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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.

The pi-systems criterion says that if ${\mathcal{I}}_1,{\mathcal{I}}_2$ are pi-systems and $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$ for all $A\in {\mathcal{I}}_1$, $B\in {\mathcal{I}}_2$, then their generated sigma-algebras are independent. Necessity is immediate. We may add $\Omega$ to either pi-system, since it does not alter the generated sigma-algebra or the product identities.

For sufficiency, fix $A\in {\mathcal{I}}_1$ and defineIt contains $\Omega$, is closed under complements by subtraction from $\mathbb{P}(A)$, and is closed under countable disjoint unions by countable additivity. Thus it is a lambda-system. It contains ${\mathcal{I}}_2$, so the pi-lambda theorem gives ${\mathcal{D}}_A={\mathcal{G}}_2$.

Now fix any $B\in {\mathcal{G}}_2$ and apply the same lambda-system argument to the sets $A\in {\mathcal{G}}_1$ satisfying the identity. The first step shows that this lambda-system contains ${\mathcal{I}}_1$, so it contains all ${\mathcal{G}}_1$. Thus the product identity holds on both entire sigma-algebras, proving their independence.

The tail sigma-algebra isFix $S\in \mathcal{T}$. independence of the variables and the pi-systems criterion imply independence of $\sigma (X_1,\dots ,X_n)$ from $\sigma (X_{n+1},X_{n+2},\dots )$. Hence $S$ is independent of every finite-head sigma-algebra ${\mathcal{H}}_n=\sigma (X_1,\dots ,X_n)$.

The increasing union ${\bigcup }_n{\mathcal{H}}_n$ is an algebra, hence a pi-system, and generates ${\mathcal{H}}_{\infty }=\sigma (X_1,X_2,\dots )$. Another pi-lambda argument extends the equality $\mathbb{P}(S\cap A)=\mathbb{P}(S)\mathbb{P}(A)$ from that union to all of ${\mathcal{H}}_{\infty }$. Since $S$ itself belongs to ${\mathcal{H}}_{\infty }$, take $A=S$:ThereforeThis is Kolmogorov zero-one law; no moment assumptions on the variables are needed.

A pi-system is a collection of subsets closed under binary intersections: $A,B\in \mathcal{I}$ implies $A\cap B\in \mathcal{I}$. Induction then gives closure under every nonempty finite intersection. A pi-system need not contain complements or unions.

For each $i$, the collection of events $\{X_i\le x\}$, together with the empty event, is a pi-system. Its generated sigma-algebra is $\sigma (X_i)$, because the half-lines generate the Borel sigma-algebra and inverse images preserve intersections and the sigma-algebra operations.

The displayed probability identity says precisely that these two pi-systems are independent. Part (ii) makes $\sigma (X_1)$ and $\sigma (X_2)$ independent, which is the definition of independent random variables. Conversely independence of the random variables applies to the half-line events and gives the identity. Thus the joint cumulative distribution function factors exactly when the variables are independent.

Independence of the full sigma-algebras immediately gives independence of their generating pi-systems. For the converse, fix $H\in {\mathcal{H}}_0$. On $\mathcal{G}$, define finite measuresThey have the same total mass $\mathbb{P}(H)$ and agree on ${\mathcal{G}}_0$ by assumption. The supplied measure-uniqueness theorem makes them equal on $\mathcal{G}$. Thus the independence formula holds for every $G\in \mathcal{G}$ and every $H\in {\mathcal{H}}_0$.

Now fix such an arbitrary $G$. The measures $H\mapsto \mathbb{P}(G\cap H)$ and $H\mapsto \mathbb{P}(G)\mathbb{P}(H)$ on $\mathcal{H}$ have equal total mass and agree on ${\mathcal{H}}_0$. A second application gives equality on all of $\mathcal{H}$. This proves independence extended from generating pi-systems.

