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Consider $\mathbb{R}$ with the $\sigma$-algebra of Borel sets, and $\mathbb{R}^\mathbb{R}$ with the product $\sigma$-algebra(see p.22 of 'Real Analysis - Gerald B. Folland'). Does $[0,1]^\mathbb{R} \subset \mathbb{R}^\mathbb{R}$ belong to the product $\sigma$-algebra?

Here is an answer I got from Kevin:

"$[0,1]^\mathbb{R}$ is not in the product $\sigma$-algebra $M$. The generators of $M$ are of the form $\pi_x^{-1}(A)$ for $x\in \mathbb{R}$ and $A\subset \mathbb{R}$, where the projection $\pi_x(f)=f(x)$. What's $\pi_x^{-1}(A)$? Well, it's the set of all functions $f:\mathbb{R}\to\mathbb{R}$ with $f(x)\in A$. We could take the intersection of countably many of these to get, for instance, the functions which map $\mathbb{Q}\to [0,1]$, but $[0,1]^{\mathbb{R}}$ could only be written as an uncountable intersection of such sets, which is not permitted in a $\sigma$-algebra."

But now how is the intersection of two functions defined? And how can I prove that $[0,1]^{\mathbb{R}}$ cannot be written as a countable intersection of such functions? Can anyone please help me?

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I think that your Kevin is a tad sloppy here (although the essence of his argument is correct) because not all elements of the product $\sigma$-algebra are countable intersections of generators.

(And when he talks about "the intersection of countably many of these" he's talking about the intersection of sets of functions - of the kind he described - not of functions.)


Anyway, here's my attempt at giving a cleaner proof:

The generators of the product $\sigma$-algebra $\Sigma$ on $X=\mathbb{R}^\mathbb{R}$ are sets of the form $\pi_x^{-1}[C]$ where $x$ is some real number and $C$ is an element of the Borel $\sigma$-algebra $\Sigma_B$ on $\mathbb{R}$ (and thus a subset of $\mathbb R$).

It is a well-known fact that if a set $A$ is in the $\sigma$-algebra generated by a set $\cal G$ of generators, then we can find a countable subset ${\cal G}'$ of $\cal G$ such that $A$ is in the $\sigma$-algebra generated by ${\cal G}'$.

So, let's assume $A=[0,1]^\mathbb{R}$ were an element of $\Sigma$. Then we could find a sequence $(x_n)_{n\in\mathbb N}$ of reals and for each $x_n$ a set $C_n$ of $\Sigma_B$ such that $A$ were already in the $\sigma$-algebra $\Sigma'$ generated by ${\cal G}'=\{\pi_{x_n}^{-1}[C_n] \mid n \in \mathbb{N} \}$. W.l.o.g. let's assume that $C_0=[0,1]$ (or otherwise we just add it) and that all $C_n$ are distinct from $\mathbb R$ (because otherwise $\pi_{x_n}^{-1}[C_n]$ would just be $X$).

As $\mathbb R$ is uncountable, we can find an $x^\ast \in \mathbb R$ which is none one of the $x_n$ of our sequence. Now consider $$ F=\{ f \in X \mid f(x^\ast) \in [0,1] \land f(x_n) \notin C_n \text{ for all } n \in \mathbb{N} \} $$ and let $$ {\cal F} = \{ G \subseteq X \mid F \subseteq G \lor F \cap G = \emptyset \}. $$ Then ${\cal F}$ is a $\sigma$-algebra on $X$ and a superset of ${\cal G}'$ (because each $\pi_{x_n}^{-1}[C_n]$ is by definition disjoint from $F$) which implies $\Sigma' \subseteq {\cal F}$. But $A$ is neither a superset of $F$ (because $f(x_0)\notin C_0=[0,1]$ for all $f\in F$) nor is $A$ disjoint from $F$ (because $f(x^\ast)\notin[0,1]$ for all $f\notin F$), so $A \notin {\cal F}$ which is a contradiction.


NB: Although (I think) Kevin didn't want to intersect functions, there's no technical problem in doing that. Functions are sets of ordered pairs and you can of course intersect two sets. But if $f$ and $g$ are two functions with the same domain $A$, then $f\cap g$ will either be $f$ (if $f=g$) or a function the domain of which is a proper subset of $A$. This is most likely not what you want or expect.

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