I have some difficulties doing this problem: Show that a function $f : \mathbb{R}^n \rightarrow \mathbb{R}^m, x \mapsto (f_1(x), . . . , f_m(x))$ is measurable if, and only if, all coordinate maps $f_i : \mathbb{R}^n \rightarrow \mathbb{R}$ for $i=1,2,...,m$ are measurable.
In order to prove this, I can use: Let $X$ be a set, let $(X_i, \mathcal{A}_i), i \in I$ be arbitrarily many measurable spaces, and let $T_i : X\rightarrow X_i$ be a family of maps. Show that a map $f$ from a measurable space $(F, \mathfrak{F})$ to $(X, \sigma(T_i : i \in I))$ is measurable if, and only if, all maps $T_i \circ f$ are $\mathfrak{F}/\mathcal{A}_i$-measurable.
Now my approach where I'm not sure if this makes sense:
Let's consider the projection map $P_{r_i}: \mathbb{R}^m \rightarrow \mathbb{R}$, $P_{r_i}(x)=x_i$, where $i \in I=\{1,...,m\}$ (the projections are the $T_i's$). Then each $P_{r_i}$ is continuous. Hence $\mathcal{B}(\mathbb{R}^m) / \mathcal{B}(\mathbb{R})$ measurable. Thus $\sigma(P_{r_1},...,P_{r_m}) \subset \mathcal{B}(\mathbb{R}^m)$ and by the statement above we get $f:(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n)) \rightarrow (\mathbb{R}^m, \mathcal{B}(\mathbb{R}^m))$ is measurable $\Leftrightarrow$ $P_{r_i} \circ f = f_i$ is $\mathcal{B}(\mathbb{R}^n)/ \mathcal{B}(\mathbb{R})$ measurable for all $i\in I$.