# chain rule for the conditional probability from measure-theoretic point of view

I am studying the probability theory from Durrett's book. Chapter 5 begins with the discussion on the conditional probability. I have a difficulty in understanding the relationship between measure-theoretic view of conditional expectation and my undergrad-level knowledge of probability.

Assume we have three random variables X, Y, and Z. In the undergrad-level probability course, we had chain rule for the conditional probability which says $$P(X\vert Y) = \int P(Z=z\vert Y) P(X\vert Y,Z=z) dz.$$ Now, I am trying to understand the above equality with measure theory tools. Specifically, we can write $$P(X\in A \vert \sigma(Y)) = \mathbb{E}[1_{X\in A}\vert \sigma(Y)].$$ I do not know how to proceed after this point to recover the chain rule for conditional probability. I would really appreciate it if you could help me on this issue.

$$P(X\in A \vert \sigma(Y)) = \mathbb{E}[1_{X\in A}\vert \sigma(Y)]$$

$$\overset{Tower \ Property}{=}E\bigg(\mathbb{E}[1_{X\in A}\vert \sigma(Y)] \bigg|\sigma(Y,Z)\bigg)$$ Tower property since $$\sigma(Y)\subset \sigma(Y,Z)$$ Conditional_expectation

$$\overset{Tower \ Property}{=}E\bigg(\mathbb{E}[1_{X\in A}\vert \sigma(Y,Z) ] \bigg|\sigma(Y)\bigg)$$ $$=E\bigg(\mathbb{E}[1_{X\in A}\vert (Y,Z) ] \bigg|Y\bigg)$$

$$=E\bigg(g(Y,Z) \bigg|Y \bigg)$$

$$=\int g(Y,Z=t) f(Z=t|Y) dt$$ $$=\int \mathbb{E}\bigg(1_{X\in A}\vert (Y,Z=t)\bigg) f(Z=t|Y) dt$$ $$=\int P\bigg\{X\in A\vert(Y,Z=t) \bigg\}f(Z=t|Y) dt$$

so
$$P(X\in A \vert \sigma(Y)) =\int P\bigg\{X\in A\vert Y,Z=t \bigg\}f(Z=t|Y) dt$$ so $$P(X\leq x \vert \sigma(Y)) =\int P\bigg\{X\leq x\vert Y,Z=t \bigg\}f(Z=t|Y) dt$$

if $$X$$ is continues you can derive by $$x$$ in both side and get conditional density.