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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?

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Hello. You might be interested in Proposition 4 of this paper: – Stéphane Laurent Sep 28 '15 at 21:35
up vote 13 down vote accepted

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 $1$.

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I suppose I should have said the answer was "almost surely yes" to allow myself a humble retreat. Thanks for the elegant, simple counterexample, Nate. It's back to the drawing board for me. – Tom LaGatta Feb 1 '11 at 23:31
What if the probability is non-atomic? – Andrés E. Caicedo Feb 1 '11 at 23:35
Incidentally, in my second example, $L^2(\mathcal{G}) = L^2(\mathcal{F})$, pointing out a problem with your proposed argument. $L^2(\mathcal{G})$ is always a closed subspace of $L^2(\mathcal{F})$, but it need not be proper. – Nate Eldredge Feb 2 '11 at 0:38
@Nate: Yes. The difference between that and my example is that I have $L^2(\mathcal{G})\not=L^2(\mathcal{F})$, but there are still no sets in $\mathcal{F}$ independent of $\mathcal{G}$ other than those with probability 0 or 1. That was what I was aiming to achieve with the construction. – George Lowther Feb 2 '11 at 1:02
@George: Indeed, and I like your example better for that reason. I had been thinking about my example when you posted yours, and I thought I might as well post it anyway. – Nate Eldredge Feb 2 '11 at 14:46

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.

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