TITLE: Does there exist an event independent of a given sigma-algebra?
QUESTION [12 upvotes]: 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?

REPLY [9 votes]: No. If the probability space contains atoms then you can easily construct sub-$\sigma$-algebras which are not independent to any nontrivial events.
However, we can still construct examples, even for measures without atoms. Suppose that $\Omega=[0,1]$ is the unit interval, $\mathcal{F}=\mathcal{B}([0,1])$ is the Borel sigma-algebra and $\mathbb{P}=\lambda$ is the Lebesgue measure. Let $\mathcal{G}$ be the sub-$\sigma$-algebra generated by sets of the form $[0,x)\cup[\frac23,\frac23+\frac34x^2)$ for $0\le x\le\frac23$. I claim that all events in $\mathcal{F}$ independent to $\mathcal{G}$ have probability zero or one.
In fact, if $f\in L^1(\Omega,\mathcal{F},\mathbb{P})$ then we can calculate the conditional expectation $g=\mathbb{E}[f\mid\mathcal{G}]$ directly. Writing $y=\frac23+\frac34x^2$ for $0\le x\le\frac23$, we can use $\frac{dy}{dx}=\frac32x$ to deduce that
$$
g(x)=g(y)=\frac{f(x)+3xf(y)/2}{1+3x/2}
$$
for almost every $x\in[0,\frac23)$. If $A\in\mathcal{F}$ is independent to $\mathcal{G}$, so that $\mathbb{P}(A\mid\mathcal{G})=c$ is constant, then consider taking $f$ to be the indicator function of $A$. Then, $f(x)+3xf(y)/2=(1+3x/2)c$ almost everywhere. However, the left hand side can only take the values $0,1,3x/2,1+3x/2$, from which we can deduce that $c=0$ or $c=1$.
This example only works because the conditional measure (or disintegration) $\mathbb{P}(\cdot\mid\mathcal{G})$ is atomic, even though $\mathbb{P}$ is not.