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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?

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    $\begingroup$ I guess it does not belong since $[0,1]$ is a closed interval. But no idea how to prove it. $\endgroup$ – user174318 Sep 14 '14 at 2:11
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Claim: every set $A$ in $\mathbb{R}^\mathbb{R}$ is of the form $$A = \{f : (f(x_1),f(x_2),\dots, f(x_n), \dots) \in \mathcal{B}(\mathbb{R}^\mathbb{N})\},$$ where $\mathcal{B}(\mathbb{R}^\mathbb{N})$ is the product $\sigma$-algebra on $\mathbb{R}^\mathbb{N}$ and $\{x_i\}_{i=1}^\infty$ is countable.

In other words, we can only specify a measurable set at countably many points. This will immediately yield your question by a contradiction argument: if $[0,1]^\mathbb{R}$ is measurable and non-empty, it contains some $f : \mathbb{R} \rightarrow \mathbb{R}$ and is specified only at countably many points. Change $f$ at one of the points not specified so that it does not lie in $[0,1]$ to obtain a contradiction to $f \in [0,1]^{\mathbb{R}}$.

To see the claim, notice that if $\mathcal{E}$ is a family of sets, the collection $\mathcal{C}(\mathcal{E})$ obtained from $\mathcal{E}$ through finitely many applications of complementation and countable union and intersection is a $\sigma$-algebra. In other words, if we assume w.l.o.g. that $\mathcal{E}$ contains all complements (i.e. $A \in \mathcal{E} \Rightarrow A^c \in \mathcal{E})$, then any $A \in \mathcal{C}(\mathcal{E})$ can be written as $$\{A : A = \cap_{i_1 \geq 1}\cup_{i_2\geq 1} \dots \cap_{i_n \geq 1} E_{i_1, \dots, i_n}, 1 \leq n < \infty, E_{i_1,\dots, i_n} \in \mathcal{E}\}.$$

In our case, the product $\sigma$-algebra on $\mathbb{R}^\mathbb{R}$ is $\sigma(\mathcal{E})$ with $\mathcal{E} = \{\pi^{-1}_x(A) : x \in \mathbb{R}, A \in \mathcal{B}(\mathbb{R})\}$. Since $\pi_x^{-1}(A) = \{f : f(x) \in A\}$ and $\sigma(\mathcal{E}) \subset \mathcal{C}(\mathcal{E})$, any $E \in \sigma(\mathcal{E})$ is specified by at most countably many points, as in the claim.

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This has nothing to do with the closedness of $[0,1]$, but you're right that $[0,1]^\mathbb{R}$ 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.

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  • $\begingroup$ How do you define the intersection of two functions from $\mathbb{R}$ to $\mathbb{R}$? $\endgroup$ – user174318 Sep 14 '14 at 4:28
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    $\begingroup$ And also how can I prove that $[0,1]^\mathbb{R}$ could only be written as an uncountable intersection? Or in other words how can I prove that $[0,1]^\mathbb{R}$ cannot be written as a countable intersection? $\endgroup$ – user174318 Sep 14 '14 at 4:47
  • $\begingroup$ You don't define the intersection of two functions, but two sets of functions. The countable intersection of generating sets $\{f:f(x_i\in A_i\}$ is just the functions sending $x_i$ into $A_i$ for all i. But there are only countably many of these conditions, so these $f$ can do anything off of a countable set. You should be able to extend this argument to include unions with no trouble, and the generators are already closed under complement $\endgroup$ – Kevin Carlson Sep 14 '14 at 19:48

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