Let $S$ be a sample space with event space $F$ and probability function $P$. Let $B \in F$ such that $P(B) > 0$. Define the conditional probability function as $P(A|B)$.
The question then asks you to prove that $P(A|B)$ is a proper probability function by stating a new sample space, event space, and proving that the event space and $P(A|B)$ follow the appropriate axioms.
I've defined the new event space as $F^{'} = \{{E \cap B} : E \in F\}$, but I'm having trouble proving that for $E \in F^{'}$, $E^c \in F^{'}$.
I've started by showing that $E = A \cap B$ for some $A \in F$ and thus $E^c = (A \cap B)^c=A^c \cup B^c$, but I'm not sure where to go from there. Since A and B are in F, so are their compliments, but I'm not sure how I go from there to showing that $A^c \cup B^c \in F^{'}$
Here We Can See E is an event over Sample Space B, the complement of E in Sample Space B is not equal to the complement of E over Sample Space S
To Prove Let us assume $$E \in F'$$ Now There Must exist A such that$$ E = A \cap B \ \ \ where \ A \in F $$
Then there must be $$ A^c \in F $$ so $$ A^c \cap B \in F' $$
Now we can say $E^c$ over sample space B is $A^c \cap B$
We can show $(A^c \cap B)$ is the complement of $(A \cap B)$ over B by taking the union of Both and proving it to be equal to B i.e $$To\ Prove : E^c \cup E = B \\ (A^c \cap B) \cup (A \cap B) \\ \bigl(A^c \cup (A \cap B)\bigr) \cap \bigl(B \cup (A \cap B)\bigr) \\ \bigl(S \cap (A^c \cup B)\bigr) \cap (B) \\ (A^c \cup B) \cap B = B \ \ \ \ \ Proved $$