Let me know if I'm talking nonsense but it looks like if you just consider the usual entropy $H(P)=\sum_k P(k)\log P(k)$ for probability measures $P$ on $\mathbb N$, then
$$
E[H(X_*Q_{n+1})]-E[H(X_*Q_{n})]= E I_n
$$
simply because of the identity
$$
p\log p-q\log q=p\log \frac pq+(p-q)\log q
$$
(the linear in $p$ part just gets killed by the conditional expectation, as usual). Then the answer is that you are fine as long as the entropy of $X_*\mathcal P$ is finite (which is always the case when you have finite range) but not necessarily if it is not (there is a difference between the finiteness of the expectation and the finiteness a.e., of course, but you said that you have a counterexample for the infinite range yourself, so I'll not elaborate).

Does this make sense?

**Edit: a counterexample without finite range requirement**.

Let us take $[0,1]$ with the Borel $\sigma$-algebra and the Lebesgue measure for the probability space and the usual dyadic partitions as the filtration.
To create the distribution of $X_*P$, just split the integers into chunks of lengths $2^n$, assign some probability $c_n$ to each chunk so that $\sum_n c_n=1$, $\sum_n nc_n=+\infty$. Now we need to construct $X$ itself. Start with the first chunk ($\{0,1}\}$ of length $2$) and assign the conditional probabilities so that $P(0)=2c_1, P(1)=0$ on the left half, the reverse holds on the right half, and the rest is split equally. That gives the information gain $I_1$ of $1\cdot c_1\log 2$ uniformly throughout the whole interval. Now take the second chunk $\{2,3,4,5\}$ and first split so that $2,3$ go to the left and $4,5$ to the right, after which separate them again at the next level. Again, the corresponding gain $I_2+I_3$ will be spread uniformly and equal to $2\cdot c_2\log 2$. Now take care of the third chunk and so on. Note that we do not have $X$ itself yet, but we have the martingales for $P(X=k|\omega\in I)$ for each dyadic $I$ and can lift the sum of gains to any level we want uniformly throughout the interval. The only little problem is that those martingales converge to some fractional values instead of $0,1$, so we do not get true set partitions. To compensate for that, we need to start making slightly non-even partitions now and then allowing the probability of every number whose fate we already decided (in the sense that once we know $I$, we know if it is either present on each further subinterval, or is absent on each further subinterval to spread or squeeze a bit. Just do it rarely enough and only once after you accumulate a uniform gain of $1$. This operation is fairly harmless because all it does to the tails is to multiply them by some positive numbers, which doesn't change the divergence of the series and, thereby, doesn't affect our gain accumulation abilities. However now, from the perspective of each integer, you start running a random walk for its conditional probability, so if your step is, say, half of the distance to the endpoint at least once in a while, then you'll end up with $0$ or $1$ almost surely and full measure will be honestly partitioned.

slightly differentdefinitions of push-forward measures in probability you are referring to? $\endgroup$2more comments