I know that from the Kolmogorov's Axioms of Probability that the probability of a finite collection of sets is equal to the sum of the probability of of each sets. What I am not sure is how the expression below can be proven.
Assume that $A_1$, $A_2$, ... , $A_n$ denote n events in $B$ (Borel set). Show that $$ P( \bigcup_{n=1}^n A_i) \le n\,\max \{P(A_i),...,P(A_n)\}$$
By a "Graham-Schmidt" type argument (if you haven't already seen this) you can show that $\mathbb{P}(\cup A_i)\leq \sum \mathbb{P}(A_i)$, but this implies that $$\mathbb{P}(\cup_{i=1}^n A_i) \leq \sum_{i=1}^n \mathbb{P}(A_i) \leq \sum_{i=1}^n \max\mathbb{P}(A_j) = n\max\{\mathbb{P}(A_i)\}$$