I'm reading about induced measure and independence from here.

Definition 1.6.5. Given two measurable spaces $\left(\Omega_{1}, \mathcal{F}_{1}\right)$ and $\left(\Omega_{2}, \mathcal{F}_{2}\right)$, we define the product space $\Omega_{1} \times \Omega_{2} \equiv\left\{\left(\omega_{1}, \omega_{2}\right): \omega_{1} \in \Omega_{1}, \omega_{2} \in \Omega_{2}\right\}$. The product $\sigma$-algebra on $\Omega_{1} \times \Omega_{2}$ is defined as $\mathcal{F}_{1} \times \mathcal{F}_{2} \equiv \sigma\left\{\rho_{1}, \rho_{2}\right\}$, where $\rho_{i}: \Omega_{1} \times \Omega_{2} \rightarrow \Omega_{i}$ are the projection maps $\rho_{1}:\left(\omega_{1}, \omega_{2}\right) \mapsto \omega_{1}$ and $\rho_{2}:\left(\omega_{1}, \omega_{2}\right) \mapsto \omega_{2}$.

Theorem 1.6.6. Let $\left(\Omega_{1}, \mathcal{F}_{1}, \mathbb{P}_{1}\right)$ and $\left(\Omega_{2}, \mathcal{F}_{2}, \mathbb{P}_{2}\right)$ be two probability spaces, and denote by $\rho_{1}: \Omega_{1} \times \Omega_{2} \rightarrow \Omega_{1}$ and $\rho_{2}: \Omega_{1} \times \Omega_{2} \rightarrow \Omega_{2}$ the projection maps.

  1. There exists a unique probability measure $\mathbb{P}_{1} \times \mathbb{P}_{2}$ on $\mathcal{F}_{1} \times \mathcal{F}_{2}$, called the product measure, under which the law of $\rho_{1}$ is $\mathbb{P}_{1}$, the law of $\rho_{2}$ is $\mathbb{P}_{2}$, and $\rho_{1}$ and $\rho_{2}$ are independent $\Omega_{1}$- and $\Omega_{2}$-valued random variables, respectively.

  2. The construction is in some sense unique: suppose that on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$ are defined an $S$-valued random variable $X$ (with law $\mu_{X}$ ) and a $T$-valued random variable $Y$ (with law $\mu_{Y}$ ). Then the law of the $S \times T$-valued random variable $(X, Y)$ is the product measure $\mu_{X} \times \mu_{Y}$.

My confusion: Let $\mathcal S, \mathcal T$ be the $\sigma$-algebra associated with $S$ and $T$ respectively.

In 2., the law of the $S \times T$-valued random variable $(X, Y)$ is said to be the product measure $\mu_{X} \times \mu_{Y}$, i.e., $\mu_{(X,Y)} = \mu_{X} \times \mu_{Y}$. This means $$\mu_{(X,Y)}(S \times T) = \mu_X (S) \times \mu_Y(T), \quad S\in \mathcal S, T \in \mathcal T,$$ or $$\mathbb P(X^{-1}(S) \cap Y^{-1}(T)) = \mathbb P(X^{-1}(S)) \times \mathbb P(Y^{-1}(T)), \quad S\in \mathcal S, T \in \mathcal T.$$

However, we have no information that $X$ and $Y$ are independent. Could you elaborate on my confusion?

  • 1
    $\begingroup$ Your last equation is exactly the definition of indepence. $\endgroup$
    – user940347
    Oct 13 at 22:26
  • $\begingroup$ @user940347 I take it from here. It's not hard to show that it's equivalent to the definition of product measure by the author. As you can see, there is no hypothesis related to the independence in 2. $\endgroup$
    – Akira
    Oct 13 at 22:37
  • 2
    $\begingroup$ Then the statement 2 is obviously wrong. Probably it is a very bad typo that the author forgot the word "independent" $\endgroup$
    – user940347
    Oct 13 at 22:44

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