Independence Borel Cantelli and Zero One Laws

Fix a natural number $n\ge 1$, distinct indices $j_0,\dots ,j_{n-1}\in J$, and events $A_k\in {\mathcal{C}}_{j_k}$. Since $J\subseteq I$, this is also a valid finite choice inside the original family.

Applying [L1] to that same finite choice gives $\mathbb{P}\left({\bigcap }_{k<n}{A}_k\right)={\prod }_{k<n}\mathbb{P}(A_k)\mathrm{.}$ Hence the restricted family is independent.

It is enough to prove the claim for a fixed finite subfamily $A_{i_0},\dots ,A_{i_{n-1}}$. We argue by induction on the number of complemented coordinates among $B_{i_0},\dots ,B_{i_{n-1}}$.

If no coordinate is complemented, the required factorization is exactly [L1].

Assume the factorization is known whenever at most $m$ coordinates are complemented, and suppose exactly $m+1$ are. Reindex so that $B_{i_{n-1}}=A_{i_{n-1}}^c$, and put $C:={\bigcap }_{k<n-1}{B}_{i_k}$. Then $C\cap A_{i_{n-1}}\subseteq C$, so [L2] gives $\mathbb{P}(C\cap A_{i_{n-1}}^c)=\mathbb{P}(C)-\mathbb{P}(C\cap A_{i_{n-1}})\mathrm{.}$ By the induction hypothesis, both terms on the right factor: $\mathbb{P}(C)={\prod }_{k<n-1}\mathbb{P}(B_{i_k})$ and $\mathbb{P}(C\cap A_{i_{n-1}})=({\prod }_{k<n-1}\mathbb{P}(B_{i_k}))\mathbb{P}(A_{i_{n-1}})$. Therefore $\mathbb{P}(C\cap A_{i_{n-1}}^c)=({\prod }_{k<n-1}\mathbb{P}(B_{i_k}))(1-\mathbb{P}(A_{i_{n-1}}))=({\prod }_{k<n-1}\mathbb{P}(B_{i_k}))\mathbb{P}(A_{i_{n-1}}^c)\mathrm{.}$ This is the desired factorization for the current finite family.

Steps 1.2 and 2.1 prove the inductive claim for every finite subfamily, so the family $(B_i{)}_{i\in I}$ is independent.

By [L1], it suffices to fix a finite list of distinct indices $i_0,\dots ,i_{n-1}$ and prove that $\sigma ({\mathrm{\Pi }}_{i_0}),\dots ,\sigma ({\mathrm{\Pi }}_{i_{n-1}})$ are independent.

Fix $B_k\in {\mathrm{\Pi }}_{i_k}$ for $k<n-1$ and define ${\mathrm{\Lambda }}_{n-1}:=\{A\in \sigma ({\mathrm{\Pi }}_{i_{n-1}}):\mathbb{P}({\bigcap }_{k<n-1}{B}_k\cap A)=({\prod }_{k<n-1}\mathbb{P}(B_k))\mathbb{P}(A)\}\mathrm{.}$ Because $\mathrm{\Omega }\in {\mathrm{\Pi }}_{i_{n-1}}$, the class ${\mathrm{\Lambda }}_{n-1}$ contains $\mathrm{\Omega }$. It is closed under relative differences of nested sets and under increasing countable unions because both sides of the defining identity are countably additive in $A$. Since the original pi-systems are independent, every $A\in {\mathrm{\Pi }}_{i_{n-1}}$ lies in ${\mathrm{\Lambda }}_{n-1}$. Therefore [L2] gives $\sigma ({\mathrm{\Pi }}_{i_{n-1}})\subseteq {\mathrm{\Lambda }}_{n-1}$.

Repeat the construction of step 1.2 for the coordinates $n-2,n-3,\dots ,0$, each time freezing already-promoted later coordinates in $\sigma ({\mathrm{\Pi }}_{i_j})$ and keeping the earlier coordinates inside the original pi-systems. Each stage produces a lambda-system containing the relevant pi-system, so [L2] successively replaces every ${\mathrm{\Pi }}_{i_j}$ by $\sigma ({\mathrm{\Pi }}_{i_j})$. Hence $\mathbb{P}\left({\bigcap }_{k<n}{A}_k\right)={\prod }_{k<n}\mathbb{P}(A_k)(A_k\in \sigma ({\mathrm{\Pi }}_{i_k}))\mathrm{.}$

Since the finite choice of indices was arbitrary, the full family $(\sigma ({\mathrm{\Pi }}_i){)}_{i\in I}$ is independent.

