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.

  • $\begingroup$ You may need to specify if the collection is finite or infinite. $\endgroup$ Feb 21, 2011 at 2:02
  • $\begingroup$ @Arturo: the collection is arbitrary. But if the treatment is different for finite and for infinite cases, I would like to hear for both. $\endgroup$
    – Tim
    Feb 21, 2011 at 2:03
  • $\begingroup$ @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. $\endgroup$ Feb 21, 2011 at 2:04
  • 1
    $\begingroup$ @Tim: Yes, according to planetmath (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. $\endgroup$ Feb 21, 2011 at 2:07
  • $\begingroup$ @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. $\endgroup$
    – Tim
    Feb 21, 2011 at 2:13

2 Answers 2


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.

  • $\begingroup$ @Arturo: Thanks! When there exist, how to construct the component $\sigma$-algebras? $\endgroup$
    – Tim
    Feb 21, 2011 at 2:45
  • $\begingroup$ @Arturo: is there some necessary and/or sufficient condition to tell if a sigma algebra on a Cartesian product can be decomposed this way? $\endgroup$
    – Tim
    Feb 21, 2011 at 2:53
  • $\begingroup$ @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. $\endgroup$ Feb 21, 2011 at 3:00
  • $\begingroup$ @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? $\endgroup$
    – Tim
    Feb 21, 2011 at 4:05
  • $\begingroup$ @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 $(X_i-A)\times\prod_{j\neq i}X_j$. $\endgroup$ Feb 21, 2011 at 4:13

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?

  • $\begingroup$ Never mind -- yes, I did misunderstand you. You're talking about $\sigma$-algebras, not about measures. Sorry. $\endgroup$ Feb 21, 2011 at 2:21
  • $\begingroup$ 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 $\{(a_0, b_0), (a_1, b_1)\}$. $\endgroup$ Feb 21, 2011 at 2:28
  • $\begingroup$ Yes, you can. You were right in the first place. Just replace "measure" with "$\sigma$-algebra". $\endgroup$ Feb 21, 2011 at 2:32
  • $\begingroup$ 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? $\endgroup$
    – Tim
    Feb 22, 2011 at 15:21
  • $\begingroup$ @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. $\endgroup$ Feb 23, 2011 at 23:01

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