Proof of Increments Are Independent of the Natural Filtration Past

Fix real numbers $0\le s<t$. By the single-random-variable description in Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras, every member of $\sigma(X_t-X_s)$ has the form $E=(X_t-X_s)^{-1}(B_0)$ for a Borel set $B_0$ (differences of random variables are random variables by the argument recorded in Stochastic Process, Independent Increments, and Inhomogeneous Poisson Process).

Step 1 (A generating $\pi$-system). Let $p$ range over the natural numbers, let $0\le u_1<\dots<u_p\le s$ be real numbers, and let $C_1,\dots,C_p$ be Borel sets; let $\mathcal{C}$ be the family of all sets of the form

with such data. $\mathcal{C}$ is closed under pairwise intersection: given two such sets, take the union of their time sets, and where a time $u$ occurs in both use $X_u^{-1}(C)\cap X_u^{-1}(C')=X_u^{-1}(C\cap C')$. Every generator $X_u^{-1}(C)$ ($u\le s$, $C$ Borel) of the natural filtration $\sigma$-algebra $\mathcal{F}^X_s$ belongs to $\mathcal{C}$ (take $p=1$), and $\mathcal{C}\subseteq\mathcal{F}^X_s$ since $\sigma$-algebras are closed under finite intersections; hence, by minimality of generated $\sigma$-algebras, $\sigma(\mathcal{C})=\mathcal{F}^X_s$.

Step 2 (Independence on the $\pi$-system). Fix $A\in\mathcal{C}$ as above and $E=(X_t-X_s)^{-1}(B_0)$. Let $0=w_0<w_1<\dots<w_q=s$ be the strictly increasing enumeration of $\{0,s\}\cup\{u_1,\dots,u_p\}$ (if $s=0$, then $q=0$, the grid is the single point $0$, and every $u_i=0$). Consider the increments $D_j=X_{w_j}-X_{w_{j-1}}$ for $1\le j\le q$, and set $D_{q+1}=X_t-X_s$. These are increments over the consecutive intervals of the grid $w_0<w_1<\dots<w_q<t$, so they are mutually independent by the independent increments hypothesis. If $q\ge1$, the grouping lemma applied to the independent family $(D_j)_{j\in\{1,\dots,q+1\}}$ with the disjoint blocks $I_1=\{1,\dots,q\}$ and $I_2=\{q+1\}$ shows that $\sigma(D_1,\dots,D_q)$ and $\sigma(D_{q+1})$ are independent.

For $0\le j\le q$ define $S_j=c+D_1+\cdots+D_j$ (so $S_0=c$, a constant). Each $S_j$ with $j\ge1$ is $\sigma(D_1,\dots,D_q)$-measurable: sums of measurable functions are measurable by the rational-decomposition argument recorded in the preliminaries of Square-Integrable Random Variables and the Mean-Square Inner Product, applied on the measurable space $(\Omega,\sigma(D_1,\dots,D_q))$; and $S_0$ is measurable with respect to every $\sigma$-algebra. Each $u_i$ equals $w_{j(i)}$ for a unique index $j(i)\in\{0,\dots,q\}$; set

Then $A'\in\sigma(D_1,\dots,D_q)$ if $q\ge1$, while for $q=0$ we have $A'\in\{\emptyset,\Omega\}$.

Telescoping gives $X_{w_j}=X_0+D_1+\cdots+D_j$ exactly, so on the event $\{X_0=c\}$ we have $X_{u_i}=S_{j(i)}$ for every $i$; hence the symmetric difference satisfies $A\,\triangle\,A'\subseteq\{X_0\ne c\}$, an event of probability $0$ by the almost sure hypothesis. For any events $F,F',G$ with $P(F\triangle F')=0$ we get $P(F\cap G)=P(F'\cap G)$: indeed $F\cap G\subseteq(F'\cap G)\cup(F\triangle F')$, so $P(F\cap G)\le P(F'\cap G)+P(F\triangle F')$ by monotonicity and finite subadditivity of the measure $P$ (both consequences of countable additivity), and symmetrically. Taking $G=E$ and $G=\Omega$:

where the middle equality holds by the independence of $\sigma(D_1,\dots,D_q)$ and $\sigma(D_{q+1})\ni E$ established above when $q\ge1$, and trivially when $A'\in\{\emptyset,\Omega\}$ (if $A'=\Omega$ then $P(A'\cap E)=P(E)$; if $A'=\emptyset$ both sides vanish).

Step 3 (Dynkin argument). Fix $E\in\sigma(X_t-X_s)$ and let

Then $\Omega\in\Lambda_E$; if $A_1\subseteq A_2$ both lie in $\Lambda_E$ then $P((A_2\setminus A_1)\cap E)=P(A_2\cap E)-P(A_1\cap E)=(P(A_2)-P(A_1))P(E)=P(A_2\setminus A_1)P(E)$ by finite additivity, so $A_2\setminus A_1\in\Lambda_E$ (in particular $\Lambda_E$ is closed under complements); and if $A_m\uparrow A$ with all $A_m\in\Lambda_E$, then by continuity of $P$ from below (countable additivity applied to the disjoint sets $A_{m+1}\setminus A_m$) we get $P(A\cap E)=\lim_m P(A_m\cap E)=\lim_m P(A_m)P(E)=P(A)P(E)$. Thus $\Lambda_E$ is a $\lambda$-system in the sense of Dynkin's Pi-Lambda Theorem, and by Step 2 it contains the $\pi$-system $\mathcal{C}$. Dynkin's theorem gives $\mathcal{F}^X_s=\sigma(\mathcal{C})\subseteq\Lambda_E$.

Since $E\in\sigma(X_t-X_s)$ was arbitrary, $P(A\cap E)=P(A)P(E)$ for all $A\in\mathcal{F}^X_s$ and $E\in\sigma(X_t-X_s)$, which is exactly independence of the two $\sigma$-algebras in the sense of Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras (for two $\sigma$-algebras the finite-subfamily condition there reduces to this product identity).

Step 4 (Consequence). If $Y$ is $\mathcal{F}^X_s$-measurable, then $\sigma(Y)=\{Y^{-1}(B):B\ \text{Borel}\}\subseteq\mathcal{F}^X_s$ by Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras, so every pair of events from $\sigma(X_t-X_s)$ and $\sigma(Y)$ satisfies the product identity; by the final paragraph of Sigma-Algebra Generated by Random Variables and Independence of Sigma-Algebras, the random variables $X_t-X_s$ and $Y$ are independent. $\square$