For each $r<m$, let ${\mathrm{\Pi }}_r$ be the class consisting of $\mathrm{\Omega }$ together with all finite intersections ${\bigcap }_{i\in F}{A}_i$, where $F\subseteq J_r$ is finite and $A_i\in {\mathcal{F}}_i$ for every $i\in F$. Each ${\mathrm{\Pi }}_r$ contains $\mathrm{\Omega }$ and is closed under finite intersections, so it is a pi-system. Moreover $\sigma ({\mathrm{\Pi }}_r)={\mathcal{G}}_r$ by definition of ${\mathcal{G}}_r$.

Fix $C_r\in {\mathrm{\Pi }}_r$ for each $r<m$. Because the index sets $J_r$ are pairwise disjoint, the event ${\bigcap }_{r<m}{C}_r$ is a finite intersection of events taken from distinct members of the original independent family. Hence [L1] gives $\mathbb{P}\left({\bigcap }_{r<m}{C}_r\right)={\prod }_{r<m}\mathbb{P}(C_r)\mathrm{.}$

Step 1.2 says that the pi-systems ${\mathrm{\Pi }}_0,\dots ,{\mathrm{\Pi }}_{m-1}$ are independent. Applying [L2] and using step 1.1 yields independence of $\sigma ({\mathrm{\Pi }}_r)={\mathcal{G}}_r$ for every $r<m$.

If $(X_i{)}_{i\in I}$ is independent, then by [L1] the sigma-algebras $\sigma (X_i)$ are independent. Since $X_i^{-1}(B)\in \sigma (X_i)$ for every $B\in {\mathrm{\Sigma }}_i$, the displayed rectangle identity follows immediately.

Conversely, for each $i$ let ${\mathrm{\Pi }}_i:=\{X_i^{-1}(B):B\in {\mathrm{\Sigma }}_i\}\subseteq \mathcal{F}$. Preimages preserve finite intersections, so each ${\mathrm{\Pi }}_i$ is a pi-system containing $\mathrm{\Omega }$. The hypothesis in clause 2 says exactly that the family $({\mathrm{\Pi }}_i{)}_{i\in I}$ is independent. Since $\sigma ({\mathrm{\Pi }}_i)=\sigma (X_i)$ by definition, [L2] implies that the sigma-algebras $\sigma (X_i)$ are independent.

Step 1.2 proves clause 2 implies clause 1, and step 1.1 proves the reverse implication. Therefore the two conditions are equivalent.

By [L2] and [L1], the variables $X_i$ are genuine random elements on the full-power-set probability space, so [L3] applies to them.

Conversely, assume [L4]. For a finite subfamily $(X_i{)}_{i\in J}$ and measurable sets $B_i\subseteq \mathbb{R}$, only finitely many values in $B_i\cap X_i(\mathrm{\Omega })$ can occur. The event $\{X_i\in B_i\text{ for all }i\in J\}$ is the disjoint union of the attained-value events $\{X_i=x_i\text{ for all }i\in J\}$ over those finitely many tuples. Summing the factorized singleton probabilities from [L4] gives ${\mathbb{P}}_w(X_i\in B_i\text{ for all }i\in J)={\prod }_{i\in J}{\mathbb{P}}_w(X_i\in B_i)\mathrm{.}$ So the rectangle criterion holds.

If the rectangle criterion holds, apply it to singleton target sets $B_i=\{x_i\}$. This gives ${\mathbb{P}}_w(X_i=x_i\text{ for all }i\in J)={\prod }_{i\in J}{\mathbb{P}}_w(X_i=x_i)$ for every finite $J\subseteq I$, which is exactly [L4].

Steps 2.1 and 1.2 prove that the finite published definition and the general rectangle criterion agree exactly on finite-valued variables over a finite probability space.

By [L1], each composite $g_i\circ X_i$ is again a random element.

Fix a finite list of distinct indices $i_0,\dots ,i_{n-1}$ and measurable sets $C_k\in {\mathcal{T}}_{i_k}$. Then $g_{i_k}^{-1}(C_k)\in {\mathrm{\Sigma }}_{i_k}$ for every $k$, so [L2] applied to the independent family $(X_i)$ gives $\mathbb{P}(g_{i_0}(X_{i_0})\in C_0,\dots ,g_{i_{n-1}}(X_{i_{n-1}})\in C_{n-1})={\prod }_{k<n}\mathbb{P}(g_{i_k}(X_{i_k})\in C_k)\mathrm{.}$

Step 1.2 is exactly the rectangle criterion for the family $(g_i\circ X_i{)}_{i\in I}$, so [L2] shows that this family is independent.

