sigma-algebra of the $X_j$; instead use tail measurability of limits of sample averages directly. Define the random variable valued in the extended real numbers $L^{\ast }={\mathrm{limsup}}_n{S}_n/n$. For every fixed $N$,because $S_N/n\to 0$ pointwise. Thus $L^{\ast }$ is measurable with respect to $\sigma (X_{N+1},X_{N+2},\dots )$ for every $N$, and hence to the tail sigma-algebra. Also $L^{\ast }=L$ almost surely.

By the Kolmogorov zero-one law, each tail event $\{L^{\ast }\le q\}$, for a rational number $q$, has probability zero or one. Since $L^{\ast }$ is finite almost surely, its distribution function can only be that of a constant $c$: taking $c=\inf \{q\in \mathbb{Q}:\mathbb{P}(L^{\ast }\le q)=1\}$ and using the density of the rational numbers gives $L^{\ast }=c$ almost surely. Its expected value identifies $c=m$. This proves the strong law of large numbers:

For a finite partition $\xi$, putThe system is K-mixing when every measurable $A$ becomes uniformly asymptotically independent of this remote future:Taking $\xi =\{B,X\setminus B\}$ makes $T^{-n}B\in {\mathcal{F}}_n(\xi )$, soThus K-mixing implies mixing.

The tail sigma-algebra of a measurable partition isBy the reverse martingale convergence theorem,in $L^1$. The uniform independence in the definition of K-mixing is equivalent to the limit being the constant $\mu (A)$ for every $A$. This holds exactly when every $\mathcal{T}(\xi )$-measurable set has measure zero or one. Hence the system is K-mixing if and only if every finite partition has trivial tail sigma-algebra.

Let $\mathcal{P}=\{A:h_{\mu }(T,{\xi }_A)=0\}$. Complements preserve the binary partition. If $A,B\in \mathcal{P}$, then ${\xi }_{A\cup B}$ is coarser than ${\xi }_A\vee {\xi }_B$, so subadditivity gives zero entropy rate. Thus $\mathcal{P}$ is an algebra. For $A={\bigcup }_{j\ge 1}{A}_j$, let $A^{(m)}={\bigcup }_{j\le m}{A}_j$. The entropy metric continuity boundtends to zero because $\mu (A△A^{(m)})\to 0$. Hence $A\in \mathcal{P}$, proving that $\mathcal{P}$ is a sigma-algebra: the Pinsker sigma-algebra.

If $A$ belongs modulo null sets to $\mathcal{T}(\xi )$ for a finite $\xi$, remote-future approximations make the entropy rate of ${\xi }_A$ zero. Conversely, if $h_{\mu }(T,{\xi }_A)=0$, the conditional-entropy formula for entropy rate givesThus $A$ is measurable modulo null sets from its strict future. Iterating this fact makes it measurable from every remote future, so $A\in \mathcal{T}({\xi }_A)$ modulo null sets. ThereforeThis is the Tail characterization of the Pinsker sigma-algebra.

The tail sigma-algebra of a finite partition $\xi$ isBy the reverse martingale convergence theorem, K-mixing is equivalent to triviality of $\mathcal{T}(\xi )$ for every finite partition $\xi$.