We have (S, $\mathcal{A}$) is a measurable space.
For $j=1,...,d$ we have $f = (f_{1},...,f_{d})$ where $f_{j}: S \rightarrow \mathbb{R}$ so $f: S \rightarrow \mathbb{R}^{d}$.
Now what I want to proof is that $f$ is measurable if and only if $f_{j}$ is measurable for all $j = 1,...,d$.
I think I understand it intuitively with the definition of a measurable function, but I have trouble with getting a formal proof on paper.
Suppose each $f_i$ is measurable. Consider $\{A \subseteq \mathbb R^{d}: f^{-1}(A) \text {is measurable }\}$. Verify that this is a sigma algebra. When $A=A_1\times A_2\times...\times A_d$ where each $A_i$ is a Borel set in $\mathbb R$ $f^{-1}(A)=\cap_i f_i^{-1}(A_i)$ which is measurable. Now sets of this type generate the Borel sigma algebra of $\mathbb R^{d}$ so the above sigma algebra contains all Borel sets and $f$ is measurable.
Converse is easy. The projection maps on $\mathbb R^{d}$ are measurable and composition of measurable maps is measurable.
