Clarification on the proof for Borel Cantelli lemma part $1$

Question

I need some clarification about the proof for the BC lemma part $1$. I am about to writing down the proof as it's written in my professor's notes and I assume BC lemma is known so I won't write it. So here it is.

First, note that the assumption $\sum_{n = 1}^{+\infty} P(A_n) < +\infty$ implies $\lim_{n\to +\infty} \sum_{m = n}^{+\infty} P(A_m) = 0$

1. I'm confused here already. Is this a sort of sum properties, or is it just a trick since when $n\to +\infty$ the sum runs from infinity to infinity?

Then we go on.

Using continuity of probability we get

$$ \begin{align} P\left(\bigcap_{n = 1}^{+\infty} \bigcup_{m = n}^{+\infty}A_m\right) & = P\left(\bigcap_{n = 1}^{+\infty} B_n\right) \\\\ & = \lim_{n\to +\infty} P(B_n) \\\\ & = \lim_{n\to +\infty} P\left(\bigcup_{m = n}^{+\infty} A_m\right) \\\\ & \leq \lim_{n\to +\infty} \sum_{m = n}^{+\infty} P(A_m) \\\\ & = 0 \end{align} $$

2. Why does he uses the $\leq$ sign? I thought that a probability measure satisfied the countable additivity property.

3. Is it legit to step from the probability of intersection to just the limit of $P(B_n)$? Why?

In the end he wrote:

Since $P\left(\bigcap_{n = 1}^{+\infty} \bigcup_{i = n}^{+\infty} A_i\right) \geq 0$ we conclude that $P\left(\bigcap_{n = 1}^{+\infty} \bigcup_{i = n}^{+\infty} A_1\right) = 0$

4. Is this necesary?

Thank you!

Answer 1

(1): The sum of a series $\sum_{n}a_{n}$ is finite if and only if you have $\lim_{n\to\infty}\sum_{k=n}^{\infty}a_{k}\to 0$. See here for more justification.

(2): This is what is known as the union bound. $P(A\cup B)\leq P(A)+P(B)$ . In particular, for infinite unions, $P(\bigcup_{k=1}^{\infty}A_{k})\leq \sum_{k=1}^{\infty}P(A_{k})$. Countable additivity is for disjoint sets. If $A_{k}$'s were disjoint then $P(\bigcup_{k=1}^{\infty}A_{k})=\sum_{k=1}^{\infty}P(A_{k})$.

But the bound can be derived using countable additivity. Consider $B_{1}=A_{1}$, $B_{2}=A_{2}\setminus B_{1}$ , $B_{3}=A_{3}\setminus B_{2}$ and so on. Then, $B_{k}$'s are disjoint but $\bigcup_{k=1}^{\infty}B_{k}=\bigcup_{k=1}^{\infty}A_{k}$

So by $P(\bigcup_{k=1}^{\infty}A_{k})=P(\bigcup_{k=1}^{\infty}B_{k})=\sum_{k=1}^{\infty}P(B_{k})$

But, notice that $P(B_{k})\leq P(A_{k})$ as $B_{k}\subseteq A_{k}$ and hence $\sum_{k=1}^{\infty}P(B_{k})\leq \sum_{k=1}^{\infty}P(A_{k})$ .

Thus $P(\bigcup_{k=1}^{\infty}A_{k})\leq \sum_{k=1}^{\infty}P(A_{k})$

(3): Suppose $A_{n}$ is a sequence of increasing sets, i.e. $A_{n-1}\subseteq A_{n}$ for all $n$. Then $P(\bigcup_{k=1}^{\infty}A_{k})=\lim_{k\to\infty}P(A_{k})$. This is again a consequence of countable additivity.

i.e. you write $B_{1}=A_{1}$, $B_{2}=A_{2}\setminus A_{1}$ , $B_{3}=A_{3}\setminus A_{2}$ and so on to get disjoint collections. i.e. $B_{k}=A_{k}\setminus A_{k-1}$ (set $A_{0}=\emptyset$ for notational consistency)

Then $P(\bigcup_{k=1}^{\infty}B_{k})=\lim_{n\to\infty}\sum_{k=1}^{n}P(B_{k})=\lim_{n\to\infty}\sum_{k=1}^{n}P(A_{k})-P(A_{k-1})$

Now observe that the sum telescopes.

i.e. $\sum_{k=1}^{n}P(A_{k})-P(A_{k-1})=P(A_{n})$

Hence $P(\bigcup_{k=1}^{\infty}A_{k})=\lim_{n\to\infty}P(A_{n})$

Now, if $C_{n}$ is a sequence of decreasing sets, then

$P(\bigcap_{k=1}^{\infty}C_{k})=P\bigg(\bigg(\bigcup_{k=1}^{\infty}C_{k}^{c}\bigg)^{c}\bigg)=1-P\bigg(\bigcup_{k=1}^{\infty}C_{k}^{c}\bigg)$ due to De-Morgan's Laws.

But $C_{k}$'s are decreasing implies that $C_{k}^{c}$ are increasing. So by previous result.

$P(\bigcap_{k=1}^{\infty}C_{k})=1-\lim_{n\to\infty}P(C_{n}^{c})=\lim_{n\to\infty}(1-P(C_{n}^{c}))=\lim_{n\to\infty}P(C_{n})$

(4): This is just a logical trick. if you have something which is positive $\geq 0$ and you also manage to show that it is $\leq 0$ then it must be $0$. In particular, if the "probabilities" were not necessarily non-negative valued (this is ridiculous but still you can consider this for argument's sake) then you cannot conclude that it is $0$. In reality, one should always state this above explanation but we rarely ever do because it is obvious. So don't worry about it too much. Unless there's a Third Reich in mathematics, I don't see someone penalizing a student for not writing the above explanation.

Warning: This answer should be taken just for understanding. In particular, one should write $\sum_{k=1}^{\infty}a_{k}$ in the limit form as $\lim_{n\to\infty}\sum_{k=1}^{n}a_{k}$ and write everything else in the form of such limits to make perfect rigorous sense.