The following question came up in a discussion with my advisor:
Let $(\Omega, \mathcal F, \mathbb P)$ be a non-trivial probability space, and suppose that $\mathcal G$ is a proper sub-$\sigma$-algebra of $\mathcal F$. Does there exist an event $U \in \mathcal F$ such that
$U$ is independent of $\mathcal G$, and
$0 < \mathbb P(U) < 1$ ?
The answer is surely yes, but we can't seem to prove it. Our initial approach was to consider the space $L^2(\Omega, \mathcal G)^\perp$ of finite-variance random variables orthogonal to $\mathcal G$ (i.e. $\mathbb E(X|\mathcal G) = 0$ a.s.). Since $\mathcal G$ is a proper sub-$\sigma$-algebra of $\mathcal F$, this space is non-trivial. Choose a non-constant $X \in L^2(\mathcal G)^\perp$ and let $U = \{X < c\}$. For some $c$ the event $U$ has non-trivial probability.
However, orthogonality does not imply independence. For a simple example, take independent random variables $Y \sim \operatorname{Bernoulli}(1,p)$ and $Z \sim N(0,1)$, then set $\mathcal G = \sigma(Y)$ and $X = YZ$. Certainly $X$ is not independent of $\mathcal G$ though one easily can check that $\mathbb E(X|\mathcal G) = 0$ a.s. Our strategy above might find $X$; could we modify our strategy to find the independent $Z$ instead?
Best Answer
No. For a very simple example, take $\Omega = \{a,b,c\}$ consisting of three points, with $\mathcal{F} = 2^\Omega$ and $P(A) = |A|/3$ the uniform measure. Let $\mathcal{G} = \{\{a\}, \{b,c\}, \Omega, \emptyset\}$. Then $\mathcal{G}$ is a proper sub-$\sigma$-algebra but no nontrivial event in $\mathcal{F}$ is independent of it.
Edit: Andres Caicedo asks for a non-atomic example. George Lowther gave one. Another is supplied by taking $\Omega = [0,1]$, $\mathcal{G} = \mathcal{B}_{[0,1]}$ the Borel $\sigma$-field, $\mathcal{F} = \mathcal{L}$ the Lebesgue $\sigma$-field which is the completion of $\mathcal{G}$, and $P = $ Lebesgue measure. Now by definition of $\mathcal{L}$, for any $A \in \mathcal{F}$ we have $A = B \cup N$ where $B \in \mathcal{G}$ and $P(N) = 0$. Then $P(A \cap B) = P(B) = P(A)$ so $A$ and $B$ are independent iff $P(A) = 0$ or $1$.