If $\boldsymbol{f}:X\to\mathbb{R}^n$ is a $\mu$-measurable function, I think it is quite easy to see that its components $f_i$ also are. In fact, the projection $\pi_i:\mathbb{R}^n\to\mathbb{R}$, $\boldsymbol{x}\mapsto x_i$ is continuous and therefore Borel measurable, and $f_i=\pi_i\circ\boldsymbol{f} $. Therefore the counterimage through $\pi_i$ of a Borel subset of $\mathbb{R}$ is a Borel subset of $\mathbb{R}^n$, whose counterimage through $\boldsymbol{f}$ is a $\mu$-measurable subset of $X$.
I cannot prove the converse, i.e. that, if $f_1,\ldots,f_n$ are $\mu$-measurable functions, then $\boldsymbol{f}=(f_1,\ldots,f_n)$ is, but I would not be amazed if it were true. Is it and, if it is, how can it be proved? I thank any answerer very much.
Best Answer
The case $n=2$ generalizes by induction so this is what we'll show. Let $U \subset \mathbb{R}^2$ be open. As $\mathbb{R}^2$ is separable and open balls in $\mathbb{R}^2$ are countable union of open intervals in $\mathbb{R}$ we may write $$U = \cup_{j=1}^\infty ( (a_{j1},b_{j1}) \times (a_{j2}, b_{j2}))$$ for some family of intervals. Here its understood that the endpoints could be $\infty$ or $-\infty$. Now, its quick to verify $$f^{-1}(U) = \cup_{j=1}^\infty (f_1^{-1}(a_{j1},b_{j1}) \cap f_2^{-1}(a_{j2},b_{j2}))$$ which is a Borel set as $f_1$ and $f_2$ are both Borel measurable.
Note this argument holds for any countable product of separable metric spaces.