3
$\begingroup$

Given a $\sigma$-algebra on a Cartesian product of a collection of sets, do there always exist a $\sigma$-algebra on each set, so that their product $\sigma$-algebra on their Cartesian product will be the same as the given $\sigma$-algebra?

If yes, how to construct the $\sigma$-algebras on the sets from the $\sigma$-algebra on their Cartesian product?

If no, when they exist, how to construct the component $\sigma$-algebras?

Thanks and regards! Any reference is appreciated as well.

  • 0
    You may need to specify if the collection is finite or infinite.2011-02-21
  • 0
    @Arturo: the collection is arbitrary. But if the treatment is different for finite and for infinite cases, I would like to hear for both.2011-02-21
  • 0
    @Tim: I don't know off the top of my head, I'd need to try to remember what the $\sigma$-algebra on an infinite product is defined to be. Remember, for example, the issues with infinite products in topology: the product topology in the infinite product is not the infinite product of the topologies.2011-02-21
  • 1
    @Tim: Yes, according to planetmath (http://planetmath.org/encyclopedia/InfiniteProductMeasure.html) you define the product $\sigma$-algebra much like you do in topology: the generating sets have to have almost all components equal to the entire thing. So it's not the "box" $\sigma$-algebra.2011-02-21
  • 0
    @Arturo: Thanks! I understood how to generate product $\sigma$-algebra from component ones. Here I would like to know the construction in the reverse direction.2011-02-21
  • 0
    @Tim: Yes; I was just wary of the fact that even if it worked for finitely many, it might not work for infinitely many even if it worked for finitely many (e.g., if you had a lot of spaces, the "box" $\sigma$-algebra would have too many sets to be a product $\sigma$-algebra). But in fact, it doesn't even work for finitely many, so it doesn't matter.2011-02-21

2 Answers 2

4

Let $A=\{0,1\}$, $B=\{0,1\}$, $\mathbb{S}=\{\emptyset, \{(0,0),(1,1)\}, \{(1,0),(0,1)\}, A\times B\}$. Then $\mathbb{S}$ is a $\sigma$-algebra on $A\times B$.

Are there $\sigma$-algebras on $A$ and on $B$ whose product is $\mathbb{S}$? Well, there aren't that many $\sigma$-algebras on the two-element set. There's the total $\sigma$-algebra, $\mathcal{P}(\{0,1\})$, and the $\sigma$-algebra $\{\emptyset,\{0,1\}\}$, and that's it (if it contains any proper subset, then it contains its complement, and it is the entire thing). Call these $\mathbb{S}_1$ and $\mathbb{S}_2$, respectively.

Is $\mathbb{S}$ the product of two of these? No.

It's not equal to $\mathbb{S}_1\times\mathbb{S}_i$, because this $\sigma$-algebra contains $\{1\}\times \{0,1\}$, but $\mathbb{S}$ does not. Symmetrically, it's not equal to $\mathbb{S}_i\times \mathbb{S}_1$.

So the only possibility would be that it is equal to $$\mathbb{S}_2\times\mathbb{S}_2 = \Bigl\{\emptyset,\{0,1\}\Bigr\} \times \Bigl\{\emptyset,\{0,1\}\Bigr\} = \Bigl\{ \emptyset, A\times B\Bigr\}$$ which is not equal to $\mathbb{S}$.

So $\mathbb{S}$ is a $\sigma$-algebra on $A\times B$ which is not a product of a $\sigma$-algebra on $A$ and a $\sigma$-algebra on $B$.

If it doesn't work for two, it has no hope of working for an infinite number of factors, so the potential distinction I suggested in the comments is in fact irrelevant.

