Show that $P_B(A)=\sum\limits_{\omega\in A\cap B}\frac{P(\{\omega\})}{P(B)}$ In our lecture the professor wrote the follwoing statement:

Let be $(\Omega,P)$ a discrete probability space and $B\subset \Omega$ where $P(B)>0$. Let be $P_B:\Omega\supseteq A\mapsto P_B(A)$ an additive mapping (you can add the images of disjoint sets) with the properties
$1.)$ $P_B(B)=1$.
$2.)$ If $A\subset B$ then $P_B(A)=c_B P(A)$, where $c_B$ is some constant.
Show that $P_B(A)=\sum\limits_{\omega\in A\cap B}\frac{P(\{\omega\})}{P(B)}$.

I don't think that this statement is true/can be proven and lacks further assumptions but my professor didn't pay attention to my concerns...
As a counter-example we can take the additive mapping $\frac{P(A)}{P(B)}$ where $A\in\Omega$ and some $B$ with $P(B)>0$. It obviously satisfies both properties. If we choose some set $C\subset \Omega$ which is "bigger" than $B$, i.e. $B\subset C$, then $P_B(C)=\frac{P(C)}{P(B)}=\frac{P(C\setminus B)+P(B)}{P(B)}=\frac{P(C\setminus B)}{P(B)}+1>\sum\limits_{\omega\in C\cap B}\frac{P(\{\omega\})}{P(B)}=1$.
Am I missing something? Or do you agree that we need further assumptions to prove the statement?
 A: The point here is really, that
$$P_B(A):=\frac{\mathbb P(A\cap B)}{\mathbb P(B)}$$
is also a probability measure. In some sense, one replaces the sample space $\Omega$ with a smaller one, namely $B$. The above definition restricts $\mathbb P$ to the sub sample space $B$ and $P_B$ measures the probability of $A$ in that new space.
The statement from the OP then provides uniquess (no other choice than using conditional probability) and the fact that $P_B$ is again a probability measure that fulfills the Kolmogoroff Axioms.
Point 2 actually says for $A\subset B$:
$$P_B(A)=\frac{\mathbb P(A\cap B)}{\mathbb P(B)}=\underbrace{\frac{1}{\mathbb P(B)}}_{=:c_B}\mathbb P(A)=c_B\,\mathbb P(A).$$
EDIT: But this is the opposite implication of what you should prove. Starting with 1. gives $$1=P_B(B)=c_B\mathbb P(B).$$ and thus $c_B=\mathbb P(B)^{-1}$.
Now, if $A\supset B$
$$c_B\mathbb P(A)=c_B\cdot\sum_{\omega\in A\cap B=B}\mathbb P(\omega)+c_B\cdot\sum_{\omega\in A\setminus B}\mathbb P(\omega)
=P_B(B)+c_B\cdot\sum_{\omega\in A\setminus B}\mathbb P(\omega)>1$$
Therefor the sum can only add the $\omega\in A\cap B$ to keep (1) valid. Taking this all together gives
$$P_B(A)=\frac{1}{\mathbb P(B)}\cdot\sum_{\omega\in A\cap B}\mathbb P(\omega)=\frac{\mathbb P(A\cap B)}{\mathbb P(B)}=:\mathbb P(A|B)
$$
So, the result is indeed that the conditional probability is the only measure that fulfills (1) and (2) from the OP.