First prove that all-coordinate rectangles generate the recursively defined finite product sigma-algebra. This is immediate for one factor. Suppose it holds for $k$ factors, write $T_k={\prod }_{i<k}{S}_i$ and ${\mathcal{A}}_k={⨂}_{i<k}{\mathrm{\Sigma }}_i$, and let $\mathcal{H}$ be the sigma-algebra generated by the $(k+1)$-coordinate rectangles. These rectangles belong to ${\mathcal{A}}_k\otimes {\mathrm{\Sigma }}_k$, so $\mathcal{H}\subseteq {\mathcal{A}}_k\otimes {\mathrm{\Sigma }}_k$. Fix $B\in {\mathrm{\Sigma }}_k$. The class ${\mathcal{C}}_B=\{A\in {\mathcal{A}}_k:A\times B\in \mathcal{H}\}$ contains $T_k$, since $T_k\times B$ is an all-coordinate rectangle. It is closed under complements relative to $T_k$, because $(T_k\setminus A)\times B=(T_k\times B)\setminus (A\times B)$, and under countable unions because product with $B$ commutes with them. Hence it is a sigma-algebra containing all $k$-coordinate rectangles. The induction hypothesis gives ${\mathcal{C}}_B={\mathcal{A}}_k$. As this holds for every $B$, every binary rectangle $A\times B$ generating ${\mathcal{A}}_k\otimes {\mathrm{\Sigma }}_k$ lies in $\mathcal{H}$, proving equality and completing the finite induction.

Let $\mathcal{D}:=\{E\subseteq {\prod }_{i<n}{S}_i:X^{-1}(E)\in \mathcal{F}\}$. Preimages preserve complements and countable unions, so $\mathcal{D}$ is a sigma-algebra. If $R={\prod }_{i<n}{B}_i$ is a measurable rectangle, then $X^{-1}(R)={\bigcap }_{i<n}{X}_i^{-1}(B_i)\in \mathcal{F}$. By step 1.1, the product sigma-algebra is generated by these all-coordinate rectangles, so $X$ is measurable for ${⨂}_{i<n}{\mathrm{\Sigma }}_i$.

For every measurable rectangle $R={\prod }_{i<n}{B}_i$, [L1] gives ${\mathbb{P}}_X(R)=\mathbb{P}(X_0\in B_0,\dots ,X_{n-1}\in B_{n-1})={\prod }_{i<n}\mathbb{P}(X_i\in B_i)={\prod }_{i<n}{\mathbb{P}}_{X_i}(B_i)\mathrm{.}$

By step 2.1, the law ${\mathbb{P}}_X$ is defined, and [L2] makes it a probability measure on the product sigma-algebra.

For every $n\ge 1$, put ${\nu }_1:={\mathbb{P}}_{X_0}$ and, for $1\le m<n$, define recursively ${\nu }_{m+1}:={\nu }_m\otimes {\mathbb{P}}_{X_m}$. For $n=1$, step 3.1 already identifies ${\mathbb{P}}_X$ with ${\nu }_1$. For $n\ge 2$, repeated use of the rectangle formula in [L3] shows that ${\nu }_n\left({\prod }_{i<n}{B}_i\right)={\prod }_{i<n}{\mathbb{P}}_{X_i}(B_i)$ for every measurable rectangle. Therefore ${\mathbb{P}}_X$ and ${\nu }_n$ agree on all measurable rectangles.

All-coordinate rectangles are closed under intersections, because intersections are taken coordinatewise; they contain the whole product and generate its sigma-algebra by step 1.1. By step 4.1, ${\mathbb{P}}_X$ and ${\nu }_n$ agree on this pi-system. Both have total mass one: [L2] gives this for the joint law, and the rectangle formula [L3] gives it at every stage of the product-law recursion. The constant exhaustion by the whole product therefore satisfies the finite-measure hypothesis of [L5], which gives ${\mathbb{P}}_X={\nu }_n={⨂}_{i<n}{\mathbb{P}}_{X_i}\mathrm{.}$ This is the claimed product joint law.

Put $Y_i:=g_i(X_i)$. By [L1], the family $Y_0,\dots ,Y_{n-1}$ is independent. On ${\mathbb{R}}^n$ with the finite product sigma-algebra, each coordinate projection ${\pi }_i$ is measurable because ${\pi }_i^{-1}(B)={\mathbb{R}}^i\times B\times {\mathbb{R}}^{n-1-i}$ is a measurable rectangle. Repeated use of [L6] therefore makes the product map $m(y_0,\dots ,y_{n-1})={\prod }_{i<n}{y}_i$ measurable.

