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For mutually independent random variables, every tail event has $\mathbb{P}(A)\in \{0,1\}$. Identical probability distributions are unnecessary. The tail sigma-algebra is trivial in probability: its events have probability zero or one, though it can contain nonempty proper sets.^[ref]

For a proof, set ${\mathcal{F}}_m=\sigma (X_1,\dots ,X_m)$ and ${\mathcal{F}}_X=\sigma (X_1,X_2,\dots )$. Independence extended from generating pi-systems makes ${\mathcal{F}}_m$ independent of $\sigma (X_{m+1},X_{m+2},\dots )$, hence of each tail event $A$. The increasing union ${\bigcup }_m{\mathcal{F}}_m$ is a generating pi-system for ${\mathcal{F}}_X$. For fixed $A$, the events $B$ satisfying $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$ form a Dynkin system. The pi-lambda theorem extends this equality to all $B\in {\mathcal{F}}_X$. Taking $B=A$ givesThus $A$ is independent of itself, forcing the two stated values.^[ref]

The useful feature is the absence of intermediate probabilities for infinite properties that ignore every finite prefix. It reduces some calculations to showing that a tail event has positive probability, or that its complement does. The theorem supplies no finite observation deadline, and an extra argument such as the calculation in infinitely many heads as a tail event is needed to decide which value occurs.

A simple predictable process can be writtenwhere each ${\xi }_j$ is bounded and ${\mathcal{F}}_{t_j}$-measurable. Define its integral to start at zero:Refining the partition leaves this sum unchanged. Every summand is a martingale, by conditioning its future increment on the information at $t_j$. Bounded coefficients and the $L^2$ bound on $M$ make the integral an L2-bounded continuous martingale. Its terminal value is the same sum with $M_{\infty }$ substituted at any infinite endpoint; $L^2$ martingale convergence supplies this limit.

We need the conditional square identity, rather than assuming independent increments. The Doob L2 maximal inequality gives $\mathbb{E}{\sup }_t|M_t{|}^2<\infty$. Localize the square identity for $M^2-[M]$ at level and bracket stopping times. The stopped identity and this maximal bound give $\mathbb{E}[M{]}_{\infty }<\infty$ by monotone convergence. Nowis integrable, so localization passes to conditional expectations and $M^2-[M]$ is a uniformly integrable martingale. Expanding a squared increment and using $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$ provesThe infinite endpoint follows from $L^2$ convergence and integrability of the bracket. The initial value $M_0$ need not be zero: the integral itself is zero-starting.

In the square of the terminal integral, all cross terms vanish. For $j<k$, the $j$th increment and both coefficients in its cross term are measurable at $t_k$, whereas the conditional mean of the $k$th increment there is zero. For the diagonal terms, the preceding conditional identity can be multiplied by the bounded ${\xi }_j^2$. ThereforeThis proves the Itô isometry for these simple integrands. Ordered stopping-time intervals also work, replacing the conditional identity by optional stopping theorem, but deterministic intervals suffice for the approximation requested here.

For that approximation, put $\mu (A)=\mathbb{E}{\int }_0^1{1}_A(\omega ,s)ds$ on the predictable sigma-algebra. This is a finite measure. That sigma-algebra is generated, apart from the zero-time slice of measure zero, by rectangles $A\times (s,t]$ with $A\in {\mathcal{F}}_s$. The indicator of every such rectangle is a bounded simple predictable process.

Here is a density proof. Let $V$ be the closed linear span of these indicators in $L^2(\mu )$, and let $\mathcal{D}$ contain the predictable sets whose indicators belong to $V$. The whole space belongs to $\mathcal{D}$; complements stay in $\mathcal{D}$; and disjoint countable unions do also, because their finite partial indicator sums converge in $L^2$ under the finite measure. Thus $\mathcal{D}$ is a Dynkin system containing the generating rectangle pi-system. The Pi-lambda theorem makes it the entire predictable sigma-algebra. Bounded predictable simple functions therefore belong to $V$. Truncating any $L^2(\mu )$ function and then approximating its truncated range by finitely many levels shows that every such function belongs to $V$. A finite sum of rectangle indicators can be rewritten on one common deterministic time partition, with the coefficient on each interval measurable at its left endpoint; it is exactly an allowed simple process.

Consequently choose $H^n$ of this form with $\Vert H^n-H{\Vert }_{L^2(\mu )}<1/n$. This provesIt is density of simple predictable processes for finite measures, with a proof of the actual approximation step.

