Help understanding conditional expectation identity

[psie]

TL;DR

I'm reading a proof on conditional probabilities and there is an identity involving conditional expectation which I'm stuck on.

Let $(\Omega,\mathcal{F},P)$ be a probability space, and let us define the conditional expectation ${\rm E}[X\mid\mathcal{G}]$ for integrable random variables $X:\Omega\to\mathbb{R}$, i.e. $X\in L^1(P)$, and sub-sigma-algebras $\mathcal{G}\subseteq\mathcal{F}$.

Definition 1: The conditional expectation ${\rm E}[X\mid\mathcal{G}]$ of $X$ given $\mathcal{G}$ is the random variable $Z$ having the following properties:
(i) $Z$ is integrable, i.e. $Z\in L^1(P)$.
(ii) $Z$ is ($\mathcal{G},\mathcal{B}(\mathbb{R}))$-measurable.
(iii) For any $A\in\mathcal{G}$ we have $$\int_A Z\,\mathrm dP=\int_A X\,\mathrm dP.$$

Definition 2: If $X\in L^1(P)$ and $Y:\Omega\to\mathbb{R}$ is any random variable, then the conditional expectation of $X$ given $Y$ is defined as $${\rm E}[X\mid Y]:={\rm E}[X\mid\sigma(Y)],$$ where $\sigma(Y)=\{Y^{-1}(B)\mid B\in\mathcal{B}(\mathbb{R})\}$ is the sigma-algebra generated by $Y$.

If $\mathcal{G}=\sigma(Y)$, then (iii) in definition 1 says that $${\rm E}[\mathbf{1}_A{\rm E}[X\mid Y]]={\rm E}[\mathbf{1}_AX],\quad \forall A\in\sigma(Y).\tag1$$

Now, in a proof I'm reading currently, there are three random variables $U,S,T$ and the following computation appears in the proof: $$\int_{T^{-1}(B)} U\,\mathrm dP={\rm E}[\mathbf{1}_B(T)U]={\rm E}[\mathbf{1}_B(T){\rm E}[U\mid S,T]].$$I simply do not comprehend the last equality, that is ${\rm E}[\mathbf{1}_B(T)U]={\rm E}[\mathbf{1}_B(T){\rm E}[U\mid S,T]]$. How does this follow from the definitions above and the identity $(1)$? I'm grateful for any help on this.

 

[psie]

I think I understand the identity in question now. First, from Durret's book, we have

If $X$ is $\mathcal G$-measurable and $E|Y|,E|XY|<\infty$, then $$E[XY|\mathcal{G}]=XE[Y|\mathcal{G}].$$

Second, we need the Tower property or the law of the iterated expectation, that is $E[Y|\mathcal{H}]= E\big[E[Y\mid \mathcal G]\mid \mathcal H\big]$, where $\mathcal H\subset\mathcal G$.

By the latter property, we have $$E[\mathbf1_B(T)U]=E[E[\mathbf1_B(T)U\mid S,T]].$$ Now, by the theorem in Durret, $\mathbf1_B(T)=\mathbf1_{T^{-1}(B)}$ is $\sigma(T)$-measurable, and this is a subset of $\sigma(S,T)=\sigma(\sigma(S)\cup\sigma(T))$. So we can "pull it out", and we are left with $$E[\mathbf1_B(T)U]=E[\mathbf1_B(T)E[U\mid S,T]],$$which is the desired identity.