I'm trying to understand the proof that the Borel $\sigma-$algebra generated by the rectangles in $\mathbb{R^2}$, $\mathcal{B}(\mathbb{R}^2)$ is equal to the $\sigma-$algebra generated by product of borel sets $\mathcal{B}(\mathbb{R}) \times \mathcal{B}(\mathbb{R}):= \sigma ( {B_1 \times B_2}, B_i \in \mathcal{B}(\mathbb{R})) $.

I get that since every rectange is a Borel set $\mathcal{B}(\mathbb{R}^2) \subset \mathcal{B}(\mathbb{R}) \times \mathcal{B}(\mathbb{R}) $.

It's the other inclusion that I don't understand:

In my notes from the lecture our teacher defined these collection of sets,

$\mathcal{S}_1=\{ (a_1,b_1] \times (a_2,b_2]\}$

$\mathcal{S}_2=\{B_1 \times B_2; B_i \in\mathcal{B}(\mathbb{R}) \}$

$\widetilde{B}_1=B_1 \times \mathbb{R}, B_1 \in \mathcal{B}(\mathbb{R})$

$\widetilde{B}_2= \mathbb{R} \times B_2, B_2 \in \mathcal{B}(\mathbb{R})$

Then $\mathcal{B}(\mathbb{R}^2 )=\sigma(\mathcal{S}_1)$ and $\mathcal{B}(\mathbb{R}) \times \mathcal{B}(\mathbb{R})=\sigma(\mathcal{S}_2)$

Here is the first thing that I don't get: In my notes I've written $\widetilde{B}_1=B_1\times\mathbb{R} \in \sigma(\mathcal{S}_1) \times \mathbb{R} \stackrel{\text{this is the equality I don't understand}}{=} \sigma(\mathcal{S}_1 \times \mathbb{R})=\sigma(\widetilde{\mathcal{S}}_1)$

Then $\widetilde{\mathcal{S}}_1 \cap \widetilde{\mathcal{S}}_2=\mathcal{S}_1 \times \mathcal{S}_1=\mathcal{S}$ (I think the last equallity is by definition).

And then the last thing I don't understand is the inclusion: $\sigma(\widetilde{\mathcal{S}}_1\cap \widetilde{\mathcal{B}}_2) \subseteq \sigma(\widetilde{\mathcal{S}}_1\cap \widetilde{\mathcal{S}}_2) $.

If somebody could help me clarify these things I'd be very greatful!

  • 1
    $\begingroup$ For a more detailed proof, take a look at this post $\endgroup$ Commented Sep 9, 2023 at 8:05

2 Answers 2


The inclusion $\mathscr B(\mathbb R^2)\subseteq \mathscr B(\mathbb R)\times \mathscr B(\mathbb R)$ is clear. For the other direction, note that if $A,B\in \mathscr B(\mathbb R)$, then $A\times B=(A\times \mathbb R)\cap (\mathbb R\times B)=\pi_1^{-1}(A)\cap \pi_2^{-1}(B)\in \mathscr B(\mathbb R^2)$ because the projections are continuous, hence measurable. Now since the $\sigma-$algebra generated by the rectangles $A\times B$ is precisely $\mathscr B(\mathbb R)\times \mathscr B(\mathbb R)$, we are done.

  • 2
    $\begingroup$ Slick!! That was much simpler! Thanks a lot! $\endgroup$
    – user202542
    Commented Dec 23, 2016 at 16:20
  • $\begingroup$ I just have one question; doesn't the measurability of the projections depend on the topology on $\mathbb{R}$ and $\mathbb{R}^2$? $\endgroup$
    – user202542
    Commented Dec 24, 2016 at 9:13
  • $\begingroup$ The Borel $\sigma-$algebra on a space $X$ is the smallest one that contains the open sets in $X$ so yes it does. A function is Borel-measurable if the preimage of every open set is Borel. So, for example, if we take $\mathbb R^2$ with the indiscrete topology, then $\pi_1^{-1}((-\infty, ])$ is not Borel. $\endgroup$ Commented Dec 24, 2016 at 16:16

Matematleta just did it all. But I wanted to provide a more complete answer for people who are beginning measure theory (such as myself).

We say a function is measurable if it takes pre-image of measurable sets to measurable sets.

Lemma 1: $f:X\rightarrow Y $ is a continuous function in topological spaces, then it is measurable (provided $X$ and $Y$ are endowed with the borel measure).

Proof: If $f$ is continuous, we define $Z=\{E\subset X\:: \: f^{-1}(E)\in \mathcal{B}(X) \}$.

Clearly, every open set is in $Z$, hence $\emptyset , X\in Z$.

Furthermore, if $E\in Z$, we have $f^{-1}(E)\in \mathcal{B}(X)$ which implies $f^{-1}(E)^C=f^{-1}(E^C)\in \mathcal{B}(X)$ and $E^C\in Z$.

If $f^{-1}(E_i)\in \mathcal{B}(X)$, then $\cup_{i\in \mathbb{N}}f^{-1}(E_i)=f^{-1}(\cup_{i\in \mathbb{N}}E_i)\in \mathcal{B}(X)$ from which we conclude $\cup_{i\in \mathbb{N}}E_i\in Z$.

Thus, $Z$ is a sigma algebra it contains open sets of $X$. Hence, $\mathcal{B}(X)\subset Z$. $\square$

$\mathcal{B}(\mathbb{R}\times \mathbb{R})=\mathcal{B}(\mathbb{R})\times \mathcal{B}(\mathbb{R})$

Proof: $\mathbb{R}^2$ is second countable with $(q_{11},q_{12})\times (q_{21},q_{22})$ as a countable basis with rational entries. If $U$ is open, we need $U=\cup_{i\in \mathbb{N}}I_{1i}\times I_{2i}\in \mathcal{B}(\mathbb{R})\times \mathcal{B}(\mathbb{R})$ because it is union if countable rectangles with entries in the borel sigma algebra. Because $\mathcal{B}(\mathbb{R})\times \mathcal{B}(\mathbb{R})$ contains all open sets and $\mathcal{B}(\mathbb{R}\times \mathbb{R})$ is the smallest such sigma algebra, $\mathcal{B}(\mathbb{R}\times \mathbb{R})\subset \mathcal{B}(\mathbb{R})\times \mathcal{B}(\mathbb{R})$.

Reciprocally, if $A_1\times A_2$ is a measurable rectangle, we have $A_1\times A_2 =\pi^{-1}_1(A_1)\cap \pi^{-1}_2(A_2)$. But projections are continuos so by Lemma $1$ they are measurable if $\mathbb{R}^2$ is endowed with the $\mathcal{B}(\mathbb{R}\times \mathbb{R})$ measure. But this means $A_1\times A_2 =\pi^{-1}_1(A_1)\cap \pi^{-1}_2(A_2)\in \mathcal{B}(\mathbb{R}\times \mathbb{R})$ and because this is a general rectangle and $\mathcal{B}(\mathbb{R})\times \mathcal{B}(\mathbb{R})$ is the smallest sigma algebra that contains rectangles, $\mathcal{B}(\mathbb{R})\times \mathcal{B}(\mathbb{R})\subset \mathcal{B}(\mathbb{R}\times \mathbb{R})$. $\square$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .