First of all, when we say that $X_n \sim \text{Unif}(0,X_{n-1})$, what does that mean, rigorously? Does it mean that for every $\omega \in \Omega$, $X_n(\omega)\sim \text{Unif}(0,X_{n-1}(\omega))$? This doesn't make much sense as $X_n(\omega)$ is a number and not a random variable. Or does it mean that for every $\omega \in \Omega$ there is some auxiliary $X'_n \sim \text{Unif}(0,X_{n-1}(\omega))$, and then $X_n(\omega) = X'_n(\omega)$? Once we have that cleared up, I have another question:

Let's say that we have $X_n \sim \text{Unif}(0,X_{n-1})$ where $X_0=1$. I've been told that defining $\mathcal A_n = \sigma[X_0, \ldots, X_{n}]$, $$\mathbb E(X_{n+1}|\mathcal A_n) = \mathbb E[\text{Unif}(0,X_{n-1})] = \frac{0+X_{n-1}}{2}$$ Similarly, for $Z_{n+1} \sim \text{Poisson}(Z_n)$ and $\mathcal A_n$ defined similarly as in the above case, I've been told that $$\mathbb E(Z_{n+1}|\mathcal A_n) = \mathbb E[\text{Poisson}(Z_{n})] = Z_n$$ and also for $Y_{n+1} \sim2 \text{Binom}(Y_n, p)$ $$\mathbb E(Y_{n+1}|\mathcal A_n) = 2\mathbb E[\text{Binom}(Y_{n}, p)] = 2Y_n p$$

Why is this true? And is this generally true? The only formula from conditional expectations that I think would apply here is that $\mathbb E(Y|\mathcal D) =_\text{a.s.} \mathbb E[Y]$ if $\mathcal F(Y)$ and $\cal D$ are independent. However, in the above case I don't think the sigma fields are independent, so I don't know how to go about proving the nice equations above.