# How to prove that conditional probability is bounded to [0,1]?

I am trying to solve the following question from Sheldon Ross, Introduction to Probability Models.

I am struggling to prove the first of the 3 conditions,

I thought of using the inclusion exclusion theorem, but I can't think of how to prove these bounds for $$P(A_i \cup B)$$ . I was thinking that showing $$P(\phi | B) = 0$$ and $$P(\Omega |B) = 1$$ should be enough since $$\phi \subseteq A_i \subseteq \Omega$$. But I want rigor in my proof.

Now I think my proof for condition 2 is correct and straightforward. For condition 3, I used $$n=2$$ as the base case and proved it via induction. But I am not sure it is a valid proof since $$\infty$$ is not a natural number and stackexchange seems to agree that it is not valid. I was wondering if induction is enough to show that countable additivity holds for any countably infinite sequence of events $${A_1, A_2, \cdots }$$ which are mutually exclusive. When can I use induction to prove that it holds for an infinite of cases? What other tools are there?

My attempts -

• Maybe I have misunderstood your question, but it seems much simpler than you are making it. As I understand it, you are trying to prove that $0 \le P(A \mid B) \le 1$ for all events $A,B$. Well, just write the definition: $P(A \mid B) = P(A \cap B) / P(B)$. This is a ratio of two nonnegative numbers, so it is nonnegative. And $A \cap B \subset B$, so the numerator is less than or equal to the denominator. Done. – Nate Eldredge Aug 12 '19 at 14:34
• @NateEldredge is there any proof for the fact that if A is a subset of B then P(A) must be less than P(B)? It seems obvious, but I thought the proof must be only depending on axioms and logic – Aditya P Aug 12 '19 at 14:43
• Sure. Write $B = A \cup (B \setminus A)$. This is a disjoint union so by axiom (iii) we have $P(B) = P(A) + P(B \setminus A)$. But by axiom (i) we have $P(B \setminus A) \ge 0$. This implies $P(B) \ge P(A)$. – Nate Eldredge Aug 12 '19 at 15:19
• ππ thanks I can't believe I overcomplicated that. What about condition 3? Do you think the induction proof is good enough? – Aditya P Aug 12 '19 at 15:21
• No, induction can prove your claim for finite unions, but not for infinite unions. – Nate Eldredge Aug 12 '19 at 16:15

Since $$0\leq P(A\cap B)\leq P(B)$$, it follows that
$$0\leq P(A\mid B)= \frac{P(A\cap B)}{P(B)}\leq 1$$
$$P(\bigcup_n A_n\mid B) =\frac{P(\bigcup_n A_n\cap B)}{P(B)}$$ $$= \sum_n \frac{P(A_n\cap B)}{P(B)} = \sum_n P(A_n\mid B)$$
• The statement $B\cap \bigcup_{i\in I} A_i=\bigcup_{i\in I}(B\cap A_i)$ holds for every index set $I$, so not only finite ones. This is an easy set theory exercise. – ε-δ Aug 12 '19 at 16:19