Let $\{X_{n}\}_{n\geq 1}$ be a sequence of random variables on a probability space, $(\Omega, \mathcal{A},\mathbb{P})$. Define the following mapping:

$X : \Omega \rightarrow \mathbb{R}^{\infty}$

where $X(\omega) = (X_{1}(\omega), X_{2}(\omega),....)$. Let $\mathcal{R}$ be any rectangle in $\mathbb{R}^{n}$. Show that $X^{-1}(\mathcal{R}) \in \mathcal{A}$.

Does this follow because each coordinate $X_{i}$ is measurable (by definition of a random variable)?

  • 1
    $\begingroup$ Yes, let $R= \times_{i=1}^n (a_i.b_i]$, and then $X^{-1}(R)=\cap_{i=1}^{n} X_i^{-1}((a_i,b_i])$, which is in the sigma-algebra of $\Omega$. I assume that your $X^{-1}(R)= \{ (X_1,X_2,...,X_n)\in R\}$. $\endgroup$ – Brian Ding Feb 23 '15 at 6:46
  • $\begingroup$ Thank you. Why does the expression $X^{-1}(\mathcal{R}) = \bigcap\limits_{i=1}^{n}X_{i}^{-1}[a_{i},b_{i})$ hold? Is this true even if we're not assuming independence? $\endgroup$ – Dayman Feb 23 '15 at 6:51
  • $\begingroup$ $X^{-1}(R)=\{ (X_1,X_2,...,X_n)\in \times _{i=1}^n(a_i,b_i] \}=\{ X_i\in (a_i,b_i],\forall i\}=\cap_{i=1}^n X_i^{-1}(a_i,b_i].$. You don't need independence in this case. $\endgroup$ – Brian Ding Feb 23 '15 at 7:05
  • $\begingroup$ Thank you again. Can I extend this argument to any set $A$ in $\sigma(\mathcal{R})$, i.e. the sigma algebra generated by all rectangles? $\endgroup$ – Dayman Feb 23 '15 at 7:33
  • $\begingroup$ To clarify the above, can I show that: $X^{-1}(A) \in \mathcal{F}$ for every $A \in \sigma(\mathcal{R}) = \mathcal{B}(\mathbb{R}^{n})$ using the same argument above? $\endgroup$ – Dayman Feb 23 '15 at 7:40

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