If $M$ is a nonempty Borel set in $\mathbb{R}^n$ and $M=M_1\times M_2\times\cdots \times M_n$, then are $M_1,M_2,\ldots,M_n$ are Borel sets in $\mathbb{R}$?

I think the answer is yes. Using definition "Borel set is formed from open sets by any operations of countable union, countable intersection and relative complement" (definition from Wikipedia), my proof can be summerized as below:

$\bullet$ Firsr, $M$ is formed from opens sets in $\mathbb{R^n}$ through those operations.

$\bullet$ Every open set in $\mathbb{R}^n$ is a countable union of $n$-cells, those are "boxes" in space which have the form $(a_1,b_1)\times (a_2,b_2)\times \cdots \times (a_n,b_n)$.

$\bullet$ Thus $M_1$ are formed by (countable) $(a_1,b_1)$'s through such operations, and thus it is Borel.

Is my proof correct? Thanks in advance.

  • $\begingroup$ @bof Thank you, I added it in my post. $\endgroup$ Jun 19, 2015 at 2:22
  • $\begingroup$ No, your proof is still incorrect. The statement in your last bullet is imprecise; what does it even mean? $\endgroup$
    – shalop
    Jun 19, 2015 at 3:00
  • $\begingroup$ @Shalop Yes that' s the point I'm not sure at. Let me take an example. Suppose $M=M_1\times M_2$ is in $\mathbb{R}^2$ and $M=A\cap B$ where $A$ and $B$ are open sets. By the second bullet, suppose $A=((a_{A1},b_{A1})\times (a_{A2},b_{A2}))\cup ((c_{A1},d_{A1})\times (c_{A2},d_{A2})) $ and the same for $B$. Thus $M_1$ can be formed as $((a_{A1},b_{A1})\cup (c_{A1},d_{A1}))\cap ((a_{B1},b_{B1})\cup (c_{B1},d_{B1}))$, so it is also Borel. Did I misunderstand something? $\endgroup$ Jun 19, 2015 at 3:11

1 Answer 1


With all due respect, your proof seems a little fuzzy to me, but this is usually an issue if one tries “attacking” a proof of measurability directly. Measure theory is tricky, so we must outwit it: More effective tricks, in general, include generating-class arguments or working with measurable functions. Let me propose an argument that involves the latter kind of trick.

Pick $x_2^{\star}\in M_2\,\ldots\,x_n^{\star}\in M_n$ arbitrarily, which can be done due to the non-emptiness assumption (which, as bof pointed out, is crucial). Define the function $f:\mathbb R\to\mathbb R^n$ as $$f(x)\equiv(x,x_2^{\star},\ldots,x_n^{\star})\quad\forall x\in \mathbb R.$$ It is easy to see that $f$ is continuous, so it is a fortiori $(\mathscr B_{\mathbb R},\mathscr B_{\mathbb R^n})$-measurable. You can check also that $f(x)\in M$ if and only if $x\in M_1$ for any $x\in\mathbb R$, so that $f^{-1}(M)=M_1$. Since $M$ is a Borel subset of $\mathbb R^n$ by assumption and $f$ is measurable, $M_1$ is a Borel subset of $\mathbb R$. The proof for any coordinate other than $1$ is analogous.

  • $\begingroup$ Your proof is really nice, thanks a lot! May I have one more question: what does the terminology "fortiori" mean? $\endgroup$ Jun 19, 2015 at 4:22
  • 1
    $\begingroup$ Your very welcome! A fortiori is latin for “all the more so.” This expression can be used to introduce a corollary implied by a previous, stronger claim. $\endgroup$
    – triple_sec
    Jun 19, 2015 at 4:25
  • 1
    $\begingroup$ Thanks again. You're really nice. $\endgroup$ Jun 19, 2015 at 5:01
  • 2
    $\begingroup$ And “your” is sloppy English for “you're.” I'm embarrassed. $\endgroup$
    – triple_sec
    Jun 19, 2015 at 5:12

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