4

Motivation

Let $(\Omega, \mathcal F, \mathbb P)$ be a standard probability space. For two $\mathcal H, \mathcal G \subset \mathcal F$, we say $\mathcal H = \mathcal G$ mod 0, if they are same up to null sets (i.e. if for each $H \in \mathcal H$, there is a counterpart $G \in \mathcal G$ with $H \Delta G$ being a null set and vice versa.). Let $X$ be a $L^1$ random variable on $\Omega$. Let $\mathcal H$ be a sub-sigma-algebra of $\mathcal F$. Then the conditional expectation of $X$ given $\mathcal H$, denoted by $E(X | \mathcal H)$, is another random variable and is unique up to null sets.

I wonder if reasoning about quantities like $E(X | \mathcal H)$ reduces to reasoning about quantities like $E(X | \sigma(Y))$ (where $Y$ is another random variable).

Question

Is it true that every sub-sigma-algebra $\mathcal H$ (of $\mathcal F$) is (up to null sets) generated by a random variable? In other words, can we always find a random variable $Y$ such that $\sigma(Y) = \mathcal H$ mod 0?

Cautions

Recall that $(\Omega, \mathcal F, \mathbb P)$ is assumed to be a standard probability space. This assumption is there to avoid pathological cases.

The question does not require that $\sigma(Y)$ and $\mathcal H$ be exactly the same, only that they be the same up to null sets. This weak requirement is because of the motivation, and because of:

Is every sigma-algebra generated by some random variable?

Some thoughts

Counterexamples cannot be formed by taking many random variables: Any $\sigma$-algebra of the form $\sigma(Y_1, Y_2, \cdots)$ can be generated by a single random variable $Y$ because $\mathbb R^{\mathbb N}$ is isomorphic to $\mathbb R$ as measurable spaces.

The invariant sigma algebra (from the $L^1$ ergodic theorem for non-ergodic measures) cannot be a counterexample: if $f_1, f_2, \cdots$ forms a countable dense subset of the $L^1$ space associated with the system, then their time averages $f_1^*, f_2^*, \cdots$ generate the invariant sigma algebra up to null sets.

Community
  • 1
Jisang Yoo
  • 1,047

1 Answers1

6

Every countably generated $\sigma$-algebra is generated by a real random variable, for if $G_0, G_1, G_2\ldots$ is a sequence of generators, we can take the random variable to be the Marczewski function given by $$g(x)=\sum_{n=0}^\infty\frac{2}{3^{n+1}}I_{G_n}(x).$$

It remains to show that every sub-$\sigma$-algebra of a standard probability space is equal to a countably generated $\sigma$-algebra mod $0$. On a standard probability space, the pseudometric $d:\mathcal{F}\times\mathcal{F}\to\mathbb{R}$ is separable, so if $\mathcal{H}\subseteq\mathcal{F}$ is a sub-$\sigma$-algebra, there is a countable set $\mathcal{C}\subseteq\mathcal{H}$ such that $\mathcal{H}$ is included in the closure of $\mathcal{C}$ and hence the closure of $\mathcal{G}=\sigma(\mathcal{C})$. But $\mathcal{G}$ is a complete subspace, since a Cauchy sequence of indicator functions is also Cauchy in convergence in measure, which is complete. So $\mathcal{H}=\mathcal{G}$ mod $0$.

Michael Greinecker
  • 34,865