For the discussion below, let $(X,\mathcal{B},\mu)$ be a probability measure. Let $\mathcal{A}\subset\mathcal{B}$ be a sub-sigma-algebra. Let $\alpha$ be a partition of $X$ to countably-many disjoint, measurable sets.
I'm reading W. Parry's "Topics in Ergodic Theory". The author introduces the concepts of "Information of a partition of a probability space" and "Entropy of a partition". I've attached the relevant definition in the end of the post.
A few sentences afterward, he mentions:
As easily seen, $H(\xi|\mathcal{A})=0$ (or $I(\xi|\mathcal{A}) \equiv 0$) if and only if $\xi$ consists of $\mathcal{A}$ sets.
I'm having two issues with the last quote.
- I'm not entirely sure what "Consist of $\mathcal{A}$ sets" means. I think it means $\xi\subset \mathcal{A}$, but I'm not confident with it.
- Under the assumption that $\xi \subset \mathcal A$ is indeed the author's intention, I don't see why this is correct. If I'm not mistaken, we'd have: $$ \begin{align} I(\xi|\mathcal{A})(x) &= - \sum_{A\in \mathcal{A}}1_A (x) \cdot \log (\mu (A|\mathcal{A})) \\ &= - \log(\mu (A_0 | \mathcal{A})) \end{align} $$ Where $x\in A_0$. It is not clear to me why the conditional measure is one in that case. I'd be glad for clarifications on that. Edit - There's a mistake in the last expression. See correction bellow.
Definition - Conditional Information:
We define $I(\alpha | \mathcal{A})$ to be the measurable function: $$ \begin{align} I(\alpha | \mathcal{A}) &: X \to \mathbb{R} \ \ \ s.t. \\ &I(\alpha | \mathcal{A}) = - \sum_{A\in \alpha} 1_A \cdot \log(\mu(A|\mathcal{A})) \end{align} $$
EDIT:
In the calculation for $I(\xi | \mathcal{A}) $ I was using the wrong definition. We should have:
$$ \begin{align} I(\xi|\mathcal{A})(x) &= - \sum_{A\in \xi}1_A (x) \cdot \log (\mu (A|\mathcal{A})) \\ &= - \log(\mu (A_0 | \mathcal{A})) \end{align} $$
Where $A_0 \in \xi $, and not $A_0 \in \mathcal A$. Under this correction, I think the desired claim is now reduced to: $$ \begin{align} &\mu (A_0 | \mathcal{A}) = 1 \\ \iff &A_0 \in \mathcal A \end{align} $$
I still can't see why is it true.