Assume each $g_i$ is nonnegative. If some $\mathbb{E}[g_j(X_j)]=0$, [L7] gives $Y_j=0$ almost surely. The finite product ${\prod }_i{Y}_i$ is then zero almost surely, so [L7] gives expectation zero, agreeing with the Statement's explicit zero-factor convention even if another expectation is infinite.

Otherwise every $\mathbb{E}[g_i(X_i)]$ is positive (possibly infinite). Let ${\mu }_i$ be the law of $Y_i$ and let $\mu ={⨂}_{i<n}{\mu }_i$. By [L2], the joint law of $Y=(Y_0,\dots ,Y_{n-1})$ is $\mu$. For each integer $k\ge 1$, put $Y_i^{(k)}=Y_i\wedge k$. The finite product ${\prod }_i{Y}_i^{(k)}$ is bounded, so [L3] and repeated [L4] give $\mathbb{E}[{\prod }_i{Y}_i^{(k)}]={\prod }_i\mathbb{E}[Y_i^{(k)}]$ with only finite factors. As $k\uparrow \mathrm{\infty }$, the products increase pointwise to ${\prod }_i{Y}_i$; [L8] passes the limit through the left integral and each factor on the right. Since all limiting factors are strictly positive, the finite product of their increasing approximants tends to their ordinary extended product without an indeterminate zero-times-infinity expression.

Now assume every $g_i(X_i)$ is integrable. Applying steps 2.1–2.2 to the nonnegative functions $\mathrm{\mid }{g}_i\mathrm{\mid }$ gives $\mathbb{E}[{\prod }_{i<n}\mathrm{\mid }{g}_i(X_i)\mathrm{\mid }]={\prod }_{i<n}\mathbb{E}[\mathrm{\mid }{g}_i(X_i)\mathrm{\mid }]<\mathrm{\infty }$. So ${\prod }_{i<n}{g}_i(X_i)$ is integrable.

Let ${\mu }_i$ be the law of $Y_i=g_i(X_i)$ and $\mu ={⨂}_{i<n}{\mu }_i$ as in step 2.2. Step 3.1 shows that the product map $m$ is $\mu$-integrable. By [L2], [L3], and repeated use of [L5], $\mathbb{E}[{\prod }_{i<n}{g}_i(X_i)]={\int }_{{\mathbb{R}}^n}md\mu ={\prod }_{i<n}{\int }_{\mathbb{R}}yd{\mu }_i={\prod }_{i<n}\mathbb{E}[g_i(X_i)]$.

Steps 2.1–2.2 prove the nonnegative case, and step 4.1 proves the integrable case.

In the original finite-valued case, let $J\subseteq I$ and let $B_i\subseteq \mathbb{R}$ be Borel for $i\in J$. Write $A_i$ for the finite attained-value set of $X_i$. The event ${\bigcap }_{i\in J}\{X_i\in B_i\}$ is the disjoint union of the joint atoms indexed by ${\prod }_{i\in J}(A_i\cap B_i)$. By [L3] and [L4], its probability is ${\sum }_{(x_i)\in {\prod }_{i\in J}(A_i\cap B_i)}{\prod }_{i\in J}\mathbb{P}(X_i=x_i)={\prod }_{i\in J}{\sum }_{x_i\in A_i\cap B_i}\mathbb{P}(X_i=x_i)={\prod }_{i\in J}\mathbb{P}(X_i\in B_i)\mathrm{.}$ Factoring finite sums proves the middle equality, including empty factors. Thus the family is independent in the sense of [L1]. Each variable is bounded because its attained-value set is finite, hence integrable on a probability space.

Suppose $I\mathrm{\ne }\mathrm{\varnothing }$ and every $X_i$ is integrable. Enumerate the finite set $I$ by $0,\dots ,n-1$ and apply [L2] with $g_i(x)=x$, which is Borel. It gives product integrability and the asserted real factorization. Selecting one enumeration of one finite set uses no axiom of choice. Step 1.1 supplies the required independence and integrability in the original finite-valued case.

Suppose $I\mathrm{\ne }\mathrm{\varnothing }$ and every $X_i$ is nonnegative. Apply [L2] to the same enumeration with $g_i(x)=\max (x,0)$, a continuous nonnegative Borel function. Since $g_i(X_i)=X_i$, [L2] gives the asserted extended factorization with precisely its zero-factor convention.

For $I=\mathrm{\varnothing }$, the random product is the constant $1$, whose expectation is $1$ on a probability space, and the product of expectations is the empty product $1$. Steps 2.1 and 3.1 and this identity prove all stated cases.

