In statistics class (very basic level, at that point we had only covered Kolmogorov axioms and some of its consequences) we saw that $$P\Bigl(\bigcup_{n\in\mathbb N}^\infty A_n\Bigr)\le\sum_{n\in\mathbb N}^\infty P(A_n)$$ I'm trying to prove it, but without much success; can someone help me puzzle this out?
For $n = 1$ it is trivial, since $P(A_ 1) = P(A_1)$.
Now suppose it is true for a natural number $k$, then it must be true for $k + 1$. We have $A_1, \dots, A_{k+1}$, i.e. $k + 1$ events.
This is where I get stuck. How am I supposed to use the third axiom, the monotonicity property (I guess) and take it from natural number $k$ to infinity and prove this? Also, the third axiom requires pairwise disjoint events, while what I'm trying to prove does not.
Hypothesis: $$P\Bigl(\bigcup_{i=1}^k A_i \cup A_{k+1} \Bigr)\le\sum_{i = 1}^k P(A_i) + P(A_{k+1}) = P\Bigl(\bigcup_{i=1}^{k+1} A_i\Bigr)\le\sum_{i = 1}^{k+1} P(A_i) $$
I guess I can take a $B$ partition from $A$ (view $A$ set as a set of disjoint events) such that $B_1, ..., B_n$ form a disjoint subset of $A$, i.e. $B_1 = A_1, B_2=A_2 \backslash A_1, B_3 = A_3\backslash(A_1 \cup A_2), ...$ This would result in $$P\Bigl(\bigcup_{i=1}^k A_i\Bigr)= P\Bigl(\bigcup_{i=1}^k B_i \Bigr)$$
Since $B$ is composed by pairwise disjoint events, by the third axiom $$P\Bigl(\bigcup_{i=1}^k B_i \Bigr) = \sum_{i = 1}^\infty P(B_i)$$
But $B \subseteq A$, so $$\sum_{i = 1}^\infty P(B_i) \leq \sum_{i = 1}^\infty P(A_i)$$ (by monotonicity).
By transitivity, $$P\Bigl(\bigcup_{i=1}^k A_i\Bigr) \leq \sum_{i = 1}^\infty P(A_i)$$
I don't know if I'm allowed to take this B disjoint subset and whether jumping to infinity by the third axiom is correct. Also, I didn't use induction.
You are correct that you do not have pairwise disjoint events currently.
Thus it may then be useful to consider $B_1, B_2, B_3, \ldots$ where $$B_k = A_k \setminus \left(\bigcup_{i=1}^{k-1} A_i\right).$$ Are these disjoint? What is their union?