Confusion regarding conditional probability distributions Let's say we condition a probability distribution with respect to some sub-sigma algebra $F$, forming $P(\cdot\mid F)$. This object has to satisfy three properties:


*

*For any $x$, $P(\cdot\mid F)(x)$ has to be a probability distribution.

*For any $A$, $P(A\mid F)(\cdot)$ has to be an $F$-measurable function.

*For any $A$, $\int P(A\mid F)(x) \, dP=P(A)$.


Well, can't I just set $P(A\mid F)(x)=P(A)$ for all $A$, $x$? What am I missing?
 A: Point 3 isn't quite correct. It should be this:

For any $A$ and $B \in F$, $\int_B P(A \mid F)(\omega) dP(\omega) = P(A \cap B)$,

as Did pointed out (I was still typing my answer when he did so!).
We can see this from the general formulation as follows. I'll use slightly different notation, but hopefully it should be clear!

Let $X$ be an integrable random variable and let $\mathcal{G} \subseteq \mathcal{F}$ be a $\sigma$-\algebra. Then there exist a random variable $Y$ such that
  
  
*
  
*$Y$ is $\mathcal{G}$-measurable,
  
*$Y$ is integrable and $E(X 1_B) = E(Y 1_B)$ for all $B \in \mathcal{G}$.
  
  
  We write $Y = E(X \mid \mathcal{G})$.

Now let's take $X = 1_A$; we then write $Y = E(1_A \mid \mathcal{G}) = P(A \mid \mathcal{G})$. You are asking to define $Y = P(A)$, so not a random variable. Now, for any $B \in \mathcal{G}$, we have these:
$$ P(A)P(B) = P(A) E(1_B) = E( P(A) 1_B ) = E(Y 1_B); $$
$$ E(X 1_B) = E(1_A 1_B) = E(1_{A \cap B}) = P(A \cap B).$$
We see that these two are the same if and only if $A$ and $B$ are independent! We notice, though, that it does tell us what the criterion should be:
$$ E(Y 1_B) = P(A \cap B) \quad \text{for all } B \in \mathcal{G}. $$
I'm not sure how much more illuminating this is than Did's comment, but I'd written almost all of it by the time I saw his comment -- hopefully it does give some further understanding!

Here's a way (that I used) to think about it is the following. Imagine a random variable $X$ with some pdf $f$ on $[0,1]$ (not necessarily a step function -- so we're considering conditional expectation, not just probability), and take the $\sigma$-algebra generated by, say $[0,1/3]$, $[1/3,2/3]$ and $[2/3,1]$. Then the conditional expectation of $X$ given this $\sigma$-algebra  -- call it $Y$ -- is just obtained by averaging: the pdf $g$ of $Y$ is a step function, and at a point $y$ it is the average of the pdf of $X$ on that interval; eg $g(1/4) = \int_0^{1/3} f(z) dz / (1/3)$. I find this a helpful viewpoint -- you may not!
A: I'm going to write $P^F_\omega(A)$ for $P(A|F)(\omega)$.
A sigma algebra $F$ represents "partial information" about the result of an experiment. Specifically, if $F$ represents the information that we're able to observe, then when a point $\omega\in\Omega$ is sampled, the information that we have access to is precisely which sets of $F$ contain $\omega$. The probability distribution $P^F_\omega$ is then in some sense the conditional distribution of $\omega$ given that information.
Now, since we "know" which sets in $B\in F$ contain $\omega$, we can certainly deduce one thing: if $B\in F$ and $\omega\in B$, then certainly $\omega$ is not in any set disjoint from $B$. Thus for $\omega\in B$, $B\in F$, we should have $P_\omega^F(A)=0$ for any $A$ disjoint to $B$. Or, equivalently and more simply:
$$\forall B\in F\ \forall \omega\in B,\ P_\omega^F(\Omega\setminus B)=0$$
If we add this statement to the list of three axioms in the OP, we get an equivalent definition for the conditional probability. As pointed out by other members, the correct definition is axioms 1-3 in the OP, but with axiom 3 replaced with the stronger assumption that for all $B\in F$ and for any event $A$:
$$\int_BP^F_\omega(A)dP=P(A\cap B)$$
I'll prove that this last identity follows from the axioms in the OP plus the extra assumption I wrote above. Simply write $A$ as $(A\cap B)\cup (A\cap B^c)$, and we have:
$$\int_BP^F_\omega(A)dP=\int_BP^F_\omega(A\cap B)dP+\int_BP^F_\omega(A\cap B^c)dP$$
The second integral is zero, since the integrand is identically zero on $B$ by our assumption. And the first integral can be rewritten as
$$\int_BP^F_\omega(A\cap B)dP=\int_\Omega P^F_\omega(A\cap B)dP-\int_{\Omega\setminus B}P^F_\omega(A\cap B)dP=\int_\Omega P^F_\omega(A\cap B)dP$$
Which equals $P(A\cap B)$ by axiom 3 in the OP.