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 collection closed under finite intersections. The uniqueness theorem says that two probability measures agreeing on a generating pi-system agree on its generated sigma-field. For finite measures one also fixes the total mass; for a sigma-finite version one requires a covering sequence from the pi-system with finite common masses. Adding the whole space to a pi-system of events causes no change to its generated sigma-field and preserves the stated independence condition.

Fix $H\in \mathcal{H}$. On $\sigma (\mathcal{G})$ the finite measures $A\mapsto P(A\cap H)$ and $A\mapsto P(A)P(H)$ agree on $\mathcal{G}$ and on the whole space. Uniqueness makes them agree for every $A\in \sigma (\mathcal{G})$. Now fix such an $A$. On $\sigma (\mathcal{H})$ the measures $B\mapsto P(A\cap B)$ and $B\mapsto P(A)P(B)$ agree on $\mathcal{H}$ and the whole space; a second application of uniqueness provesThus the two generated sigma-fields are independent. The extension is done one argument at a time, not by assuming the required independence for arbitrary unions.

For the independent random variables, let $\mathcal{G}$ be the rectangles ${\bigcap }_{i=1}^m\{Y_i\in B_i\}$ and let $\mathcal{H}$ be the analogous rectangles in the $Z_j$. Both are pi-systems. Joint independence of the full list factors each combined rectangular probability into the product of the individual probabilities, hence into $P(G)P(H)$. Their generated sigma-fields are exactly $\sigma (Y_1,\dots ,Y_m)$ and $\sigma (Z_1,\dots ,Z_n)$. The proved result therefore gives independence of the two vector-generated sigma-fields.

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 lambda-system contains the whole space and is closed under complements and countable disjoint unions. Let $\mathcal{D}$ be the smallest lambda-system containing a pi-system $\mathcal{P}$. For $A\in \mathcal{P}$, the family of $B\in \mathcal{D}$ with $A\cap B\in \mathcal{D}$ is a lambda-system containing $\mathcal{P}$, hence is all of $\mathcal{D}$. Fixing $B\in \mathcal{D}$ and repeating the argument shows $\mathcal{D}$ is closed under intersections. A lambda-system closed under intersections is a sigma-algebra, so $\mathcal{D}=\sigma (\mathcal{P})$. This proves the pi-lambda theorem. If two probability measures agree on $\mathcal{P}$, the sets on which they agree form a lambda-system; thus they agree on $\sigma (\mathcal{P})$.

The nonnegative form of Fubini's theorem, also called Tonelli theorem, states that for a nonnegative product-measurable function on two sigma-finite measure spaces, the product integral equals either iterated integral, allowing the value infinity.

For Lebesgue measure on the plane, a bounded interval rectangle $I\times J$ has measure $|I||J|$. Its preimage under $f$ is $(\lambda I)\times ({\lambda }^{-1}J)$, which has the same measure. Under $g$, its vertical slice at $x\in I$ is $J-sx$, whose one-dimensional Lebesgue measure is $|J|$ by translation invariance. Tonelli therefore gives $\mu (g^{-1}(I\times J))=|I||J|$. Horizontal slices likewise prove the equality for $h$. The interval rectangles form a generating pi-system. Sigma-finite measure uniqueness, obtained by the finite-measure pi-lambda argument on an increasing rectangular exhaustion, extends these equalities to all Borel sets. Completion extends them to all Lebesgue-measurable sets, since Borel null sets and their subsets remain null.

Writing ${\lambda }^2=c$, matrix multiplication givesThus rotations through angles in $[0,\pi /2)$ preserve measure. Their inverses and compositions do too; every plane rotation is a finite composition of these, including a right angle as two rotations through $\pi /4$. Lebesgue measure is invariant under every rotation.

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 $\mathcal{F}$,

Use finite cutoff windows to avoid subtracting two infinite masses. Write $D$ for the unit disc, $D_U={\mathbf{L}}_D\setminus {\mathbf{L}}_U$ on loops surrounding $0$, and $A(U)=\nu (D_U)$. For $0<r<1$, let $H_r=D_{rD}$. These increase to all loops in ${\mathbf{L}}_D$ surrounding $0$ as $r\downarrow 0$.

