Let $f_1,\dots,f_n:X\rightarrow \mathbb{R}$ be measurable functions, where $X$ is endowed with $\sigma$-algebra $\mathcal{F}$ and $\mathbb{R}$ with Borel $\sigma$-algebra $\mathcal{B}^1$. Is it true that $(f_1,\dots,f_n)$ is measurable if $\mathbb{R}^n$ is also endowed with Borel $\sigma$-algebra $\mathcal{B}^2$?
I can see that it holds for sets in $\mathcal{B}^2$ which are product of sets in $\mathcal{B}^1$, but I couldn't prove it for the remaining ones.
Let $\left(X_{1}, \Sigma_{1}\right)$ and $\left(X_{2}, \Sigma_{2}\right)$ be two measurable spaces. The $\sigma$ -algebra for the corresponding product space $X_{1} \times X_{2}$ is called the product $\sigma$ -algebra and is defined by $\Sigma_{1} \times \Sigma_{2}=\sigma\left(\left\{B_{1} \times B_{2}: B_{1} \in \Sigma_{1}, B_{2} \in \Sigma_{2}\right\}\right)$
The Borel $\sigma$-algebra for $\mathbb{R}^{n}$ is generated by half-infinite rectangles and by finite rectangles. For example, $$ \mathcal{B}\left(\mathbb{R}^{n}\right)=\sigma\left(\left\{\left(-\infty, b_{1}\right] \times \cdots \times\left(-\infty, b_{n}\right]: b_{i} \in \mathbb{R}\right\}\right)=\sigma\left(\left\{\left(a_{1}, b_{1}\right] \times \cdots \times\left(a_{n}, b_{n}\right]: a_{i}, b_{i} \in \mathbb{R}\right\}\right) $$
As a result, you see that if $f_1,\dots,f_n:X\rightarrow \mathbb{R}$ are all measurable functions, then the preimage of a Borel set in $\mathbb{R}^{n}$ is sent in a measurebale set in the sigma algebra of $X$.
