Question on the notion of a $\sigma$-algebra generated by a function

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 $\{f \in B \} = \{\omega \in \Omega : f(\omega) \in B\}$

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 $\sigma(f)$.

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 $\sigma(p_{1},\cdots,p_{n})$.

The answer to (1) is no (sorry, Morning... :-)) since the only structure one needs is a $\sigma$-algebra on the image set.

More in details, let $\Omega$ denote any set and $(E,\mathcal{E})$ any measurable space (this means only that $E$ is a set and that $\mathcal{E}$ is a $\sigma$-algebra on $E$). For any function $f:\Omega\to E$, the $\sigma$-algebra generated by $f$ is defined as $\sigma(f)=\{\{f\in A\},A\in\mathcal{E}\}$, where $\{f\in A\}=f^{-1}(A)$.

Of course, if $\Omega$ is itself a measurable space, that is, if one is given a $\sigma$-algebra $\mathcal{F}$ on $\Omega$, then $f$ is a random variable from $(\Omega,\mathcal{F})$ to $(E,\mathcal{E})$ iff $\sigma(f)\subset\mathcal{F}$. This condition means exactly (surprise, surprise...) that $f^{-1}(A)\in\mathcal{F}$ for every $A\in\mathcal{E}$.

• I think when one talks about a function, then `by default' it takes values in $\mathbb R$, equipped with ${\cal B}(\mathbb R)$ (surely, one can equip $\mathbb R$ with a different $\sigma$-algebra... but this would always be specified). In general, in the case you discussed here I believe $f$ is referred to as a (measurable) mapping to $(E,{\cal E})$. This, of course, is a more general case. – Morning Feb 26 '11 at 23:48
• @Morning Not sure I agree with your terminology, see en.wikipedia.org/wiki/Function_%28mathematics%29 for example (or Theo's answer...). Thanks for your comment. – Did Feb 27 '11 at 6:51

1) By default, yes to the first question. As for the second one, note that $\{0,1\}\subset\mathbb R$, hence $p_i$ is a mapping from $\Omega$ to $\mathbb R$. No need to introduce $(a,b)$ here.

2) If you write out $p_i:\Omega\ni \omega = \{\omega_n\}_{n\in\mathbb N} \mapsto \omega_i\in\{0,1\}\subset\mathbb R$, then $\sigma(p_i)$ should be clearly defined as before. Note that here, $\Omega$ is usually equipped with the product $\sigma$-algebra. For example, $p_1^{-1}(\{0\}) = \{0\}\times\{0,1\}^{\mathbb N}$.

Actually, you don't need to write $\sigma(p_i)$ out to solve the exercise, since by definition, $\sigma(f)\subset\sigma(f,g)$ for arbitrary measurable functions $f,g$.

• Ah you're right, since $\mathbb{R}$ is open ofcourse. But for my own understanding I'd still like to understand what $\sigma(p_i)$ looks like. – Stijn Feb 26 '11 at 15:45