I've started learning about measure theory and I'm trying to get some intuitive grasp of the basic concepts. This is only succeeding partially so far.
There is an exercise which I don't quite understand. Here it is:
Let $\Omega$ be the set of all sequences $\omega = (\omega_1,\omega_2,\ldots)$ where $\omega_n \in \{0,1\}$ $\forall n \geq 1$. Define for all $n$ the projections $p_n:\Omega \rightarrow \{0,1\}$ and let $\mathcal{F}_n = \sigma(p_1,\ldots,p_n)$. Prove that $\mathcal{F}_1 \subseteq \mathcal{F}_2 \subseteq \ldots$
The course material that I'm using defines the $\sigma$-algebra of a function in the context of borel sets $B \in \mathcal{B}(\mathbb{R})$, as in $\sigma(f) = \{\{f \in B\} : B \in \mathcal{B}(\mathbb{R})\}$ where $\{f \in B \} = \{\omega \in \Omega : f(\omega) \in B\}$
I have two questions about this exercise:
1) Is a $\sigma$-algebra generated by some function $f$ always defined in the context of Borel-algebras? Ie, in the case of our functions $p_i$, should we think of $p_i$ as a function mapping some sequence $\omega$ to $\{0,1\} \subseteq (a,b)$ for some $a,b \in \mathbb{R}$?
2) How should I read $\sigma(p_1)$? Because I'm having trouble connecting the nature of $p_i$ with the aformentioned definition of $\sigma(f)$.
Best Answer
If $f: X \to (Y,\Sigma)$ is a function from a set to any measurable space (a space equipped with a $\sigma$-algebra) then it is easy to check that $\{f^{-1}(S)\,:\,S \in \Sigma\} = \sigma(f)$ is a $\sigma$-algebra (sorry, I can't bring myself to writing $\{f \in \Sigma\}$ for the pre-image $f^{-1}(S)$ of $S$ even if there's some justification). This is because taking pre-images commutes with unions, intersections and complements. By definition of measurability, $\sigma(f)$ is the smallest $\sigma$-algebra $\mathcal{A}$ on $X$ making $f$ measurable, i.e. $\sigma(f)$ is the smallest $\sigma$-algebra $\mathcal{A}$ such that $f^{-1}(S) \in \mathcal{A}$ for all $S \in \Sigma$. So the answer to your first question is no.
Now if $f,g: X \to (Y,\Sigma)$ are two functions, then $\sigma(f,g)$ is the smallest $\sigma$-algebra making both $f$ and $g$ measurable. It doesn't have such an easy description in general, but it is clear that $\sigma(f), \sigma(g) \subset \sigma(f,g)$ and actually $\sigma(f,g)$ is the intersection of all $\sigma$-algebras containing both $\sigma(f)$ and $\sigma(g)$. In fact, if $\mathcal{S}$ is any collection of subsets of $X$ then it generates a $\sigma$-algebra $\sigma(\mathcal{S}) = \bigcap_{\mathcal{S} \subset \mathcal{A}} \mathcal{A}$, where the intersection is taken over all $\sigma$-algebras containing $\mathcal{S}$. Since the power set of $X$ contains $\mathcal{S}$ and is a $\sigma$-algebra, this intersection is non-empty. You should convince yourself that an arbitrary intersection of $\sigma$-algebras is again a $\sigma$-algebra.
In your example, you have $X = Y^{\mathbb{N}}$ and in this situation the $\sigma$-algebra $\sigma(p_{1},\ldots,p_{n})$ has a semi-concrete description. Note that a pre-image of $p_{i}$ is of the form $\underbrace{Y \times \cdots \times Y}_{(i-1)\;\text{times}} \times S_{i} \times Y \times Y \times \cdots$, where $S_{i} \subset Y$ is arbitrary. By taking the intersection of the sets $p_{i}^{-1}(S_{i})$ this means that all the sets of the form $S_{1} \times \cdots \times S_{n} \times Y \times Y \times \cdots$ with $S_{1},\ldots,S_{n} \in \Sigma$ must belong to $\sigma(p_{1},\cdots,p_{n})$ and in fact, these so-called cylinder sets generate the $\sigma$-algebra $\sigma(p_{1},\cdots,p_{n})$.