2
$\begingroup$

Let

  • $(\Omega,\mathcal A)$ and $(\Omega_i,\mathcal A_i)$ be measurable spaces
  • $X_i:\Omega\to\Omega_i$ be $(\mathcal A,\mathcal A_i)$-measurable

Are we able to show that $\sigma((X_1,X_2))=\sigma(X_1,X_2)$?

By definition, $$\sigma((X_1,X_2))=(X_1,X_2)^{-1}(\mathcal A_1\otimes\mathcal A_2)\tag1$$ and we know that this is equal to $$\sigma((X_1,X_2)^{-1}(\mathcal E_1\times\mathcal E_2),\tag2$$ if $\mathcal E_i\subseteq\mathcal A_i$ with $\sigma(\mathcal E_i)=\mathcal A_i$. On the other hand, $$\sigma(X_1,X_2)=\sigma(\sigma(X_1)\cup\sigma(X_2))=\sigma(X_1^{-1}(\mathcal A_1)\cup X_2^{-1}(\mathcal A_2))\tag3$$ and $$X_i^{-1}(\mathcal A_i)=\sigma(X_i^{-1}(\mathcal E_i)).\tag4$$ But I don't see that the claim follows from these equalities.

$\endgroup$
3
$\begingroup$

$X_1^{-1} (A)=(X_1,X_2)^{-1}(A\times \Omega_2)$, $X_2^{-1} (A)=(X_1,X_2)^{-1}( \Omega_1 \times A)$ and $(X_1,X_2)^{-1} (A\times B)= X_1^{-1}(A) \cap X_2^{-1}(B)$. Using these identities prove that each side is contained in teh other.

$\endgroup$
2
$\begingroup$

Observe that $\Omega_1\times\Omega_2$ is equipped with the product $\sigma$-algebra which means exactly that:$$(X_1,X_2)\text{ is measurable iff }X_i=\pi_i\circ(X_1,X_2)\text{ are measurable for }i=1,2\tag1$$ where $\pi_i:\Omega_1\times\Omega_2\to\Omega_i$ denotes the projection for $i=1,2$

$\sigma((X_1,X_2))$ is by definition the smallest $\sigma$-algebra on $\Omega_1\times\Omega_2$ such that $(X_1,X_2):\Omega\to\Omega_1\times\Omega_2$ is measurable.

$\sigma(X_1,X_2)$ is by definition the smallest $\sigma$-algebra on $\Omega_1\times\Omega_2$ such that $X_i:\Omega\to\Omega_i$ is measurable for $i=1,2$.

So $(1)$ tells us directly that they are the same.


Actually we have $\mathcal A_1\otimes\mathcal A_2=\sigma\left(\pi_1^{-1}(\mathcal A_1)\cup\pi_2^{-1}(\mathcal A_2)\right)$ so that: $$\sigma((X_1,X_2))=(X_1,X_2)^{-1}(\mathcal A_1\otimes\mathcal A_2)=(X_1,X_2)^{-1}\left(\sigma\left(\pi_1^{-1}(\mathcal A_1)\cup\pi_2^{-1}(\mathcal A_2)\right)\right)=$$$$\sigma\left((X_1,X_2)^{-1}\left(\pi_1^{-1}(\mathcal A_1)\cup\pi_2^{-1}(\mathcal A_2)\right)\right)=\sigma(X_1^{-1}(\mathcal A_1)\cup X_2^{-1}(\mathcal A_2))=\sigma(X_1,X_2)$$where the third equality is based on the rule:$$f^{-1}\left(\sigma\left(\mathcal{V}\right)\right)=\sigma\left(f^{-1}\left(\mathcal{V}\right)\right)$$For a proof of that rule see here.

$\endgroup$
  • $\begingroup$ Are we able to show $\sigma(\sigma(X,Y)\cup\sigma(Z))=\sigma(X,Y,Z)$ in the same way? $\endgroup$ – 0xbadf00d Aug 17 '18 at 11:35
  • 1
    $\begingroup$ Evidently $\sigma(X,Y)\cup\sigma(Z)\subseteq\sigma(X,Y,Z)$ so that $\sigma(\sigma(X,Y)\cup\sigma(Z))\subseteq\sigma(X,Y,Z)$. Conversely $X,Y,Z$ are measurable wrt $\sigma(\sigma(X,Y)\cup\sigma(Z))$ so that $\sigma(X,Y,Z)\subseteq\sigma(X,Y)\cup\sigma(Z)$. $\endgroup$ – drhab Aug 17 '18 at 11:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.