Let $f_i:\Omega\to\mathbb{R}$, $i\in I=\{1,2,...,n\}$, then \begin{align} \sigma(f_i,\, i\in I) = \sigma \{f_i^{-1}(B_i):B_i\in \mathcal{B},\, i\in I\}, \end{align} where $\mathcal{B}$ is the Borel algebra on $\mathbb{R}$.
Can we replace the $\mathcal{B}$ with the its generators? That is, can we write \begin{align} \sigma(f_i,\, i\in I) = \sigma \{\{f_i\leq x\}:x\in\mathbb{R},\, i\in I\}? \end{align}
Yes: let $$\mathcal A_1:= \{\{f_i\leq x\}:x\in\mathbb{R},\, i\in I\}$$ and $$ \mathcal A_2:= \{f_i^{-1}(B_i):B_i\in \mathcal{B},\, i\in I\}. $$ Then $\mathcal A_1\subset\mathcal A_2$ by definition hence $\sigma\left(\mathcal A_1\right)\subset \sigma\left(\mathcal A_2\right)$. Now, we notice that $\sigma\left(\mathcal A_1\right)$ is a $\sigma$-algebra which contains $\mathcal A_2$. Indeed, for all $i\in I$, $$ \sigma\left(\mathcal A_1\right)\supset \sigma\left(\{\{f_i\leq x\}:x\in\mathbb{R}\} \right)= f_i^{-1}\sigma\left(\left\{(-\infty,x] :x\in\mathbb{R}\right\} \right)=\{f_i^{-1}(B_i):B_i\in \mathcal{B}\}, $$ where we used the equality $f^{-1}(\sigma(\mathcal C))=\sigma(f^{-1}(\mathcal C))$.