Finally apply the preceding isometry to $M_t=B_{t\wedge 1}$, whose bracket is $t\wedge 1$. The elementary integrals $I_n={\int }_0^1{H}_s^{n}d{B}_s$ satisfyDefine the Itô integral as the $L^2$ limit of these elementary integrals. The same identity shows independence of the approximating sequence, linearity, mean zero, and $\mathbb{E}({\int }_0^1{H}_{s}d{B}_s{)}^2=\mathbb{E}{\int }_0^1{H}_s^{2}ds$. Applying the Doob inequality to the difference processes also gives uniform-in-time $L^2$ convergence on $[0,1]$, with a continuous martingale version of the integral.

A pi-system is a family closed under finite intersections. A Dynkin system contains the whole underlying set, is closed under complements in it, and is closed under countable pairwise-disjoint unions.

Every sigma-algebra is both a pi-system and a Dynkin system. Conversely, in a family having both properties, intersections and complements give finite unions and differences. For any sequence $(A_n)$ in it, replace $A_n$ by the disjoint sets $B_n=A_n\setminus {\bigcup }_{j<n}{A}_j$, which still belong to the family. Their disjoint union equals ${\bigcup }_n{A}_n$, proving closure under arbitrary countable unions. Thus the family is a sigma-algebra.

${\mathcal{E}}_1$ is a Dynkin system but not a pi-system. The whole ten-point set has even cardinality, complements preserve parity, and disjoint unions of even sets are even. Two even sets can intersect in a singleton, for example $\{1,2\}$ and $\{2,3\}$.

${\mathcal{E}}_2$ is neither. The infinite set $\mathbb{N}\setminus \{1\}$ belongs to it, but its singleton complement does not, so it is not a Dynkin system. The two infinite sets $\{1\}\cup \{2,4,6,\dots \}$ and $\{1,3,5,\dots \}$ intersect in the excluded singleton, so it is not a pi-system.

${\mathcal{E}}_3$ is a pi-system but not a Dynkin system. Intersections of bounded open intervals are bounded open intervals or empty. However, the family does not contain the whole underlying set $\mathbb{R}$. None of the three families is a sigma-algebra.

The uniqueness theorem for measures says that two measures agreeing on a generating pi-system agree on its generated sigma-algebra, provided they are finite with the same total mass, or there is a common increasing exhaustion by pi-system sets of finite measure.

For the finite case let $\mathcal{D}=\{A:\mu (A)=\nu (A)\}$. Equality of total masses gives closure under complements, and countable additivity gives closure under disjoint countable unions. Thus $\mathcal{D}$ is a Dynkin system containing the generating pi-system. The pi-lambda theorem puts the entire generated sigma-algebra inside $\mathcal{D}$.

For the exhaustion version, let $E_n\uparrow E$ with $E_n$ in the pi-system and $\mu (E_n)=\nu (E_n)<\infty$. The finite measures $A\mapsto \mu (A\cap E_n)$ and $A\mapsto \nu (A\cap E_n)$ agree on the generating pi-system, since it is intersection-closed. The finite proof applies, and continuity from below of a measure as $n\to \infty$ gives $\mu (A)=\nu (A)$ for every measurable $A$. A finiteness hypothesis is essential to this standard extension-uniqueness statement.

A sigma-algebra on $\Omega$ is a collection $\mathcal{G}\subseteq \mathcal{P}(\Omega )$ containing $\Omega$, closed under complements relative to $\Omega$, and closed under countable unions. These properties also give closure under countable intersections by De Morgan laws.

Let $\mathcal{G}={\bigcap }_{r\in R}{\mathcal{F}}_r$. Every constituent sigma-algebra contains $\Omega$, so $\Omega \in \mathcal{G}$. If $A\in \mathcal{G}$, its complement belongs to every ${\mathcal{F}}_r$, hence to $\mathcal{G}$. Likewise, if $A_j\in \mathcal{G}$ for all $j$, then ${\bigcup }_j{A}_j$ belongs to every ${\mathcal{F}}_r$. Thus $\mathcal{G}$ is a sigma-algebra. For an empty indexing set, the usual intersection convention gives $\mathcal{G}=\mathcal{P}(\Omega )$, which is also a sigma-algebra.

A union need not work. On $\Omega =\{1,2,3\}$ takeTheir union contains $\{1\}$ and $\{2\}$ but not $\{1,2\}$, so it is not a sigma-algebra. Put $\mathcal{A}={\bigcup }_r{\mathcal{F}}_r$ and let $\mathfrak{C}$ be the collection of all sigma-algebras on $\Omega$ containing $\mathcal{A}$. This collection is nonempty because it contains the power set $\mathcal{P}(\Omega )$. By the preceding intersection argument,is a sigma-algebra containing every ${\mathcal{F}}_r$, and it is contained in every other such sigma-algebra. This proves existence and minimality.

