Consider two events $G$, $A \in E(\Omega)$, where $E(\Omega)$ is an event space.
Consider P(G|A) is the posterior probability distribution over $G$. Now if you sum over $G$, then why is it equal to $1$?
In short, Why $\sum_{G} P(G|A) = 1$?
Edit 1:
I came across Kolmogorov definition for Conditional Probability and Law of Total Probability, according to which I can do following:
$\sum_{G} P(G|A) = \sum_{G} \frac{P(G \cap A)}{P(A)} = \frac{1}{P(A)} \sum_{G}P(G \cap A) = \frac{1}{P(A)}P(A) = 1$
I am not sure if the calculation above is correct or not?
Even if it is correct, Why $P(G \cap A) = P(A)$?
$\sum_G \Bbb P(G\mid A)=1$ is true when you have: some sequence of mutually disjoint events $(G)$ whose arbitrary union is a superset of the event $A$. That is that $(G)$ must form a partition of some superset of $A$ (and most usually this is the outcome set, $\Omega$).
$$\begin{align}\sum_G\Bbb P(G\mid A) &= \sum_G\dfrac{\Bbb P(A\cap G)}{\Bbb P(A)} && \text{conditional probability}\\ & =\dfrac{\Bbb P(\bigcup_G (A\cap G))}{\Bbb P(A)} && \text{probability for a union of }\textit{disjoint events}\\ &= \dfrac{\Bbb P(A\cap\bigcup_G (G))}{\Bbb P(A)}&&\text{distribution}\\& =\dfrac{\Bbb P(A)}{\Bbb P(A)}&&\text{since } A\subseteq \bigcup_G(G)\\ &= 1\end{align}$$