Added. To answer the added question: suppose that there do exist $\sigma$-algebras $\mathbb{S}_i$ on $X_i$ such that $\mathbb{S}=\prod\mathbb{S}_i$. Any $A_i\in\mathbb{S}_i$ will necessarily satisfy that $$A\times\prod_{j\neq i} X_j \in\mathbb{S}.$$ Conversely, suppose that $A\subseteq X_i$ is such that $A\times \prod_{j\neq i}X_j\in\mathbb{S}$. Looking at the projection from $\prod X_i$ to $X_i$, you get that $A\in\mathbb{S}_i$. The projection does not work; I'm fairly sure it is the case that such an $A$ will lie in $\mathbb{S}_i$, but I find myself unable to prove it right now. Edit: This gap was filled by Morning in his answer to this question.

Assuming this is the case, then $\mathbb{S}_i$ would consist exactly of the subsets of $X_i$ such that $A\times\prod_{j\neq i}X_j\in\mathbb{S}$.

You can see how this fails in the example above. The collection of all $A\subseteq \{0,1\}$ such that $A\times\{0,1\}\in\mathbb{S}$ is just $\{\emptyset, \{0,1\}\}$, but the product of these does not exhaust the $\sigma$-algebra.

If you define the $\mathbb{S}_i$ this way, then the product $\sigma$-algebra $\prod\mathbb{S}_i$ is always contained in $\mathbb{S}$, but need not be equal to it.

  • 0
    @Arturo: Thanks! When there exist, how to construct the component $\sigma$-algebras?2011-02-21
  • 0
    @Arturo: is there some necessary and/or sufficient condition to tell if a sigma algebra on a Cartesian product can be decomposed this way?2011-02-21
  • 0
    @Tim: I was adding some stuff. There is basically one possible candidate; if it works, then that's what you want, if it doesn't work, then it's not a product $\sigma$-algebra.2011-02-21
  • 0
    @Arturo: Thanks! (1) I was wondering how you got the "Conversely" in your added part. The projection is a measurable mapping, can it really always map a measurable set to another measurable set? (2) In your last paragraph, by defining $\mathbb{S}_i$ did you mean that $\mathbb{S}_i$ is the sigma algebra generated from the those $A$ of $X_i$ in the conversely part?2011-02-21
  • 0
    @Tim: Maybe I'm off on that part. Let me think about it. For the last paragraph: if I'm not mistaken, the collection of all such $A$ is *already* a $\sigma$ algebra: it is closed under countable unions, contains the empty set and all of $X_i$, and it is closed under complements because the complement of $A\times\prod_{j\neq i}X_j$ is $(X_i-A)\times\prod_{j\neq i}X_j$.2011-02-21
  • 0
    @Tim: The projection argument is wrong; I'm pretty sure the conclusion holds, but I can't figure it out right now. Sorry.2011-02-21
  • 0
    @Arturo: Thanks! That's all right. I value what I have learned from your answers, and appreciate your spending time with my questions.2011-02-21
2

Suppose $A$ and $B$ are each 2-element sets. It's easy to create a measure on the 4-element set $A\times B$ which is not the product of measures on $A$ and $B$. Or perhaps I misunderstood you?

  • 0
    Never mind -- yes, I did misunderstand you. You're talking about $\sigma$-algebras, not about measures. Sorry.2011-02-21
  • 0
    Well, I suppose you could simply consider this: let $A = \{a_0, a_1\}$ and $B = \{b_0, b_1\}$. Consider the $\sigma$-algebra on $A\times B$ consisting of the empty set, the whole set, and the two sets $\{(a_0, b_1), (a_1, b_0)\}$ and $\{(a_0, b_0), (a_1, b_1)\}$.2011-02-21
  • 0
    Yes, you can. You were right in the first place. Just replace "measure" with "$\sigma$-algebra".2011-02-21
  • 0
    Thanks! If we are now talking about measure space instead of sigma algebra, could you explain why a given measure on the product sigma algebra may not be decomposable into component measures so that their product is the given measure?2011-02-22
  • 0
    @Tim: Sorry, I didn't see your question until just now. I won't try to typeset this, but given the 4-point product of two 2-point spaces (as above, but now saying that every set is measurable), you could define a measure on the four points by giving them values 1, 1, 1, and 2 (in any order, really). You should be able to convince yourself that no measures on the 2-point spaces could generate this.2011-02-23