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Let

  • $(\Omega,\mathcal{A})$ and $(\Omega',\mathcal{A}')$ be measurable spaces
  • $X_1,\ldots,X_n$ be measurable with respect to $\mathcal{A}$-$\mathcal{A}'$
  • $Y_m:=X_n-X_m$ for $1\le m\le n$

I'm wondering whether or not we've got $$\sigma(X_1,\ldots,X_n)=\sigma(Y_1,\ldots,Y_n)\tag{1}$$


I'm unsure how exactly I need to progress. Clearly, $$\sigma(X_1,\ldots,X_n)\stackrel{\text{def}}{=}\sigma\left(\underbrace{\bigcup_{i=1}^n\left\{X_i^{-1}(A):A\in\mathcal{A}'\right\}}_{=:\mathcal{H}}\right)$$ So, given $A\in\mathcal{A}'$, we would need to show that $$Y_m^{-1}(A)\in\mathcal{H}\;\;\;\text{for all }1\le m\le n$$ This would imply "$\supseteq$" in $(1)$. However, I don't think that we can compute $Y_m^{-1}(A)$.

So, is $(1)$ wrong? If not, how can we really prove it?

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Suppose $X_1=X_2$. What is $\sigma(X_2-X_1, X_2-X_2)$ compared with $\sigma(X_1,X_2)$?

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