How can we prove $$\mathbb{E}[UV\mid \mathcal{B} ] = U\mathbb{E}[V\mid \mathcal{B} ]$$
given that both r.v's are positive and $U$ being $\mathcal{B}$-measurable?
I use this property pretty often but I don't know how to actually prove it. Either we have to show that $\forall B \in \mathcal{B}$
$$\int_{B}UVd\mathbb{P} =\int_{B}U\mathbb{E}[V \mid \mathcal{B} ] d\mathbb{P} $$
or $\forall Z \text{ that is } \mathcal{B}$-measurable and positive
$$\mathbb{E}[UVZ] = \mathbb{E}[U\mathbb{E}[V\mid \mathcal{B} ]Z ]$$
but how?
Notice that there exists a monotone sequence of simple random variables $U_n$ which are $\mathcal B$ - measurable converging point wise to $U$. The random variables $U_n$ can be written as $$U_n=\sum_{k=1}^{N_n}\mathbf{1}_{A_{k,n}}$$ where $A_{k, n}\in\mathcal B $.
Now we have using MCT \begin{align} \int_B UV\, d\mathbb P&=\int_{ B}\lim_{n\to\infty} \sum_{k=1}^{N_n}\mathbf{1}_{A_{k, n}} V\, d\mathbb P \\ &=\lim_{n\to\infty} \sum_{k=1}^{N_n}\int_{A_{k, n} \cap B} V\, d\mathbb P \\ &=\lim_{n\to\infty} \sum_{k=1}^{N_n}\int_{A_{k, n} \cap B} \mathbb E(V\mid \mathcal B)\, d\mathbb P\\ &=\lim_{n\to\infty} \sum_{k=1}^{N_n}\int_{ B} \mathbf{1}_{A_{k,n}} \mathbb E(V\mid \mathcal B)\, d\mathbb P \end{align} And this implies indeed using MCT again that...
You need to figure out how to use MCT yourself though.