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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!

Tim
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  • For measure space, $\pi$ being measurable doesn't make sense since $E_i$ is not a topological space. For topological spaces you are right. –  Jan 13 '13 at 00:06
  • @Sanchez: What I meant is measurable for sigma algebra case, and continuous for topology case, not mixing them up. – Tim Jan 13 '13 at 00:08
  • 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. –  Jan 13 '13 at 00:10
  • On the other hand, if we purely focus on whether the two sets generate the same sigma algebra, then yes. –  Jan 13 '13 at 00:12
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    @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? – Tim Jan 13 '13 at 00:17
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    @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. – Paul VanKoughnett Jan 13 '13 at 00:26
  • @Paul VanKoughnett, this definition of measurability is unfamiliar to me - when I learnt measure theory, the codomain I cared about is usually $\mathbb{C}$ or $\mathbb{R}$ with usual topology. So I always think that measurability of a function requires the codomain being a topological space. Sorry for the mistake. –  Jan 13 '13 at 00:33
  • @Sanchez: no apology necessary! – Paul VanKoughnett Jan 14 '13 at 03:16

1 Answers1

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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 $(2) = (1)$.

  • 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? – Tim Jan 13 '13 at 00:59
  • @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 $E_i$. –  Jan 13 '13 at 01:01
  • 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. – Tim Jan 13 '13 at 02:17
  • @Tim, I agree with what you say. –  Jan 13 '13 at 02:20
  • Do you know why many books use (1) without mentioning (2)? – Tim Jan 13 '13 at 03:12
  • @Tim, at least for the case of topological spaces, (1) would give you a basis rather than a subbasis, which could be more useful in various situations. –  Jan 13 '13 at 03:15
  • Thanks! What is the advantage of (1) over (2) for measurable space? – Tim Jan 13 '13 at 04:11
  • @Tim From a categorial perspective, the first definition makes it easy to show that the product(structure) is a product as defined in category theory, so people with that background might prefer it. – Michael Greinecker Jan 13 '13 at 10:11
  • @MichaelGreinecker: Thanks! Then I guess, in category theory, product does not necessarily make projections (if defined) structure-preserving? – Tim Jan 13 '13 at 15:43
  • @Tim They do. Take a look at their wikipedia page. – Michael Greinecker Jan 13 '13 at 15:44
  • @MichaelGreinecker: For product of set system of some type, I think the second definition is the one equivalent to making the projections structure-preserving. The first definition is equivalent to the second one only when the type of set system is defined to be closed under intersection. Suppose the set sytem is not sigma algebra nor topology, and not necessarily closed under intersection, which definition do you think is still same as the one making projections preserve its structure, (1) or (2)? (2) I think? (note: I made a typo in my earlier comment to Saches) – Tim Jan 13 '13 at 16:01
  • @Tim I don't know how I would define "preserving structure" in general. – Michael Greinecker Jan 13 '13 at 16:09
  • @MichaelGreinecker: By a mapping being structure-preserving, I mean like a measurable mapping and a continuous mapping. – Tim Jan 13 '13 at 16:28
  • @Tim That is not a definition. – Michael Greinecker Jan 13 '13 at 16:38
  • @MichaelGreinecker: Given two sets $X$ and $Y$, each with a set system of the same type, $F_X$ and $F_Y$, a mapping $f: X\to Y$ is structure preserving, if for every member $A \in F_Y$, $f^{-1}(A) \in F_X$. – Tim Jan 13 '13 at 16:42
  • let us continue this discussion in chat – Tim Jan 14 '13 at 15:03