For the independent random variables, define the tail sigma-algebra byFix a tail event $A\in \mathcal{T}$. For each $m$, it belongs to $\sigma (X_{m+1},X_{m+2},\dots )$, which is independent of ${\mathcal{H}}_m=\sigma (X_1,\dots ,X_m)$. To justify independence of the generated sigma-algebras, first use independence of finite collections on cylinder events, then extend twice by the pi-lambda theorem. In particular, $\mathbb{P}(A\cap C)=\mathbb{P}(A)\mathbb{P}(C)$ for every $C\in {\mathcal{H}}_m$ and every $m$.

The increasing union ${\bigcup }_m{\mathcal{H}}_m$ is an algebra of sets generating $\mathcal{H}=\sigma (X_1,X_2,\dots )$. The collection of $C\in \mathcal{H}$ satisfying this factorization is a Dynkin system: it contains $\Omega$, is closed under complements, and under countable disjoint unions. The pi-lambda theorem therefore extends the factorization to all $C\in \mathcal{H}$. Since $A\in \mathcal{H}$, taking $C=A$ gives $\mathbb{P}(A)=\mathbb{P}(A{)}^2$. Consequently,This is the Kolmogorov zero-one law, proved here directly.

For each fixed $m$, write $S_n=S_{m-1}+{\sum }_{j=m}^n{X}_j$ for $n\ge m$, with $S_0=0$. The random variable $S_{m-1}$ is finite, so $S_{m-1}/\sqrt{n}\to 0$. Adding a real sequence tending to zero changes neither the extended-real limit inferior nor the extended-real limit superior. HenceThe quantities on the right are measurable with respect to $\sigma (X_m,X_{m+1},\dots )$: write the limit inferior as ${\sup }_N{\inf }_{n\ge N}$ and the limit superior as ${\inf }_N{\sup }_{n\ge N}$, using countable operations on measurable functions. Therefore both threshold events belong to every such sigma-algebra, and hence are tail events, for every real threshold, including when either limit is infinite.

The final recurrence claim is false under the printed hypotheses. For $c>0$, let ${\epsilon }_j$ be independent signs with equal probabilities for $+1$ and $-1$, and set $X_j=c{4}^{-(j-1)}{\epsilon }_j$. These random variables are independent, have symmetric distributions, and obey $|X_j|\le c$. Nevertheless, the geometric series gives, on every sample path,Thus the proposed repeated-return event has probability zero, even though none of the increments is degenerate. The Kolmogorov zero-one law does not determine which of zero and one occurs.

An additional identical-distribution hypothesis makes the intended conclusion true. Here is a full proof of that qualified version, using the oscillation of a bounded centered iid random walk. Symmetry and integrability give $\mathbb{E}{X}_j=0$, so $S_n$ is a martingale for its natural filtration. If the common probability distribution is concentrated at zero, every $S_n$ is zero and the conclusion is immediate. Otherwise choose $\delta >0$ with $\mathbb{P}(|X_1|>\delta )>0$. The Borel-Cantelli lemmas then give $|X_n|>\delta$ infinitely often almost surely; thus $S_n$ cannot converge to a finite limit.

For $a>0$, let $T_a=\inf \{n:S_n\ge a\}$. Before $T_a$ the sum is below $a$, and at $T_a$ it is at most $a+c$. The stopped martingale $a+c-S_{n\wedge T_a}$ is therefore nonnegative. A stopped integrable martingale remains a martingale at these bounded times, and a nonnegative martingale converges finitely almost surely by the martingale convergence theorem. On $\{T_a=\infty \}$ this would force $S_n$ to converge finitely, which has just been ruled out. Hence $T_a<\infty$ almost surely. Apply the same argument to $-S_n$ and take a countable intersection over positive integer $a$. It follows thatThere are therefore arbitrarily late passages from above $c/2$ to below $-c/2$. A path with bounded increments of magnitude at most $c$ cannot make such a passage without visiting $[-c/2,c/2]$: jumping directly between its two open complementary half-lines would require an increment larger than $c$. There are infinitely many such visits. Thus with the added identical-distribution hypothesis, the intended answer is

Values at the endpoint one do not affect any of the conclusions, since $\mathbb{P}(U=1)=0$. Set $b_n(1)=1$ to define each random variable everywhere, and choose a Borel measurable function agreeing with $f$ almost everywhere if necessary. These choices leave all integrals and almost-sure assertions unchanged.

