# Help with a sigma-algebra problem with random variables (show $\sigma(X_S)\subseteq \sigma(X_T)$ if $S\subseteq T$)

My problem is as follows:

Let $$X_S$$ and $$X_T$$ be two stochastic processes where $$S,T$$ are index sets. Let $$\sigma(X_S)$$ and $$\sigma(X_T)$$ denote the sigma-algebra generated by $$X_S$$ and $$X_T$$. Show that $$\sigma(X_S)\subseteq \sigma(X_T)$$ if $$S\subseteq T$$.

Here, sigma-algebra generated by a random variable is defined as the following.

Suppose $$X:(\Omega,\cal{F}) \to$$ $$(E,\cal E)$$, then the sigma-algebra generated by $$X$$ is $$\sigma(X)=\sigma\{X^{-1}(A):A\in \cal E \}$$, i.e. the sigma-algebra generated by $$\{X^{-1}(A):A\in \cal E \}$$.

This claim is actually very intuitive, but I need help to write a formal proof. I can give an intuitive example of this claim for the case of finite index sets.

Suppose $$S=\{1\}$$ and $$T=\{1,2\}$$ and so $$X_S=\{X_1\}$$ and $$X_T=\{X_1,X_2\}$$. Let $$\Omega$$ be a set of students, and treat $$X_1,X_2$$ as some properties for each $$\omega\in \Omega$$. Now let $$X_1(\Omega)=\{male, female\}$$, $$X_2(\Omega)=\{undergrad, grad\}$$.

Clearly $$\sigma(X_S)=\{\emptyset, \Omega, m, f\}$$ where "$$m$$" denotes the set of all male students for convenience and so is "$$f$$" for the set of all female students.

Likewise, $$\sigma(X_T)=\{\emptyset, \Omega,m\&u, f\&u,m\&g,f\&g,m,f,u,g\}$$ where "$$u$$" denotes the set of all undergraduate students and "$$g$$" denotes the set of all graduate students. I use things like $$m\&g$$ to represent students carrying both properties.

Thus we can see $$\sigma(X_S)\subseteq \sigma(X_T)$$.

To write out a formal proof, we need to show for any $$A\in \sigma(X_S)$$ we have $$A\in \sigma(X_T)$$. I can intuitively explain this is true but anyone can help with a formal proof? I am new to such stuff so I really need help. Thank you!

Let $(\Omega,\mathcal{A})$ and $(E,\mathcal{F})$ be measurable spaces and $X: (\Omega,\mathcal{A}) \to (E,\mathcal{F})$ be a mapping. Then $X$ is measurable if, and only if,
$$X^{-1}(F) \in \mathcal{A}$$
for any $F \in \mathcal{F}$. Since, by definition, $\sigma(X)$ is the smallest $\sigma$-algebra containing all these sets, this means that $\sigma(X)$ is the smallest $\sigma$-algebra $\mathcal{A}'$ such that $X: (\Omega,\mathcal{A}') \to (E,\mathcal{F})$ is measurable. If we want to show the relation $\sigma(X) \subseteq \mathcal{A}'$ for some sigma-algebra $\mathcal{A}'$, it therefore suffices to check that $X: (\Omega,\mathcal{A'}) \to (E,\mathcal{F})$ is measurable.
So, suppose that $S \subseteq T$. Then, by definition of $\mathcal{A}' := \sigma(X_T)$, we know that $X_t: (\Omega,\mathcal{A}') \to (E,\mathcal{F})$ is measurable for any $t \in T$. In particular, for $s \in S \subseteq T$, we get that $$X_s: (\Omega,\mathcal{A}') \to (E,\mathcal{F})$$ is measurable. Hence, by the above considerations, $\sigma(X_s) \subseteq \sigma(X_T)$. Since this holds for any $s \in S$, this proves $\sigma(X_S) \subseteq \sigma(X_T)$.