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.

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 $X^{-1}(R)={\bigcap }_{i<n}{X}_i^{-1}(B_i)\in \mathcal{F}$. Since [L4] says the product sigma-algebra is generated by such rectangles, $X$ 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 1.1, the law ${\mathbb{P}}_X$ is defined, and [L2] makes it a probability measure on the product sigma-algebra.

For $n=1$, step 1.2 already identifies ${\mathbb{P}}_X$ with ${\mathbb{P}}_{X_0}$. For $n\ge 2$, define recursively ${\nu }_1:={\mathbb{P}}_{X_0}$ and ${\nu }_{m+1}:={\nu }_m\otimes {\mathbb{P}}_{X_m}$. 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.

The measures ${\mathbb{P}}_X$ and ${\nu }_n$ are finite, hence sigma-finite, and step 2.2 shows that they agree on the generating measurable rectangles from [L4]. The uniqueness clause of [L3], applied recursively through the finite product construction, 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. 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 $\mu$. Using [L3] for the measurable map $m$ and then applying [L4] repeatedly on the product measure space yields $\mathbb{E}[{\prod }_{i<n}{g}_i(X_i)]={\int }_{{\mathbb{R}}^n}md\mu ={\prod }_{i<n}{\int }_{\mathbb{R}}yd{\mu }_i={\prod }_{i<n}\mathbb{E}[g_i(X_i)]\mathrm{.}$

Now assume every $g_i(X_i)$ is integrable. Applying step 2.1 to the nonnegative functions $\mathrm{\mid }{g}_i\mathrm{\mid }$ gives $\mathbb{E}[{\prod }_{i<n}\mathrm{\mid }{g}_i(X_i)\mathrm{\mid }]={\prod }_{i<n}\mathbb{E}[\mathrm{\mid }{g}_i(X_i)\mathrm{\mid }]<\mathrm{\infty }\mathrm{.}$ 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 2.1. Step 3.1 shows that the product map $m$ is $\mu$-integrable. By [L2], [L3], and repeated use of [L5], $\mathbb{E}[{\prod }_{i<n}{g}_i(X_i)]={\int }_{{\mathbb{R}}^n}md\mu ={\prod }_{i<n}{\int }_{\mathbb{R}}yd{\mu }_i={\prod }_{i<n}\mathbb{E}[g_i(X_i)]\mathrm{.}$

Step 2.1 proves the nonnegative case, and step 4.1 proves the integrable case.

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 $\mathrm{Cov}(X,Y)=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]=0.$

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 $\mathbb{P}(A_n\mathrm{i}\mathrm{.}\mathrm{o}\mathrm{.})=\mathbb{P}\left({\mathrm{lim\hspace{0.17em}sup}}_{n\to \mathrm{\infty }}{A}_n\right)=0.$

For each $k$, the indicator ${\mathbf{1}}_{A_k}$ is a real random variable by [L2]. Repeated use of [L4] and [L2] gives $\mathbb{E}[S_n]={\sum }_{k=1}^n\mathbb{E}[{\mathbf{1}}_{A_k}]={\sum }_{k=1}^n\mathbb{P}(A_k)=a_n\mathrm{.}$ The divergence hypothesis makes $a_n\to +\mathrm{\infty }$. In particular, there is $N\ge 1$ with $a_n>0$ for every $n\ge N$.

For $i\mathrm{\ne }j$, step 1.1 and [L1] give $\mathbb{E}[{\mathbf{1}}_{A_i}{\mathbf{1}}_{A_j}]=\mathbb{E}[{\mathbf{1}}_{A_i\cap A_j}]=\mathbb{P}(A_i\cap A_j)=\mathbb{P}(A_i)\mathbb{P}(A_j)=\mathbb{E}[{\mathbf{1}}_{A_i}]\mathbb{E}[{\mathbf{1}}_{A_j}]\mathrm{.}$ Also ${\mathbf{1}}_{A_k}^2={\mathbf{1}}_{A_k}$ for every $k$.

By [L3], the partial sum $S_n$ and its square $S_n^2$ are measurable. Expanding $S_n^2$ and using step 2.1 together with linearity from [L4] yields $\mathbb{E}[S_n^2]={\sum }_{k=1}^n\mathbb{P}(A_k)+2{\sum }_{1\le i<j\le n}\mathbb{P}(A_i)\mathbb{P}(A_j)=a_n^2+{\sum }_{k=1}^n(\mathbb{P}(A_k)-\mathbb{P}(A_k{)}^2)\le a_n^2+a_n\mathrm{.}$ So $S_n$ is square-integrable, and [L4] gives $\mathrm{Var}(S_n)=\mathbb{E}[S_n^2]-\mathbb{E}[S_n{]}^2\le a_n\mathrm{.}$

Fix $\epsilon >0$ and $n\ge N$. Applying [L5] to $S_n$ gives $\mathbb{P}(\mid \frac{S_n}{a_n}-1\mid \ge \epsilon )=\mathbb{P}(\mathrm{\mid }{S}_n-a_n\mathrm{\mid }\ge \epsilon a_n)\le \frac{\mathrm{Var}(S_n)}{{\epsilon }^2{a}_n^2}\le \frac{1}{{\epsilon }^2{a}_n}\mathrm{.}$ Hence $S_n\mathrm{/}{a}_n\to 1$ in probability along the defined tail $n\ge N$.

