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Define the Borel sigma-algebra on $\mathbb{R}^n$ as the smallest sigma-algebra containing all $n$-rectangles $(a_1, b_1) \times \cdots \times (a_n, b_n)$.

Is it true that the Borel sigma algebra contains all sets of the form $A_1 \times \cdots \times A_n$, where each $A_i$ is some Borel set in $\mathbb{R}$?

I thought this would be trivially true, but I had a lot of trouble trying to prove it, and I'm not even sure its true anymore.

If this is a well-known result, could you please refer me to a text where it has been (dis)proved ?

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2 Answers 2

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A way to prove it: 1/ any set of the form $A_1 \times \mathbb R \ldots \times \mathbb R$, where $A_1$ is Borel, or more generally a "Borel rectangle" with only one slice not equal to the whole space, is in the Borel sigma-algebra (this is essentially a 1-dimensional Borel set, and those are generated by open intervals). 2/ any product $A_1 \times \ldots \times A_n$ (with each $A_i$ Borel) is a finite intersection of sets of the above form.

Not sure I should have answered this, it may be a homework problem... I'd have just written a comment but I'm not reputable enough to do so :)

Any standard reference on measure theory will provide a proof of the result you're asking about (say, Dudley's book).

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  • $\begingroup$ No, this wasn't a homework problem. I've been wondering whether B(R^n) should be defined as the smallest sigma algebra containing all rectangles, or the smallest sigma algebra containing all products of Borel sets in R. I was trying to prove that these are equivalent, and getting worried that I couldn't and I was getting stuck trying to do step 1 as you suggested. Ok, just need to try harder then... $\endgroup$
    – Cosmonut
    Commented Sep 15, 2010 at 11:15
  • $\begingroup$ When considering more general sigma algebras both definitions might not coincide - the reasonable definition is the first one (smallest sigma-algebra containing the open sets, so the usual definition of Borel sigma-algebra in a topological space), while the second definition would be a good way to define the product of sigma algebras. So in that case what you wanted is to show "the Borel sigma-algebra on the product space is equal to the product of the Borel sigma-algebras". $\endgroup$
    – Julien Melleray
    Commented Sep 15, 2010 at 12:35
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    $\begingroup$ Ok, done. Proved Step 1, with the standard trick of: - Consider all sets of the form A1 x R^(n-1) which belong to Borel sets of R^n, where A1 is a set in R - Showed that was a sigma algebra - Since (a, b) x R^(n-1) is in Borel sets of R^n, A1 can any Borel set of R. Thanks for the help and the addendum. $\endgroup$
    – Cosmonut
    Commented Sep 16, 2010 at 8:44
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    $\begingroup$ @Julien: As far as I can see, this is a different problem. In general, the $\sigma$-algebra generated by rectangles may be smaller than the $\sigma$-algebra generated by products of Borel sets (i.e., the product $\sigma$-algebra), since Borel (or open) sets in the original space are not necessarily generated by intervals (assuming the the space is linearly ordered so that it at least makes sense). On the other hand, the $\sigma$-algebra generated by open sets (i.e., the Borel algebra in the product) may be larger than the product $\sigma$-algebra, since ... $\endgroup$
    – Emil Jeřábek
    Commented Aug 17, 2011 at 11:09
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    $\begingroup$ ... in general an open set need not be a countable union of rectangles (or products of Borel sets, for that matter). So we are discussing three different $\sigma$-algebras. $\endgroup$
    – Emil Jeřábek
    Commented Aug 17, 2011 at 11:12
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I would like to generalize this, since I was stumped for a while by the original question too. Let $\mathcal A_i$ be any collection of sets in $X_i$ such that $X_i \in \mathcal A_i$. For any collection of sets $\mathcal A$, let $S(\mathcal A)$ be the smallest $\sigma$-algebra containing $\mathcal A$. Let $I$ be a finite index set. Then $\Pi_{i \in I} \space S(\mathcal A_i) \subset S(\Pi_{i \in I} \space \mathcal A_i)$. The proof should be clear from the answer by Julien Melleray and the 'Ok, done' comment by Cosmonut above. It follows that $S(\Pi_{i \in I} \space S(\mathcal A_i)) = S(\Pi_{i \in I} \space \mathcal A_i)$.

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