# Does product sigma algebra of n B(R) (Borel) coincide with B(R^n)?

My question comes from different modes of probability law, which I met. I see two options in equality: set belonged to $$B(\mathbb R^n)$$ and Cartesian product of B belonged to $$B(\mathbb R)$$.

First note that product sigma-algebra is not a product of sigma-algebras. The last object is not a sigma-algebra at all. Look, for instance, two sets $$B_1=(0,1)\times(0,1)$$ and $$B_2=(1,3)\times(1,3)$$. Every set belongs to $$\mathfrak B(\mathbb R)\times \mathfrak B(\mathbb R)$$ and the union $$B_1\cup B_2$$ does not since it is not a rectangle.

Product sigma-algebra is defined as the sigma-algebra generated by all sets $$B_1\times\ldots\times B_n\in\mathfrak B(\mathbb R)\times\ldots\times \mathfrak B(\mathbb R)$$ It is denoted by $$\mathfrak B(\mathbb R)\otimes\ldots\otimes \mathfrak B(\mathbb R)$$ It coincides with $$\mathfrak B(\mathbb R^n)$$.

Prove this fact for $$n=2$$ for simplicity.

First show $$\mathfrak B(\mathbb R^2)\subseteq \mathfrak B(\mathbb R)\otimes \mathfrak B(\mathbb R)$$.

Take arbitrary rectangle $$(a,b)\times (c,d)$$. It belongs to $$\mathfrak B(\mathbb R)\otimes \mathfrak B(\mathbb R)$$ since $$(a,b)\in\mathfrak B(\mathbb R)$$ and $$(c,d)\in\mathfrak B(\mathbb R)$$. Then the set of all possible rectangles belongs to $$\mathfrak B(\mathbb R)\otimes \mathfrak B(\mathbb R)$$. Then the sigma-algebra $$\mathfrak B(\mathbb R^2)$$ generated by the set of all rectangles is a subset of the sigma-algebra $$\mathfrak B(\mathbb R)\otimes \mathfrak B(\mathbb R)$$. Recall the reason: $$\mathfrak B(\mathbb R^2)$$ is a smallest sigma-algebra containing all rectangles, and $$\mathfrak B(\mathbb R)\otimes \mathfrak B(\mathbb R)$$ is some sigma-algebra which also containes all rectangles, so the first one is nested in the second one.

Next show $$\mathfrak B(\mathbb R)\otimes \mathfrak B(\mathbb R)\subseteq \mathfrak B(\mathbb R^2)$$.

Let $$\mathcal F$$ be the collection of all subsets $$A$$ of $$\mathbb R$$ such that $$A\times \mathbb R\in \mathfrak B(\mathbb R^2)$$. Note that all intervals are in $$\mathcal F$$ and also $$\mathcal F$$ is a sigma-algebra. The last fact can be checked from definitions easily. Therefore $$\mathfrak B(\mathbb R)\subseteq \mathcal F$$. So we obtained that for every $$A\in \mathfrak B(\mathbb R)$$, $$A\times \mathbb R\in \mathfrak B(\mathbb R^2)$$.

Similarly, for every $$B\in \mathfrak B(\mathbb R)$$, $$\mathbb R\times B\in \mathfrak B(\mathbb R^2)$$. Then also $$A\times B=(A\times \mathbb R)\cap (\mathbb R\times B) \in \mathfrak B(\mathbb R^2).$$ And therefore the sigma-algebra generated by the collection of all rectangles $$A\times B$$ for any $$A,B\in \mathfrak B(\mathbb R)$$ became a subset of $$\mathfrak B(\mathbb R^2)$$, so $$\mathfrak B(\mathbb R)\otimes \mathfrak B(\mathbb R)\subseteq \mathfrak B(\mathbb R^2)$$.

We prove that these sigma-algebras coincide.

• Thank you for answer. So now, if i know that the class consists of cartesian, one dimensional borel sets generates B(R^n), could you tell me does exist some theorem, which in probability context allows to induce probability measure of n-dimensional random vector (probability law) just by set it on sets in the shape of cartesian product of borels? Mar 4, 2020 at 9:27
• I do not completely sure that I understand your question. Do you mean the following theorem: en.wikipedia.org/wiki/Kolmogorov_extension_theorem ? In any case, I see that it is a new question which does not relate to this one.
– NCh
Mar 4, 2020 at 13:27
• @Mentossinho Or possibly you have in mind this theorem: en.wikipedia.org/wiki/Carath%C3%A9odory%27s_extension_theorem It is more probable.
– NCh
Mar 4, 2020 at 14:45
• @Mentossinho What do you mean by "induce" a measure? Do you want to specify what you want the measure to be in on sets in the shape of cartesian product of borels? There may be no such a mesaure, but if there is one, then it is unique, because of a "uniqueness of measure" theorem that uses the Pi-Lambda theorem. Feb 28, 2023 at 20:00

Yes, this holds more generally for separable metric spaces.

Let $$X_1, \ldots, X_n$$ be metric spaces and let $$X = \prod_1^n X_j$$, equipped with the product metric. Then, $$\bigotimes^1_n \mathcal B(X_j) \subset \mathcal B(X)$$. If the $$X_j$$'s are separable, then $$\bigotimes^1_n \mathcal B(X_j) = \mathcal B(X)$$.

Since $$\Bbb R$$ is separable (i.e., it has a countable dense subset), it follows that $$\mathcal B(\Bbb R^n) = \bigotimes^1_n \mathcal B(\Bbb R)$$