Prove $p (\bigcup A_i) \leq \sum p (A_i) $
Now i doubt that this has not been proven already as it is a common question, but i am typing latex on my tablet because i spilled water on my computer so i am having trouble finding the question in the archives. if somebody could post a link. Thanks. In case it has not been proven, i am thinking of using induction, but it just doesnt feel like the right thing to do to prove this question.
Besides the proofs based on set operations, there is another way with the help of the indicator function. Let $1_A(x)$ be the indicator function, that is, $1_A(x) = 1$ on $x \in A$ and $0$ on $A^c$. We see that $P(A) = E(1_A(X))$. Obviously, $$ 1_{\cup_n A_n}(x) \leq \sum_n 1_{A_n}(x)$$ Taking expectation on both sides, each term on the right side in the sum is $ \geq 0 $ so we can interchange the sum and taking expectation, which completes the proof.
