[proofplan] Compare the joint law with the product law on measurable rectangles. Rectangles define independence and generate the product sigma-algebra. [/proofplan] [step:Independence implies product joint law] Assume $X_1,\ldots,X_n$ are independent. For measurable rectangles $A_1\times\cdots\times A_n$ with $A_i\in\mathcal E_i$, \begin{align*} \mathcal L_X(A_1\times\cdots\times A_n) &=\mathbb P(X_1\in A_1,\ldots,X_n\in A_n) \\ &=\prod_{i=1}^n\mathbb P(X_i\in A_i) \\ &=\prod_{i=1}^n\mu_i(A_i) \\ &=(\mu_1\otimes\cdots\otimes\mu_n)(A_1\times\cdots\times A_n). \end{align*} The measurable rectangles form a pi-system that generates the product sigma-algebra, so uniqueness of finite measures gives \begin{align*} \mathcal L_X=\mu_1\otimes\cdots\otimes\mu_n. \end{align*} [/step] [step:Product joint law implies independence] Conversely suppose $\mathcal L_X=\mu_1\otimes\cdots\otimes\mu_n$. For any $A_i\in\mathcal E_i$, \begin{align*} \mathbb P(X_1\in A_1,\ldots,X_n\in A_n) &=\mathcal L_X(A_1\times\cdots\times A_n) \\ &=(\mu_1\otimes\cdots\otimes\mu_n)(A_1\times\cdots\times A_n) \\ &=\prod_{i=1}^n\mu_i(A_i) \\ &=\prod_{i=1}^n\mathbb P(X_i\in A_i). \end{align*} Thus the variables are independent. [/step]