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Suppose $(\mathfrak{B}_n)_n$ are independent with respect to a finite measure $\mu$. Show that for any $N$, the $\sigma$-algebras $\mathfrak{B}_1,\mathfrak{B}_2,...,\mathfrak{B}_N,\sigma(\bigcup_{n=N+1}^\infty\mathfrak{B}_n)$ are also independent.


It is enough to show that $\mu(A\cap B)=\mu(A) \mu(B)$ where $A\in\mathfrak{B}_i$ for some $i\in\{1,...,N \}$ and $B\in \sigma(\bigcup_{n=N+1}^\infty\mathfrak{B}_n)$. I know also that $\sigma(\bigcup_{n=N+1}^\infty\mathfrak{B}_n)=\sigma(\epsilon)$ where $\epsilon=\bigcup_{n=1}^\infty\epsilon_n$ and $\epsilon_n=\{\bigcap_{k=N+1}^n A_k: A_{N+1}\in\mathfrak{B}_{N+1},A_{N+2}\in\mathfrak{B}_{N+2},...,A_{n}\in\mathfrak{B}_{n}\}$ But I couldn't continue because I don't know how to write an element $B\in \sigma(\epsilon)$. Thanks for your helps!

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Let $\mathscr{R}$ be the class of sets $B$ such that, for all $A\in\mathfrak{B}_1\cup\cdots\cup\mathfrak{B}_N$, we have $\mu(A\cap B)=\mu(A)\mu(B)$.

Clearly, $\bigcup_{k=N+1}^\infty\mathfrak{B}_k\subseteq\mathscr{R}$. I believe you can show by yourself that $\mathscr{R}$ is a $\sigma$-algebra (you're going to use finiteness of $\mu$ when dealing with complements), thus $\sigma(\bigcup_{k=N+1}\mathfrak{B}_k)\subseteq\mathscr{R}$. This means that for all $B\in\sigma(\bigcup_{k=N+1}^\infty\mathfrak{B}_k)$ and all $A\in\mathfrak{B}_j$ for $j\leq N$, $\mu(A\cap B)=\mu(A)\mu(B)$.

Remark Note that we have to assume that $\mu$ is in fact a probability. If we have two $\sigma$-algebras $\mathfrak{B}_1$ and $\mathfrak{B}_2$ on a space $X$ which are independent with respect to a finite measure $\mu$, we have $X\in\mathfrak{B}_1$ and $X\in\mathfrak{B}_2$, so $\mu(X)=\mu(X\cap X)=\mu(X)\mu(X)$, thus either $\mu(X)=0$ or $1$.

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