Tower property of conditional expectation

Solution 1:

The last equality in your observation does not apply in general (i.e. if $X$ is not discrete). Let $U,V,W$ be random variables such that $V\in \mathcal{L}^1(P)$. In order to show that $$ E[V\mid W]=E[E[V\mid U,W]\mid W] $$ we note that the right hand side is indeed $\sigma(W)$-measurable, so we only need to check the defining equation, i.e. check that $$ \int_A V\,\mathrm{d}P=\int_A E[V\mid U,W]\,\mathrm{d}P $$ for all $A\in\sigma(W)$. Let such an $A$ be given. Then $A\in\sigma(W)\subseteq \sigma(U,W)$ and therefore $$ \int_A E[V\mid U,W]\,\mathrm{d}P=\int_A V\,\mathrm{d}P $$ and we are done.