Let $(E_i, \mathcal{B}_i)$ be measurable (or topological) spaces, where $i \in I$ is an index set, possibly infinite. Their product sigma algebra (or product topology) $\mathcal{B}$ on $E= \prod_{i \in I} E_i$ is defined to be the coarsest one that can make the projections $\pi_i: E \to E_i$ measurable (or continuous).

Many sources said the following is an equivalent definition: $$\mathcal{B}=\sigma \text{ or }\tau\left(\left\{\text{$\prod_{i \in I}B_i$, where $B_i \in \mathcal{B}_i, B_i=E_i$ for all but a finite number of $i \in I$}\right\}\right),$$ where $\sigma \text{ and }\tau$ mean taking the smallest sigma algebra and taking the smallest topology. Honestly I don't quite understand why this is the coarsest sigma algebra (or topology) that make the projections measurable (or continuous).

Following is what I think is the coarsest one that can make the projections measurable

$$\mathcal{B}=\sigma \text{ or }\tau\left(\left\{\text{$\prod_{i \in I}B_i$, where $B_i \in \mathcal{B}_i, B_i=E_i$ at least for all but one $i \in I$}\right\}\right),$$ because $\pi^{-1}_k (E_k) = \text{$\prod_{i \in I}B_i$, where $B_i=E_i$ for all $i \neq k$}$. So I was wondering if the two equations for $\mathcal{B}$ are the same?

Thanks and regards!

  • $\begingroup$ For measure space, $\pi$ being measurable doesn't make sense since $E_i$ is not a topological space. For topological spaces you are right. $\endgroup$ – user27126 Jan 13 '13 at 0:06
  • $\begingroup$ @Sanchez: What I meant is measurable for sigma algebra case, and continuous for topology case, not mixing them up. $\endgroup$ – StackExchange for All Jan 13 '13 at 0:08
  • $\begingroup$ I know what you mean. Remember that measurability is only defined for a map from a measure space to a topological space, not from a measure space to a measure space. $\endgroup$ – user27126 Jan 13 '13 at 0:10
  • 3
    $\begingroup$ @Sanchez: (1) "measurability is only defined for a map from a measure space to a topological space, not from a measure space to a measure space." I don't think this is true. (2) "if we purely focus on whether the two sets generate the same sigma algebra, then yes." How is that true? $\endgroup$ – StackExchange for All Jan 13 '13 at 0:17
  • 3
    $\begingroup$ @Sanchez: Are you reading Rudin? He only defines measurability for maps from a measure space to a topological space, but it can be defined for maps between two measure spaces: a map is measurable if the inverse image of every measurable set is measurable. Rudin's definition is equivalent to this one if we replace the topological space with the Borel measurable space it defines. $\endgroup$ – Paul VanKoughnett Jan 13 '13 at 0:26

For the comments: I retract my error for the definition of measurability. Sorry.

For the two things generating the same sigma algebra (or topology, which is similar):

We use $\langle - \rangle$ to denote the smallest sigma algebra containing the thing in the middle. We want to show that $$(1) \hspace{5mm}\langle \prod_{i} B_i \rangle$$ where $B_i \in \mathcal{B}_i$, and $B_i = E_i$ for all but finitely many $i$s, is the same as $$(2) \hspace{5mm}\langle \prod_{i} B_i \rangle$$ where $B_i \in \mathcal{B}_i$, and $B_i = E_i$ for all but one $i$.

It is clear that $(2) \subset (1)$, since the generating collection in (2) is a subset of that of (1).

On the other hand, $(2)$ contains $\prod_{i} B_i$ where $B_i \in \mathcal{B}_i$, and $B_i = E_i$ for all but finitely many $i$s, since it is the (finite) intersection of the generators. For example, $$B_1 \times B_2 = (B_1 \times E_2) \cap (E_1 \times B_2)$$ So $(2) \supset (1)$. Therefore $(2) = (1)$.

  • $\begingroup$ Thanks! When you prove (2)⊃(1) for sigma algebra, isn't (2) contains $∏_iB_i$ where $B_i=E_i$ for all but countably many many, becaue sigma algebra is closed under countable intersection, instead of finite intersection? $\endgroup$ – StackExchange for All Jan 13 '13 at 0:59
  • $\begingroup$ @Tim, That's also true. But for the purpose of showing the sigma algebra (2) contains all the generators of (1), it suffices to look at finite intersections. It is of course true that the sigma algebra (either (1) or (2)) would contain products with all but countably many pieces being $E_i$. $\endgroup$ – user27126 Jan 13 '13 at 1:01
  • $\begingroup$ Thanks! Suppose the set sytem is not sigma algebra nor topology, and it is not necessarily closed under intersection, which one do you think is still same as the one making projections preserve its structure, (1) or (2)? I think (2) is only when the set system is closed under intersection, while (1) is always. $\endgroup$ – StackExchange for All Jan 13 '13 at 2:17
  • $\begingroup$ @Tim, I agree with what you say. $\endgroup$ – user27126 Jan 13 '13 at 2:20
  • $\begingroup$ Do you know why many books use (1) without mentioning (2)? $\endgroup$ – StackExchange for All Jan 13 '13 at 3:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.