Is conditional probability actually a probability? Let $Ω$ be a sample space and $P$ a probability measure on $Ω$. Let $A ⊆ Ω$ be an event of positive probability. Let $P_A$ be defined by $P_A(E) = P(E|A) $for any event $E ⊆ Ω$. Show that $P_A$ satisfies the first two axioms for a probability measure on $Ω$. Show that it also satisfies the third and/or fourth.
My solution: The axioms of probability are: $$\text{(i) }0\leq P(A) \leq1 \text{ for each event $A$}\\ \text{(ii) }P(\Omega)=1\text{ and }P(\varnothing)=0\\ \text{(iii) If }A_1, A_2, A_3,... \text{is a sequence of pairwise disjoint events then}\\ P\left( \bigcup_{i=1}^{\infty}A_i\right) = \sum_{i=1}^{\infty} P(A_i)$$ I don't really see how I would prove this. I feel like it's just obvious that this is the case because the conditional probability is composed of regular probabilities. Am I just supposed to compute each axiom by using $P_A(E)$ instead of $P?$
Should $P(A)> 0$, then the definition of conditional probability has it that $$P_A(E)=\dfrac{P(A\cap E)}{\mathsf P(A)}$$
Use this to show that since $P()$ satisfies the axioms, then $P_A()$ shall too.
$${\text{(i) }0\leq P_A(E) \leq 1 \text{ for each event $E:E\subseteq\Omega$}\\ \text{(ii) }P_A(\Omega)=1\text{ and }P_A(\varnothing)=0\\ \text{(iii) If }E_1, E_2, E_3,... \text{is a sequence of }\textbf{pairwise disjoint}\text{ events then}\\\qquad P_A\left( \bigcup_{i=1}^{\infty}E_i\right) = \sum_{i=1}^{\infty} P_A(E_i)}$$