Q:

Could someone explain Boole's inequality proof to me?

I am currently writing a report about George Boole and need help understanding this proof, any help would be appreciated.
Thank you mathematicians.

A:

Boole's inequality states that for any events $A_1,A_2,\dots$, 
$$\mathbb{P}\left(\bigcup_{i=1}^\infty A_i\right ) \leq \sum_{i=1}^\infty \mathbb{P}(A_i).$$
The proof makes use of the fact that for any disjoint events $B_1,B_2,\dots$,
$$\mathbb{P}\left(\bigcup_{i=1}^n B_i\right) = \sum_{i=1}^\infty \mathbb{P}(B_i).$$
How does this help? If we can find a sequence of events $B_1,B_2,\dots$ such that all of the following hold:

$B_1,B_2,\dots$ are disjoint
$\bigcup_{i=1}^\infty A_i = \bigcup_{i=1}^\infty B_i$
For all $i$, $\mathbb{P}(B_i) \leq \mathbb{P}(A_i)$,

then we would have
$$\mathbb{P}\left(\bigcup_{i=1}^\infty A_i\right )=\mathbb{P}\left(\bigcup_{i=1}^\infty B_i\right )= \sum_{i=1}^\infty \mathbb{P}(B_i) \leq \sum_{i=1}^\infty \mathbb{P}(A_i).$$
(The first equality is from bullet point two, the second equality is from bullet point one, and the inequality is from bullet point three.)
The idea is that we can turn the $A_i$ into disjoint sets by "cutting away" the parts of the $A_i$ that overlap with the other sets. To do this in a systematic way, we define
\begin{align*}
B_1 &= A_1\\
B_2 &= A_2 - A_1\\
B_3 &= A_3 - (A_1 \cup A_2)\\
B_4 &= A_4 - (A_1 \cup A_2 \cup A_3)\\
&\vdots\\
B_i &= A_i - \bigcup_{j=1}^{i-1}A_j\\
&\vdots
\end{align*}
Clearly each $B_i\subseteq A_i$, since we obtain $B_i$ by taking $A_i$ and "cutting away" some part of it. So $\mathbb{P}(B_i) \leq \mathbb{P}(A_i)$. So we have satisfied the third bullet point above. 
The fact that $B_i \subseteq A_i$ for all $i$ also implies that the union of the $B_i$ is contained in the union of the $A_i$: $\bigcup_{i=1}^\infty B_i \subseteq \bigcup_{i=1}^\infty A_i$. This is half of the second bullet point above. The other half requires showing that $\bigcup_{i=1}^\infty A_i \subseteq \bigcup_{i=1}^\infty B_i$. If $x \in \bigcup_{i=1}^\infty A_i$, then $x \in A_i$ for some $i$, or possibly more than one $i$. For instance, maybe $x$ is in $A_3$, $A_4$, and $A_6$ (and none of the others). How do we know which $B_i$ it is in? By definition, $B_4$ does not contain any of the elements in $A_3$, as they are cut away. Similarly, $B_6$ does not contain any of the elements in $A_3$. So $x$ cannot be in $B_4$ or $B_6$. $B_3$ contains all the elements of $A_3$ that aren't also in $A_1$ or $A_2$. Since $x$ is not in $A_1$ or $A_2$, it doesn't get "cut away" from $A_3$ when we define $B_3$. Hence $x \in B_3$. In general, if we know $x \in A_{i_1},A_{i_2},\dots$ where $i_1<i_2<\cdots$, then we know $x \in B_{i_1}$.
Finally, we need to show that $B_1,B_2,\dots$ are disjoint, i.e., for any $i \neq j$, $B_i \cap B_j = \emptyset$. Suppose $i \neq j$, and without loss of generality we can assume $i<j$ (otherwise just re-name the indices so that this is true). Then $B_j$ is defined by taking $A_j$ and removing all of the elements of $A_1,A_2,\dots,A_{j-1}$. In particular, since $i<j$, all of the elements of $A_i$ are removed from $B_j$. And since $B_i \subseteq A_i$, we know all of the elements of $B_i$ are removed from $B_j$. So they have to be disjoint. If you want to prove this symbolically, we have
\begin{align*}
B_i \cap B_j &\subseteq A_i \cap \left(A_j - \bigcup_{k=1}^{j-1}A_k\right)\\
&=A_i \cap \left(A_j \cap \left(\bigcup_{k=1}^{j-1} A_k\right)^c \right)\\
&= A_i \cap\left(A_j \cap \left(\bigcap_{k=1}^{j-1}A_k^c\right)\right)\\
&\subseteq A_i \cap A_i^c=\emptyset.
\end{align*}
So $B_i \cap B_j \subseteq \emptyset$, which implies $B_i \cap B_j = \emptyset$.
Now we have verified all three bullet points, so the inequality follows.

