Question about sigma-algebras Assume some random variables $$X_1,\dots,X_n : \Omega \to \mathbb{R}$$ are given where $(\Omega,\Sigma)$ and $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ denote measurable spaces.
How can one proof that
$$  \sigma(X_1,\dots,X_n) =  \sigma(X_1,X_2-X_1,\dots,X_n-X_{n-1})$$
holds, where for random variables $Y_1,\dots,Y_n$ the term $$\sigma(Y_1,\dots,Y_n)$$ 
denotes the smallest sigma-algebra that contains the sets $$ \bigcup_{i=1}^n Y_i^{-1}(\mathcal{B}(\mathbb{R})) ?$$ 
 A: Show that in general, if $Y:\Omega\to\mathbb{R}$ and $u:\mathbb{R}\to\mathbb{R}$ are measurable, then $\sigma(u\circ Y)\subseteq \sigma(Y)$.
Now define $u(x_1,\ldots,x_n)=(x_1,x_2-x_1,\ldots,x_n-x_{n-1})$ with inverse $v(x_1,\ldots,x_n)=(x_1,x_2+x_1,\ldots,x_n+\cdots+x_1)$ and note that both $u$ and $v$ are measurable. Thus
$$\sigma(X_1,\ldots,X_n-X_{n-1})=\sigma( u\circ (X_1,\ldots,X_n))\subseteq\sigma(X_1,\ldots,X_n)
$$
and
$$
\sigma(X_1,\ldots,X_n)=\sigma(v\circ (X_1,\ldots,X_n-X_{n-1}))\subseteq \sigma(X_1,\ldots,X_n-X_{n-1})
$$
showing equality of the two sigma-algebras.
A: Suppose first that for all $i \in \{1,\dots, n\}$, $X_i$ is measurable with respect to $\sigma(X_1,\dots X_n)$. Then we know that $X_i - X_{i-1}$ are also measurable with respect to  $\sigma(X_1, \dots, X_n)$, for $2 \leq i \leq n$, by the algebraic properties of measurable functions. Thus, since $X_1$ and the $X_{i} - X_{i-1}$ are measurable with respect to $\sigma(X_1,\dots,X_n)$, and $\sigma(X_1, X_2 - X_1,\dots, X_n - X_{n-1})$ is the smallest $\sigma$-algebra for which $X_1, X_2 - X_1, \dots, X_n - X_{n-1}$ are measurable functions, we get that $$\sigma(X_1, X_2 - X_1, \dots, X_n - X_{n-1}) \subseteq \sigma(X_1, \dots X_n).$$ 
Conversely, suppose $X_1$ and $X_2 - X_1$ are measurable with respect to the $\sigma$-algebra on the right, we must have that $X_1 + (X_2 - X_1) = X_2$ measurable to the $\sigma$-algebra on the right. Continue this process until you've shown that $X_i$ is measurable with respect to $\sigma(X_1, X_2 - X_2, \dots X_n - X_{n-1})$ for all $i \in \{1,\dots,n\}$. Again, since $\sigma(X_1, X_2, \dots ,X_n)$ is the smallest $\sigma$-algebra for which the $X_i$ are measurable functions, we get $$\sigma(X_1, X_2 - X_1, \dots, X_n - X_{n-1}) \supseteq \sigma(X_1, \dots X_n),$$ and thus equality.  
