[proofplan] Fix a time $t\in T$ and compare the generators of the natural filtration $\mathcal F_t^X$ with the sigma-algebra $\mathcal G_t$. The hypothesis that every earlier [random variable](/page/Random%20Variable) $X_s$ is $\mathcal G_t$-measurable implies that every event of the form $X_s^{-1}(A)$, with $s\leq t$ and $A\in\mathcal E$, belongs to $\mathcal G_t$. Since $\mathcal F_t^X$ is the sigma-algebra generated by precisely these events, the minimality property of generated sigma-algebras gives $\mathcal F_t^X\subset\mathcal G_t$. [/proofplan] [step:Fix a time and name the generating family of past observation events] Fix $t\in T$. Define the family of past observation events \begin{align*} \mathcal A_t := \{X_s^{-1}(A): s\in T,\ s\leq t,\ A\in\mathcal E\}. \end{align*} By the definition of the natural filtration, \begin{align*} \mathcal F_t^X = \sigma(\mathcal A_t). \end{align*} Thus it remains to prove that the sigma-algebra $\mathcal G_t$ contains every element of $\mathcal A_t$, because any sigma-algebra containing a family of sets also contains the sigma-algebra generated by that family. [/step] [step:Use measurability into $\mathcal G_t$ to place every generator inside $\mathcal G_t$] Let $B\in\mathcal A_t$. By the definition of $\mathcal A_t$, there exist an index $s\in T$ with $s\leq t$ and a measurable set $A\in\mathcal E$ such that \begin{align*} B = X_s^{-1}(A). \end{align*} The hypothesis says that $X_s:(\Omega,\mathcal G_t)\to(E,\mathcal E)$ is measurable. By the definition of measurability, the preimage under $X_s$ of every set in $\mathcal E$ belongs to $\mathcal G_t$. Applying this to the set $A\in\mathcal E$ gives \begin{align*} X_s^{-1}(A)\in\mathcal G_t. \end{align*} Therefore $B\in\mathcal G_t$. Since $B\in\mathcal A_t$ was arbitrary, we have \begin{align*} \mathcal A_t\subset\mathcal G_t. \end{align*} [guided] We want to prove that $\mathcal F_t^X$ is contained in $\mathcal G_t$. Since $\mathcal F_t^X$ is generated by the information contained in the random variables $X_s$ for $s\leq t$, the first task is to check that every basic event determined by one of those random variables already lies in $\mathcal G_t$. Let $B\in\mathcal A_t$. By the definition of the family $\mathcal A_t$, there are an index $s\in T$ with $s\leq t$ and a measurable set $A\in\mathcal E$ such that \begin{align*} B = X_s^{-1}(A). \end{align*} The assumption of the theorem applies exactly to this pair $(s,t)$: because $s\leq t$, the map $X_s:(\Omega,\mathcal G_t)\to(E,\mathcal E)$ is measurable. Measurability means that every measurable subset of the target has a preimage in the source sigma-algebra. Since $A\in\mathcal E$, we obtain \begin{align*} X_s^{-1}(A)\in\mathcal G_t. \end{align*} Using $B=X_s^{-1}(A)$, this gives $B\in\mathcal G_t$. The event $B$ was an arbitrary element of $\mathcal A_t$, so every generator of the natural filtration at time $t$ belongs to $\mathcal G_t$: \begin{align*} \mathcal A_t\subset\mathcal G_t. \end{align*} This is the essential content of the argument: $\mathcal G_t$ already records every event observable from the process up to time $t$. [/guided] [/step] [step:Invoke minimality of the generated sigma-algebra] Because $(\mathcal G_t)_{t\in T}$ is a filtration, $\mathcal G_t$ is a sigma-algebra on $\Omega$. The previous step proved $\mathcal A_t\subset\mathcal G_t$. By the defining minimality property of the generated sigma-algebra, $\sigma(\mathcal A_t)$ is contained in every sigma-algebra on $\Omega$ that contains $\mathcal A_t$. Hence \begin{align*} \mathcal F_t^X = \sigma(\mathcal A_t)\subset\mathcal G_t. \end{align*} Since $t\in T$ was arbitrary, this holds for every $t\in T$, which proves the theorem. [/step]