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May 27, 2017 at 15:28 comment added fedja @Squark Good point. Indeed it can :-). It is just my background as an analyst that makes me to create functions on $\mathbb R^n$ instead of choosing some more convenient underlying space. I should become more flexible about that, or my brain will become too stiff and fail to see the obvious.
May 27, 2017 at 15:02 comment added Vanessa Yeah, I'm convinced, thank you. The construction can be simplified by letting $\Omega$ be the Cartesian product of $[0,1]$ (responsible for selecting a number inside each chunk) and $\mathbb{N}$ (responsible for selecting the chunk).
May 27, 2017 at 14:39 history edited fedja CC BY-SA 3.0
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May 27, 2017 at 14:02 comment added fedja @Squark OK, I'll add it to the answer. Just hold on for a while: I'm a slow typist.
May 27, 2017 at 13:56 comment added Vanessa Definitely, I'm intrigued.
May 27, 2017 at 12:55 comment added fedja @Squark -Only that my counterexample was wrong- I checked mine, it seems correct. Do you want me to post it?
May 27, 2017 at 10:55 comment added Vanessa Only that my counterexample was wrong. In fact, in the countable case you can replace $H(X_* Q_n)$ with $-D_{KL}(X_* Q_n \mid\mid X_* Q_0)$ and it remains to prove that the latter is almost surely bounded below which sounds plausible. Anyway, I reformulated the question to refer only to the finite case so never mind...
May 27, 2017 at 10:45 comment added Vanessa You are absolutely right, except that you need $E[H_n] - E[H_{n+1}]$ there because the definition of $H$ has a minus sign that is not present in the definition of $D_{KL}$.
May 27, 2017 at 10:44 vote accept Vanessa
May 27, 2017 at 10:22 history edited fedja CC BY-SA 3.0
edited body
May 27, 2017 at 10:11 history answered fedja CC BY-SA 3.0