Every dyadic interval at level $n+1$ lies in a dyadic interval at level $n$. Equivalently, $b_n(U)$ is a measurable function of $b_{n+1}(U)$, so ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$: these sigma-algebras form a filtration. Since the uniform distribution of $U$ gives $\mathbb{E}|f(U)|={\int }_0^1|f(u)|du<\infty$, the conditional expectations $X_n=\mathbb{E}[f(U)\mid {\mathcal{F}}_n]$ are integrable and adapted. The tower property of conditional expectation givesThus $(X_n)$ is a martingale, and part (a) proves that it is uniformly integrable.

For $I_{n,k}=[k{2}^{-n},(k+1)2^{-n})$, the event $\{U\in I_{n,k}\}$ is an atom of a sigma-algebra ${\mathcal{F}}_n$, with probability $2^{-n}$. The value of the conditional expectation there is thereforeConsequently $\boxed{X_n=f_n(U)\text{ almost surely}.}$ This also shows directly that ${\int }_0^1|f_n|\le {\int }_0^1|f|$.

By the uniformly integrable martingale convergence theorem, there is an integrable $Y$ such that $X_n\to Y$ almost surely and in the L1 norm. To identify this limit, put ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. Since $0\le U-b_n(U)<2^{-n}$ away from the harmless endpoint, $U={\lim }_n{b}_n(U)$, and $U$ is ${\mathcal{F}}_{\infty }$-measurable. Conversely every $b_n(U)$ is a measurable function of $U$. Thus ${\mathcal{F}}_{\infty }=\sigma (U)$, and both $Y$ and $f(U)$ have ${\mathcal{F}}_{\infty }$-measurable versions.

For $A\in {\mathcal{F}}_m$, the defining property of conditional expectation gives $\mathbb{E}[X_n{1}_A]=\mathbb{E}[f(U)1_A]$ whenever $n\ge m$. Passing to the limit in the L1 norm yieldsThe increasing union ${\bigcup }_m{\mathcal{F}}_m$ is an algebra of sets generating ${\mathcal{F}}_{\infty }$. The class of events for which this integral identity holds is a Dynkin system, so the pi-lambda theorem extends the identity to every $A\in {\mathcal{F}}_{\infty }$. Since both random variables are measurable there, uniqueness of conditional expectation gives $Y=f(U)$ almost surely. We have proved that the dyadic conditional averages recover integrable functions:because a probability-one assertion for $U$ is a full-Lebesgue measure assertion for $x$. Likewise, the uniform distribution of $U$ turns convergence in L1 intoNo pointwise continuity of $f$ is required.

For independent random variables $X_1,X_2,\dots$, let ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$, $\mathcal{F}=\sigma (X_1,X_2,\dots )$, and let the tail sigma-algebra be $\mathcal{T}={\bigcap }_n\sigma (X_{n+1},X_{n+2},\dots )$. Kolmogorov zero-one law states that every $A\in \mathcal{T}$ has probability zero or one.

Such an event is independent of each ${\mathcal{F}}_n$ by independence of the variables. The collection of events $B\in \mathcal{F}$ satisfying $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$ is a Dynkin system: it contains the whole space and is closed under complements and disjoint countable unions. It contains the pi-system ${\bigcup }_n{\mathcal{F}}_n$, which generates $\mathcal{F}$. The pi-lambda theorem therefore makes $A$ independent of all of $\mathcal{F}$. Since $A$ itself belongs to $\mathcal{F}$,

On a fixed underlying set, a pi-system is a nonempty family closed under finite intersections. A Dynkin system, also called a d-system, contains the whole underlying set, is closed under complements relative to it, and under countable disjoint unions. Equivalently it is closed under differences of nested members and increasing countable unions. A sigma-algebra contains the whole set, is closed under complements and arbitrary countable unions, hence also countable intersections. A Dynkin system need not be a sigma-algebra unless additional intersection closure is available.

If $\theta$ is measure-preserving, equality holds on every measurable set and hence on the specified $\pi$-system. Conversely, let $\mathcal{D}=\{A\in \mathcal{E}:{\mu }_{\theta }(A)=\mu (A)\}$. Both finite measures have equal total mass. Therefore $E\in \mathcal{D}$, complements of members remain in $\mathcal{D}$ by subtracting from the finite total mass, and disjoint countable unions remain in $\mathcal{D}$ by countable additivity. Thus $\mathcal{D}$ is a Dynkin system.

It contains the generating pi-system $\mathcal{A}$, so Dynkin lemma gives $\mathcal{E}=\sigma (\mathcal{A})\subseteq \mathcal{D}$. ConsequentlyThis is exactly the measure-preserving property.

