Prove $P(B|A) \le P(\bar{A} \cup B)$ provided $P(A) > 0$.

Question

In a philosophy paper I've encountered the claim that for any probability measure $P$ alongside events $A$ and $B$ it follows that $P(B|A) \le P(\bar{A} \cup B)$ provided $P(A) > 0$. I'm wondering if anyone might help me to see how to prove this claim. So far I've derived the following

$\begin{align*} P(\bar{A} \cup B) &= P(\bar{A}) + P(B) - P(\bar{A} \cap B)\\ &= 1 - P(A) + P(B) - P(\bar{A}|B) P(B)\\ &= 1 - P(A) + (1 - P(\bar{A}|B)) P(B)\\ &= 1 - P(A) + P(A|B)P(B)\\ &= 1 - P(A) + P(B|A)P(A)\\ &= 1 - P(A)(1 - P(B|A))\\ \end{align*}$

but I'm not seeing how to manipulate this further to get the inequality.

Answer 1

Note that

\begin{align*} P(\bar{A} \cup B) &= 1 - P(A)(1 - P(B|A)) \\ &\geq 1- 1(1 - P(B|A)) \\ &= 1- 1 + P(B|A) \\ &= P(B|A) \end{align*}