# Does it hold $\sigma(X_0,\ldots,X_n)=\sigma(X_1,X_1-X_0,\ldots,X_n-X_{n-1})$?

Let

• $(\Omega,\mathcal{A})$ and $(\Omega',\mathcal{A}')$ be measurable spaces
• $X_1,\ldots,X_n$ be measurable with respect to $\mathcal{A}$-$\mathcal{A}'$
• $Y_m:=X_m-X_{m-1}$ for $1\le m\le n$ and $Y_0:=X_0$

I'm wondering whether or not we've got $$\sigma(X_0,\ldots,X_n)=\sigma(Y_0,\ldots,Y_n)\tag{1}$$

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?

Note that $$\left(X_0,\ldots,X_n\right)=T\left(Y_0,\ldots,Y_n\right)$$ for a linear (continuous) and bijective map $T\colon\mathbf R^{n +1}\to \mathbf R^{n +1}$.