Law of Iterated Expectation with Inequality Conditional I am wondering if $\mathbb{E}(X\mid Y<z)=\mathbb{E}(\mathbb{E}(X\mid Y)\mid Y<z)$, for $z\in\mathbb{R}$. My intuition says that it is true, because events of the form $\{\omega:Y(\omega)<z\}\in\sigma(Y)$, however, I'm not sure how to apply the conditional expectation properties to get to the result directly. I have an attempt of proof, however, and I wish to know if it seems correct. It is as follows: Note that $\mathbb{E}(X\mid A)=\frac{\mathbb{E}(X1_A)}{\mathbb{P}(A)}$, so we have
$$\mathbb{E}(X\mid Y<z)=\frac{\mathbb{E}(X1_{Y<z})}{F_Y(z)}=\frac{\mathbb{E}(\mathbb{E}(X1_{Y<z}\mid Y))}{F_Y(z)}=\frac{\mathbb{E}(1_{Y<z}\mathbb{E}(X\mid Y))}{F_Y(z)}=\mathbb{E}(\mathbb{E}(X\mid Y)\mid Y<z)$$
Where the second equality follows from the Law of Iterated Expectation, the third follows from $1_{Y<z}$ being $\sigma(Y)$-measurable, as $A=\{\omega:Y(\omega)<z\}\in\sigma(Y)$, and the last one, from the the identity of the expectation conditional on the event A, considering $\mathbb{E}(X\mid Y)$ a random variable that only depends on $Y$.
 A: The question is more about understanding of measure-theoretic definitions than reasoning. 
Measure-theoretic definitions start with $\mathbb{E}(X\vert\mathcal{F}).$ Then it extends to $\mathbb{E}(X\vert Y)$ by treating it as an alias of $\mathbb{E}(X\vert\sigma(Y))$, and finally  $\mathbb{E}(X\vert A)$ means  $\mathbb{E}(X\vert 1_A) \equiv  \mathbb{E}(X\vert \{\phi, A, A^c, \Omega\}).$
Let $A = \{Y < z\} \in \sigma(Y)$. Further let $\mathcal{F}_1=\{\phi, A, A^c, \Omega\}$, and $\mathcal{F}_2=\sigma(Y).$ Then $\mathcal{F}_1 \subset \mathcal{F}_2$. Your result $\mathbb{E}(X\mid A)=\mathbb{E}(\mathbb{E}(X\mid Y)\mid A)$ directly follows from a known theorem $\mathbb{E}(X\mid \mathcal{F}_1)=\mathbb{E}(\mathbb{E}(X\mid \mathcal{F}_2)\mid \mathcal{F}_1)$.
PS: If you insist on directly proving instead of applying the theorem, you can make use of the fact that $A, A^c$ form a partition of $\mathcal{F}_1$ , and rules of partitioning, $\mathbb{E}(X \vert \sigma(\Omega_1, \cdots)) = \frac{E(X; \Omega_i)}{P(\Omega_i)}$ on $\Omega_i$. $\int_AXdP=\int_A\mathbb{E}(X\mid Y)dP$ since $A \in \sigma(Y)$. The left side gives values of random variable $\mathbb{E}(X\mid A)$ scaled by $P(A)$ for $\omega \in A$, and the right hand side gives values of random variable of $\mathbb{E}(\mathbb{E}(X\mid Y)\mid A)$ scaled by $P(A)$ for $\omega \in A$. Similar conclusion holds for $\omega \in A^c$. Hence the two random variables are the same.
