Several algebras in conditional and join probabilities There are several cases in combining conditional and joint probabilities, I get confused to show them if there are correct:

Given $A, B, C$ are random variables, $P(A|B)$ denotes conditional probability and $P(B, C)$ denotes joint probability.


$P((A|B)|C) = P(A|(B, C))$


We have: $P((A|B)|C) = \frac{P((A|B), C)}{P(C)}$. This requires following


$P((A|B), C) = P(A|(B, C))P(C)$


If the equality holds, there would be no ambiguity with $P(A|B, C)$
We can to understand $\Bbb{P}(A \vert B) = \Bbb{E}[1_A |\sigma(1_B)](\omega)$ for any $\omega \in B$. 

In that sense one would interpret $\Bbb{P}((A \vert B) \vert C) = \Bbb{E}\Bbb{E}[1_A |\sigma(1_B)] \vert \sigma(1_C)](\omega)$ for $\omega \in B \cap C$

$$\Bbb{P}((A \vert B) \vert C) = \Bbb{E}\Bbb{E}[1_A |\sigma(1_B)] \vert \sigma(1_C)](\omega) = \Bbb{E}\Bbb{E}[\Bbb{E}[1_A |\sigma(1_B, 1_C)] |\sigma(1_B)] \vert \sigma(1_C)](\omega) = \Bbb{E}\Bbb{E}[1_A |\sigma(1_B, 1_C)] \vert \sigma(1_C)](\omega) = \Bbb{E}[1_A |\sigma(1_B, 1_C)](\omega) = P(A \vert B,C)  $$

Note $\Bbb{E}[1_A\vert \sigma(1_B)]$ is a random variable 
$$\Bbb{E}[1_A\vert \sigma(1_B)] = P(A\vert B) 1_B + P(A\vert B^c)1_{B^c} $$

The relation we used for conditional expectations was this one.

$$\mathcal{G} \subset \mathcal{F} \Rightarrow\Bbb{E}[\Bbb{E}[1_A |\mathcal{G}] |\mathcal{F}] = \Bbb{E}[\Bbb{E}[1_A |\mathcal{G}]$$

where $\mathcal{F},\mathcal{G}$ are sigma algebras.

Note that we are understanding $P(A\vert B,C) = \Bbb{E}[1_A |\sigma(1_B, 1_C)]$ wich is different than $\Bbb{E}[1_A |\sigma(1_B 1_C)]$ to finally get a number we write

$$\Bbb{E}[1_A |\sigma(1_B, 1_C)] = P(A\vert B \cap C)1_{B \cap C} + P(A\vert B^c \cap C)1_{B^c \cap C} + P(A\vert B \cap C^c)1_{B \cap C^c}+ P(A\vert B^c \cap C^c)1_{B^c \cap C^c}$$

So you define $P(A \vert B,C) = \Bbb{E}[1_A |\sigma(1_B, 1_C)] (\omega)$  for any $\omega \in B \cap C$

As you would have defined $P(A\vert B) = \Bbb{E}[1_A |\sigma(1_B)](\omega)$ for any $\omega \in B$
We can to understand $\Bbb{P}(A \vert B) = \Bbb{E}[1_A |\sigma(1_B)](\omega)$ for any $\omega \in B$. 

In that sense one would interpret $\Bbb{P}((A \vert B) \vert C) = \Bbb{E}\Bbb{E}[1_A |\sigma(1_B)] \vert \sigma(1_C)](\omega)$ for $\omega \in B \cap C$

$$\Bbb{P}((A \vert B) \vert C) = \Bbb{E}\Bbb{E}[1_A |\sigma(1_B)] \vert \sigma(1_C)](\omega) = \Bbb{E}\Bbb{E}[\Bbb{E}[1_A |\sigma(1_B, 1_C)] |\sigma(1_B)] \vert \sigma(1_C)](\omega) = \Bbb{E}\Bbb{E}[1_A |\sigma(1_B, 1_C)] \vert \sigma(1_C)](\omega) = \Bbb{E}[1_A |\sigma(1_B, 1_C)](\omega) = P(A \vert B,C)  $$

Note $\Bbb{E}[1_A\vert \sigma(1_B)]$ is a random variable 
$$\Bbb{E}[1_A\vert \sigma(1_B)] = P(A\vert B) 1_B + P(A\vert B^c)1_{B^c} $$

The relation we used for conditional expectations was this one.

$$\mathcal{G} \subset \mathcal{F} \Rightarrow\Bbb{E}[\Bbb{E}[1_A |\mathcal{G}] |\mathcal{F}] = \Bbb{E}[\Bbb{E}[1_A |\mathcal{G}]$$

where $\mathcal{F},\mathcal{G}$ are sigma algebras.

Note that we are understanding $P(A\vert B,C) = \Bbb{E}[1_A |\sigma(1_B, 1_C)]$ wich is different than $\Bbb{E}[1_A |\sigma(1_B 1_C)]$ to finally get a number we write

$$\Bbb{E}[1_A |\sigma(1_B, 1_C)] = P(A\vert B \cap C)1_{B \cap C} + P(A\vert B^c \cap C)1_{B^c \cap C} + P(A\vert B \cap C^c)1_{B \cap C^c}+ P(A\vert B^c \cap C^c)1_{B^c \cap C^c}$$

So you define $P(A \vert B,C) = \Bbb{E}[1_A |\sigma(1_B, 1_C)] (\omega)$  for any $\omega \in B \cap C$

As you would have defined $P(A\vert B) = \Bbb{E}[1_A |\sigma(1_B)](\omega)$ for any $\omega \in B$