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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 $\sigma(f)$.

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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 $\{f^{-1}(S)\,:\,S \in S\} = \sigma(f)$ is a $\sigma$-algebra (sorry, I can't bring myself to writing $\{f \in S\}$ 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 $\sigma(p_{1},\cdots,p_{n})$.

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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 $A\in\mathcal{E}$.

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Very helpful, thanks. –  Stijn Feb 26 '11 at 16:05
    
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
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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 $f,g$.

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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
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