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 $B(\mathbb R)$.


2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$ Commented Mar 4, 2020 at 9:27
  • $\begingroup$ 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. $\endgroup$
    – NCh
    Commented Mar 4, 2020 at 13:27
  • $\begingroup$ @Mentossinho Or possibly you have in mind this theorem: en.wikipedia.org/wiki/Carath%C3%A9odory%27s_extension_theorem It is more probable. $\endgroup$
    – NCh
    Commented Mar 4, 2020 at 14:45
  • $\begingroup$ @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. $\endgroup$
    – user334639
    Commented 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 $$\mathcal B(\Bbb R^n) = \bigotimes^1_n \mathcal B(\Bbb R)$$


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