Generating the whole sigma algebra with two independent sigma algebras Let $(\Omega,\mathcal A,\mathsf P)$ be a probability space. Let $\mathcal B\subset\Omega$ be a sub-$\sigma$-algebra of $\mathcal A$. Does there always exist a sub-$\sigma$-algebra $\mathcal C$ of $\mathcal A$ such that $\mathcal B$ is independent of $\mathcal C$ and such that $\sigma(\mathcal B\cup\mathcal C)=\mathcal A$ ?
This question is motivated by the Definition of causal graphs, where a random variable $X$ is said to "cause" another random variable $Y$ if there exists a measurable function $f$ and a random variable $E$ independent of $X$ such that $Y = f(X, E)$. I was wondering if any two random variables always "cause" each other.

The answer is no: As a simple counter-example, consider
$$(\Omega,\mathcal A,\mathsf P)=(\{1,2,3\}, \mathfrak P(\{1,2,3\}), \text{Uniform}(\{1,2,3\})),$$ where $\mathfrak P(\{1,2,3\})$ denotes the power set of $\{1,2,3\}$ and $ \text{Uniform}(\{1,2,3\})$ denotes the uniform measure over $\{1,2,3\}$.
Choose now $\mathcal B=\{\emptyset,\{1\},\{2,3\},\Omega\}$. Then any sigma-algebra $\mathcal C$ such that $\sigma(\mathcal B\cup\mathcal C)=\mathcal A$ must contain either $\{2\}$ or $\{1,2\}$ so that $\{2\}\in\sigma(\mathcal B\cup\mathcal C)$. But such a $\mathcal C$ cannot be independent of $\mathcal B$, since we then have either $$\mathsf P(\{1\}\cap\{2\})=0\neq \mathsf P(\{1\})\mathsf P(\{2\})$$ or $$\mathsf P(\{1\}\cap\{1,2\})=\frac 13\neq \frac 29 = \mathsf P(\{1\})\mathsf P(\{1,2\}).$$
