Question: 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 $\mathbb R$.

Answer: 
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
$$
\mathcal{B}(\mathbb R)^m\subset \mathcal{B}(\mathbb R^m)
$$
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.