If $M$ is Borel and $M=M_1\times M_2\times\cdots \times M_n$, then $M_i$'s are Borel? 
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.
 A: 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.