Dynkin lemma, also called the pi-lambda theorem, states that if a Dynkin system $\mathcal{D}$ contains a pi-system $\mathcal{P}$, then

Let $\lambda (\mathcal{P})$ be the smallest Dynkin system containing $\mathcal{P}$. It is enough to prove that $\lambda (\mathcal{P})$ is a sigma-algebra, because then

First fix $A\in \mathcal{P}$ and defineThis is a Dynkin system. It contains $\Omega$ because $A\in \lambda (\mathcal{P})$; closure under relative complements follows fromand closure under disjoint unions follows by distributing $A\cap -$ over the union. Since $\mathcal{P}$ is closed under intersections, $\mathcal{P}\subseteq {\mathcal{G}}_A$. Minimality therefore givesThus $A\cap B\in \lambda (\mathcal{P})$ whenever $A\in \mathcal{P}$ and $B\in \lambda (\mathcal{P})$.

Now fix $B\in \lambda (\mathcal{P})$ and defineThe same argument shows that ${\mathcal{H}}_B$ is a Dynkin system, and the preceding paragraph shows that it contains $\mathcal{P}$. Hence ${\mathcal{H}}_B=\lambda (\mathcal{P})$. We have proved that $\lambda (\mathcal{P})$ is itself a pi-system.

A Dynkin system that is also a pi-system is a sigma-algebra: it is closed under arbitrary finite intersections, hence finite unions by complements, and any countable union can be disjointified before using closure under disjoint unions. Therefore $\lambda (\mathcal{P})=\sigma (\mathcal{P})$, proving the lemma.

Fix $A_1\in {\mathcal{A}}_1$ and letThis is a Dynkin system. It contains $\Omega$. If $B\in {\mathcal{D}}_{A_1}$, thenso it is closed under complements. Countable additivity gives closure under countable disjoint unions.

The hypothesis says ${\mathcal{A}}_2\subseteq {\mathcal{D}}_{A_1}$. Since ${\mathcal{A}}_2$ is a pi-system, Dynkin lemma yieldsThus the factorization holds for every $A_1\in {\mathcal{A}}_1$ and every $B\in \sigma ({\mathcal{A}}_2)$.

Now fix such a $B$ and defineThe identical calculation makes ${\mathcal{E}}_B$ a Dynkin system. The first step gives ${\mathcal{A}}_1\subseteq {\mathcal{E}}_B$, so a second application of Dynkin lemma givesTherefore the probability factorization holds for everywhich is precisely independence of the two sigma-algebras. This is independence extended from generating pi-systems.

Let $\mathcal{G}=\sigma (A_1,A_2,\dots )$. The limit $Y$ from part (c) is $\mathcal{G}$-measurable because each $X_n$ is. If $A\in {\bigcup }_n{\mathcal{F}}_n$, then $A\in {\mathcal{F}}_N$ for some $N$, and for every $n\ge N$,The $L^1$ convergence lets us pass to the limit and obtain $\mathbb{E}[Y{1}_A]=\mathbb{E}[X{1}_A]$.

The sets on which this identity holds form a Dynkin system, and ${\bigcup }_n{\mathcal{F}}_n$ is a generating pi-system. The Dynkin lemma therefore extends the identity to every $A\in \mathcal{G}$. Consequently $Y$ has both defining properties of $\mathbb{E}[X\mid \mathcal{G}]$, established here without appealing to the general existence theorem.

Fix $A\in {\mathcal{A}}_1$ and defineThis is a Dynkin system: it contains $\Omega$; taking complements preserves the equality because $\mathbb{P}(A\cap B^c)=\mathbb{P}(A)-\mathbb{P}(A\cap B)$; and countable additivity preserves it under disjoint unions. The hypothesis says that ${\mathcal{D}}_A$ contains the pi-system ${\mathcal{A}}_2$. Hence Dynkin lemma gives ${\mathcal{D}}_A={\mathcal{G}}_2$.

Now fix any $B\in {\mathcal{G}}_2$ and putThe same complement and disjoint-union calculations make this a Dynkin system. The first step shows that it contains ${\mathcal{A}}_1$, so Dynkin lemma gives ${\mathcal{E}}_B={\mathcal{G}}_1$. ThusThis is precisely the definition of independent sigma-algebras, and proves independence extended from generating pi-systems.

If a Dynkin system contains a pi-system, then it contains the sigma-algebra generated by that pi-system. This proves uniqueness of finite measures from agreement on a generating pi-system.