On the finite-mass windows, the domain-containment events form a pi-system. Indeed, if $U,V$ are simply connected and contain $0$, a simple loop surrounding $0$ that lies in both has its filled interior in both. It therefore lies in the component $W$ of $U\cap V$ containing $0$, which is simply connected. Thus, on the support of $\nu$,Inclusion-exclusion for the finite deficits givesTaking $V=rD$ and subtracting from $\nu (H_r)$ gives the useful recovery formulaThe assumed generation of the loop sigma-field by domain-containment events and the uniqueness theorem for measures now determine the finite restriction to $H_r$. Increasing these windows recovers ${\nu }_D$; conformal transport recovers its restriction to any simply connected domain containing the marked point. If a simply connected domain omits $0$, no loop in it can surround $0$.

The argument uses finite annular deficits. This is the usual local-finiteness convention for these loop measures and is explicitly ensured by the finite logarithmic formula assumed in the continuation. Bare sigma-finiteness must not be used as permission for formal $\infty -\infty$ subtraction. This is the recovery of a loop measure from conformal deficits.

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.

Let $V$ be the closure of the simple predictable processes in the Hilbert space $L^2(\mathcal{P},\mu )$. Common refinement of deterministic partitions makes these processes a vector space. To prove density of simple predictable processes for finite measures, suppose $g\in V^{\perp }$. Since $\mu$ is a finite measure, the Cauchy-Schwarz inequality gives $g\in L^1(\mu )$, sodefines a finite signed measure. The indicator of each predictable generating rectangle, including each time-zero rectangle, is a simple predictable process; hence $\nu$ vanishes on these rectangles.

Finite intersections of the rectangles are again rectangles or empty. Add the whole space to this pi-system. Its $\nu$-mass is zero because $1_{\Omega \times [0,n]}$ is simple and converges to $1$ in $L^2(\mu )$ by the dominated convergence theorem. The zero sets of $\nu$ form a lambda-system now that its total mass is zero. The Pi-lambda theorem therefore gives $\nu (E)=0$ for every $E\in \mathcal{P}$. Taking the sets where $g>0$ and $g<0$ gives $g=0$ $\mu$-almost everywhere.

Thus $V^{\perp }=\{0\}$ and the orthogonal complement criterion for a closed subspace of a Hilbert space yieldsOmitting the time-zero term would make this conclusion false when $\mu$ is concentrated at time zero.

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.

The map swaps each adjacent pair of binary digits: $({\omega }_1,{\omega }_2,{\omega }_3,{\omega }_4,\dots )\mapsto ({\omega }_2,{\omega }_1,{\omega }_4,{\omega }_3,\dots )$. Canonical expansions with no eventual all-ones tail remain canonical under this swap, and ${\theta }^2=\mathrm{id}$.

Under Lebesgue measure, any specified values of $m$ distinct binary digits have probability $2^{-m}$. Indeed they describe a union of dyadic intervals at the largest of the specified positions, with that total length. The preimage of a cylinder prescribing its first $m$ output digits prescribes $m$ distinct input digits and therefore has the same measure. The half-open dyadic intervals, with the empty set, form a pi-system generating the Borel sigma-algebra. Apply the preceding uniqueness result to obtainThe choice of binary expansion matters at dyadic endpoints for the map's pointwise definition; the chosen convention removes that ambiguity and those endpoints are null in the measure calculation.

The product sigma-algebra isThe product measure ${\mu }_1\otimes {\mu }_2$ is the probability measure on this sigma-algebra satisfyingThe measurable rectangles form a pi-system generating ${\mathcal{A}}_1\otimes {\mathcal{A}}_2$. Any two probability measures agreeing on them therefore agree on the generated sigma-algebra by the pi-lambda theorem, proving uniqueness.