Do we have two definitions for conditional probabilities? I have currently two probability books in front of me, and the more I read about conditional probabilities, the more I get confused. 
My first question is: do we have two definitions for conditional probabilities?
I mean, are these two definitions valid for conditional probabilities? given that $P(Y\in B)$ and $P(Y=y)$ will not equal 0.
$P(X\in A|Y=y)=\frac{P(X\in A,Y=y)}{P(Y=y)}$
$P(X\in A|Y\in B)=\frac{P(X\in A,Y\in B)}{P(Y\in B)}$


*

*What is the difference between these two definitions? 

*How can I use them to define the conditional distribution as a measure?


any guidance is appreciated to clear my confusion
 A: There is only one definition, and the two statements you give are valid.
The conditional probability is defined for two events.
Given a universe of possible outcomes, $\Omega$, a set of events $E$ (actually a $\sigma$-algebra of events), an event being a subset of $\Omega$, and a measure $P$ on $(\Omega,E)$, with a number of axioms that define a probability ($(\Omega,E,P)$ is then a probability space), then the conditional probability of the event $A$ conditioned by the event $B$ such that $P(B)\ne0$ ($A,B\in E$) is:
$$P(A|B)=\frac{P(A \cap B)}{P(B)}$$

Note that $X\in A$, $Y=y$, $Y\in B$ are all events. There is a common abuse of notation here. $X$ and $Y$ are random variables (that is, measurable functions from $\Omega$ to $\Bbb R$), $A$ and $B$ are subsets of $\Bbb R$, $y$ is a real number, and the preceding events are respectively


*

*$\{\omega\,|\,\omega\in\Omega\wedge X(\omega)\in A\}$

*$\{\omega\,|\,\omega\in\Omega\wedge Y(\omega)=y\}$

*$\{\omega\,|\,\omega\in\Omega\wedge Y(\omega)\in B\}$


There is no inconsistency here, you just have to be accurate about the notation.

The second part of your question: how to define a conditional probability measure?
Given an event $B$ such that $P(B)\ne0$, you can define another probability measure,
$$P_B(A)=\frac{P(A\cap B)}{P(B)}$$
You may then check the axioms of a probability on $P_B$ (for instance you have obviously $P_B(\Omega)=\frac{P(\Omega\cap B)}{P(B)}=\frac{P(B)}{P(B)}=1$). Here you can replace $B$ by any of the preceding events.
