Showing that $(f_1,f_2,\dots,f_m)$ is a measurable function from $(\mathbb R^m,\mathcal B(\mathbb R^m))$ into itself If $\{f_i, 1 \le i \le m\}$ is a set of real valued Borel functions on $\mathbb R$, how to show a vector of functions, $(f_1, f_2,..., f_m)$ is a measurable mapping from $(\mathbb R^m, \mathcal{B}(\mathbb R^m))$ to $(\mathbb R^m, \mathcal{B}(\mathbb R^m))$ itself?
It seems to be an obvious result. However, I don't know how to prove it.
I think it is equivalent to show i-th entry of projection of a open set in $\mathbb R^n$ is still open in $\mathbb R$.
 A: Since $\{f_i, 1 \le i \le m\}$ is a set of Borel measurable functions
$$
\{x:f_i(x)<c_i\}\subset \mathcal{B}(\mathbb R)
$$
for all possible $c_i\in\Bbb{R}$.
So
$$
\{x:f_1(x)<c_1\}\times \{x:f_2(x)<c_2\}\times\cdots\times\{x:f_m(x)<c_m\}\subset \mathcal{B}(\mathbb R)^m\subset \mathcal{B}(\mathbb R^m)
$$
for all possible $c_1\cdots c_m\in\Bbb{R}$. This means $(f_1, f_2,..., f_m)$ is Borel measurable.
Edit: $\quad$ We prove $\mathcal{B}(\mathbb R)^m\subset \mathcal{B}(\mathbb R^m)$. The idea is inspired by PhoemueX's comment.
Let $\overrightarrow{A_i}=\mathbb {R^i}×A_i×\mathbb {R^{m−i−1}}$, where $A_i\:(1\leqslant i \leqslant m)$ are Borel set. Then clearly $\overrightarrow{A_i}\subset \mathcal{B}(\mathbb R^m)$. 
Now for any set $B\in\mathcal{B}(\mathbb R)^m$
$$
B=A_1\times\cdots \times A_m=\bigcap_{i=1}^m \overrightarrow{A_i}\in\mathcal{B}(\mathbb R^m)
$$
for $\overrightarrow{A_i}\subset \mathcal{B}(\mathbb R^m)$ and Borel $\sigma$-algebra is closed under countable intersection. So we have
$$
\mathcal{B}(\mathbb R)^m\subset \mathcal{B}(\mathbb R^m)
$$
A: You're right. you have that the image of an open set under projection is open, then the preimage of that is measurable in $\mathbb{R}^1$. Then, the product of measurable sets is measurable, and you are done. 