In the original finite-valued atom-independence convention, summing joint atom probabilities over finite attained-value sets proves independence for Borel inverse-image events, as in the finite-valued bridge of [L2]. The map $f_t(x)=\exp (tx)$ is continuous by [L5]. The family $\{B\subseteq \mathbb{R}:f_t^{-1}(B)\in \mathcal{B}(\mathbb{R})\}$ is a sigma-algebra, since inverse images preserve complements and countable unions, and it contains the open sets by continuity. Therefore it contains every Borel set by [L5], so $f_t$ is Borel measurable. Hence $Y_i=f_t(X_i)$ is a measurable nonnegative real variable by [L3]. For any Borel set $B$, $\{Y_i\in B\}=\{X_i\in f_t^{-1}(B)\}\in \sigma (X_i)\mathrm{.}$ Thus $\sigma (Y_i)\subseteq \sigma (X_i)$, and the independence in [L4] passes to the family $(Y_i{)}_{i\in I}$. This inverse-image argument uses no choice principle.

Iterating [L3] gives $\exp (t{\sum }_i{X}_i)={\prod }_i{Y}_i$ pointwise. Apply the nonnegative branch of [L2] to the independent variables from step 1.1 and use [L1] in each factor to obtain the displayed equality in $[0,+\mathrm{\infty }]$. If every individual MGF is finite at $t$, the finite product on the right is finite. On an original finite probability space, all exponential variables are bounded, so [L1] identifies these integrals with the finite MGFs, and [L2] also covers its finite-valued atom-independence convention.

For $I=\mathrm{\varnothing }$, the sum is zero and the random product is $1$. Thus $M_0(t)=\mathbb{E}[\exp (0)]=1$ on a probability space, and the product on the right is the empty product $1$.

Since $X$ and $Y$ are square-integrable, they are integrable. Applying [L1] with $g_0(x)=x$ and $g_1(y)=y$ gives $\mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]\mathrm{.}$

Substituting step 1.1 into [L2] yields $\mathrm{Cov}(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]=0.$

Each $X_i$ is integrable because $2\mathrm{\mid }{X}_i\mathrm{\mid }\le 1+X_i^2$ and the measure has total mass one. For distinct $i,j$, pairwise independence and [L1] give $\mathbb{E}[X_i{X}_j]=\mathbb{E}[X_i]\mathbb{E}[X_j]$, so [L2] gives $\mathrm{Cov}(X_i,X_j)=0$.

The finite sum $S={\sum }_{i\in I}{X}_i$ is square-integrable: repeated use of $(a+b{)}^2\le 2a^2+2b^2$ gives an integrable bound for $S^2$. By [L2], $\mathrm{Var}(S)=\mathrm{Cov}(S,S)={\sum }_{i,j\in I}\mathrm{Cov}(X_i,X_j)={\sum }_{i\in I}\mathrm{Var}(X_i),$ where step 1.1 removes all off-diagonal terms. The empty sum is the constant zero, and the singleton case is the identity.

Since $X$ is $1$ with probability $p$ and $0$ otherwise, both $X$ and $X^2$ agree almost surely with the indicator of the measurable event $\{X=1\}$. By [L6], $\mathbb{E}[X]=\mathbb{E}[X^2]=\mathbb{P}(X=1)=p$.

Hence $\mathrm{Var}(X)=p-p^2=p(1-p)$.

For distinct $i,j$, the independent Bernoulli variables in [L1] and [L5] have $X_i{X}_j={\mathbf{1}}_{\{X_i=X_j=1\}}$ almost surely, so [L6] and independence give $\mathbb{E}[X_i{X}_j]=\mathbb{P}(X_i=X_j=1)=p^2$. Their means are $p$ by step 1.1, so [L3] gives $\mathrm{Cov}(X_i,X_j)=p^2-p^2=0$. Linearity gives $\mathbb{E}[S]={\sum }_{i<n}p=np$, and [L4] with these zero covariances gives $\mathrm{Var}(S)={\sum }_{i<n}p(1-p)=np(1-p)$.

At $p=0$ or $p=1$ the variance formula is zero, and at $n=0$ both finite sums are empty and equal zero.

By [L1], the event $\{A_n\mathrm{i}\mathrm{.}\mathrm{o}\mathrm{.}\}$ is exactly ${\mathrm{lim\hspace{0.17em}sup}}_{n\to \mathrm{\infty }}{A}_n$.

Applying [L2] to the probability measure $\mathbb{P}$ and the measurable sets $A_n$ gives $\mathbb{P}(A_n\mathrm{i}\mathrm{.}\mathrm{o}\mathrm{.})=\mathbb{P}\left({\mathrm{lim\hspace{0.17em}sup}}_{n\to \mathrm{\infty }}{A}_n\right)=0.$