Function measurable iff the components are? 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.
 A: 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 product of two 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.
A: Although your question seems to have been solved, you mentioned in a comment that you were looking for something more general than this answer. So I'll add something more general here.
Consider a measurable space $(X, \Sigma_X)$ and bunch of $n$ measurable spaces $(Y_i, \Sigma_Y^i)$ for $1 \leq i \leq n$. The $\sigma$-algebras $\Sigma_Y^i$ can be combined into a product $\sigma$-algebra denoted as follows:
$$\Sigma_Y = \otimes_{i=1}^n \Sigma_Y^i$$
Now, consider functions $f_1, f_2, \dots, f_n$ where the type of each function is $f_i : X \to Y_i$. Suppose every $f_i$ is measurable w.r.t. $\Sigma_X$ and $\Sigma_Y^i$. You want to show that the vector valued function $f = (f_1, f_2, \dots, f_n)$ is measurable w.r.t. $\Sigma_X$ and $\Sigma_Y$.
To show this, we will invoke the $\pi$-System Sufficiency lemma from below. With this, it is sufficient to show only that the preimages of all elements of a $\pi$-system generating $\Sigma_Y$ are measurable w.r.t. $\Sigma_X$.
Well, luckily for us, it's quite easy to find such a $\pi$-system. By definition, the product measure $\Sigma_Y$ is defined as the $\sigma$-algebra generated by the following $\pi$-system:
$$\{Y_1 \times Y_2 \times \dots \times Y_n : Y_i \in \Sigma_Y^i\}$$
Thus, applying our lemma, given any $Y_1, \dots, Y_n \in \Sigma_Y$, it suffices to show:
$$f^{-1}(Y_1 \times Y_2 \times \dots \times Y_n) \in \Sigma_X$$
The nice thing about this is that since our set is a "rectangle" i.e. a product of measurable sets, we can decompose the inverse image into individual components. Such a decomposition could not be done for an arbitrary set in $\Sigma_Y$, which shows why the application of the $\pi$-System Sufficiency lemma was an essential step. So let's decompose:
$$f^{-1}(Y_1 \times Y_2 \times \dots \times Y_n) = \\
f_1^{-1}(Y_1) \cap f_2^{-1}(Y_2) \cap \dots \cap f_n^{-1}(Y_n) = \\
\bigcap_{i = 1}^n f_i^{-1}(Y_i)$$
Since each $f_i$ is itself measurable, then each $f_i^{-1}(Y_i) \in \Sigma_X$ for all $i$. And since all $\sigma$-algebras are closed under finite intersections, then we have:
$$\bigcap_{i = 1}^n f_i^{-1}(Y_i) \in \Sigma_X$$
Which completes the proof.
Lemma: $\pi$-System Sufficiency
Suppose we have sets $X, Y$ with respective $\sigma$-algebras $\Sigma_X, \Sigma_Y$. Suppose $A$ is a $\pi$-system over $Y$ such that $\sigma(A) = \Sigma_Y$. Then any function $f : X \to Y$ is measurable w.r.t. $\Sigma_X$ and $\Sigma_Y$ iff $\forall E \in A : f^{-1}(E) \in \Sigma_X$.
For now, I'll leave the proof of this up to an interested reader, but may come back to this and fill in the details. Basically, the strategy is to apply the $\pi$-system lemma from here, and show that the class of $Y$ subsets whose preimage is measurable w.r.t. $\Sigma_X$ is a d-system. Or you can probably find a proof somewhere on Google :)
Notes

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*The question alludes to a measure $\mu$. This implies the existence of a measure space over which $\mu$ is defined. However, the above result only needs a measurable space, not a measure space. That is, we don't need a measure at all, just a $\sigma$-algebra. Of course, as mentioned here, any measurable space can be turned into a measure space easily enough, so this is just a minor point.