For each integer $m\ge 1$, let $n_m$ be the least index with $a_{n_m}\ge a_1+m^2$; it exists by step 1.1. Since $a_{n_m-1}<a_1+m^2\le a_{n_m}$ and $a_{n_m}-a_{n_m-1}=\mathbb{P}(A_{n_m})\le 1$, one has $a_1+m^2\le a_{n_m}<a_1+m^2+1.$ Therefore step 4.1 yields $\mathbb{P}(\mid \frac{S_{n_m}}{a_{n_m}}-1\mid \ge \epsilon )\le \frac{1}{{\epsilon }^2(a_1+m^2)},$ and the sum over $m$ is finite.

For each integer $r\ge 1$, step 5.1 with $\epsilon =1\mathrm{/}r$ gives ${\sum }_{m=1}^{\mathrm{\infty }}\mathbb{P}(\mid \frac{S_{n_m}}{a_{n_m}}-1\mid \ge \frac{1}{r})<+\mathrm{\infty }\mathrm{.}$ Applying [L6] to these deviation events shows that, for each $r$, only finitely many of them occur almost surely.

Intersect the full-probability events from step 6.1 over all $r\in {\mathbb{N}}_{>0}$. On that still full-probability event, for every $r$ there is $M_r(\omega )$ such that $\mid \frac{S_{n_m}(\omega )}{a_{n_m}}-1\mid <\frac{1}{r}(m\ge M_r(\omega ))\mathrm{.}$ Hence $S_{n_m}\mathrm{/}{a}_{n_m}\to 1$ almost surely.

Fix $\omega$ in the full-probability event from step 7.1. If $n_m\le n<n_{m+1}$, then $S_{n_m}(\omega )\le S_n(\omega )\le S_{n_{m+1}}(\omega )$ and $a_{n_m}\le a_n<a_{n_{m+1}}$, so $\frac{S_{n_m}(\omega )}{a_{n_{m+1}}}\le \frac{S_n(\omega )}{a_n}\le \frac{S_{n_{m+1}}(\omega )}{a_{n_m}}\mathrm{.}$ Since $a_{n_m}\mathrm{/}{a}_{n_{m+1}}\to 1$ and $a_{n_{m+1}}\mathrm{/}{a}_{n_m}\to 1$ by the bounds in step 5.1, step 7.1 squeezes $S_n(\omega )\mathrm{/}{a}_n$ to $1$. Therefore $S_n\mathrm{/}{a}_n\to 1$ almost surely for all sufficiently large $n$, equivalently for all $n$ with $a_n>0$.

Step 8.1 is exactly the asserted frequency law.

Let $S_n:={\sum }_{k=0}^n{\mathbf{1}}_{A_k}$ and $a_n:={\sum }_{k=0}^n\mathbb{P}(A_k)$. The divergence hypothesis makes $a_n\to +\mathrm{\infty }$, and [L1] gives $S_n\mathrm{/}{a}_n\to 1$ almost surely. Therefore on a full-probability event there is $N$ such that $S_n\ge a_n\mathrm{/}2$ for every $n\ge N$, hence $S_n\to +\mathrm{\infty }$.

The partial counts $S_n(\omega )$ diverge to $+\mathrm{\infty }$ exactly when the event $A_n$ occurs for infinitely many indices $n$. By [L2], this is precisely the event $A_n\mathrm{i}\mathrm{.}\mathrm{o}\mathrm{.}$. Since step 1.1 shows it has probability $1$, the second Borel-Cantelli conclusion follows.

The independent sequence $(X_n)$ gives an independent family of sigma-algebras $(\sigma (X_n){)}_{n\in \mathbb{N}}$. Grouping the first $m+1$ coordinates into one block and the remaining coordinates into the other, [L2] shows that ${\mathcal{H}}_m$ and ${\mathcal{G}}_m$ are independent sigma-algebras.

Because $A$ lies in the tail sigma-algebra, [L1] gives $A\in \sigma (X_n:n\ge m+1)={\mathcal{G}}_m$. Step 1.1 therefore yields $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)\mathrm{.}$ So every tail event is independent of every event in the finite initial sigma-algebra.

For $m\in \mathbb{N}$, let ${\mathcal{H}}_m:=\sigma (X_0,\dots ,X_m)$ and put $\mathcal{A}:={\bigcup }_{m\in \mathbb{N}}{\mathcal{H}}_m$. Since the family $({\mathcal{H}}_m)$ is increasing, $\mathcal{A}$ is an algebra of events. Define $\mathcal{D}:=\{B\in \sigma (\mathcal{A}):\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)\}\mathrm{.}$ Using the continuity statements from [L3], the class $\mathcal{D}$ is closed under increasing unions and decreasing intersections, so it is a monotone class.

For each $m$, [L1] gives ${\mathcal{H}}_m\subseteq \mathcal{D}$. Hence the algebra $\mathcal{A}={\bigcup }_m{\mathcal{H}}_m$ is contained in $\mathcal{D}$.

Because $\mathcal{D}$ is a monotone class containing the algebra $\mathcal{A}$, [L2] yields $\sigma (\mathcal{A})\subseteq \mathcal{D}$. But $\sigma (\mathcal{A})=\sigma (X_n:n\ge 0)$, and [L4] puts every tail event in this full-coordinate sigma-algebra. Therefore $A\in \mathcal{D}$, so $\mathbb{P}(A)=\mathbb{P}(A\cap A)=\mathbb{P}(A{)}^2\mathrm{.}$