Let $X_{1}, X_{2}, \dots, X_{k}$ be random variables defined on the measurable space $(\Omega, \mathcal{F})$. Let $\mathcal{G} = \sigma(X_{1}, X_{2}, \dots, X_{k})$ and $\mathcal{A} = \{ \bigcap\limits_{i=1}^{k}X_{i}^{-1} (A_{i}) : A_{i} \in \mathcal{B}(\mathbb{R}) \} $.
I want to show that $\mathcal{G} = \sigma(\mathcal{A})$. I find this quite intuitive, but I am not sure how to approach the proof. It is easy to show that $\mathcal{A}$ is a $\pi$-system, but I am not sure if this is helpful.
Note that $\sigma(X_i)\subset\mathcal{A}$ for all $i$. (To see this, note that $X_i^{-1}(A)$ is the same as $X_i^{-1}(A)\cap\Omega$ which is the same as $X_i^{-1}(A)\cap X_1^{-1}(\mathbb{R})\cap\cdots\in\mathcal{A}$.) Hence $\bigcup_{i=1}^n\sigma(X_i)\subset \sigma(\mathcal{A})$, so $$ \sigma(X_1, \dots, X_n) = \sigma\left(\bigcup_{i=1}^n\sigma(X_i)\right)\subset \sigma(\mathcal{A}). $$ The other direction follows from $\mathcal{A}\subset \sigma(X_1, \dots, X_n)$, which is clear since $\mathcal{A}$ consists of finite intersections of elements of $\bigcup_{i=1}^n\sigma(X_i)$.