🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

In the definition of a probability measure, for $i\in \{1,\dots ,n\}$ we set $E_i=A_i$ and then for any $j>n$ we set $E_j=\varnothing$, then we can see that $E_1,E_2,\dots$ is pairwise disjoint, and thus

$P(A_1\cup A_2\cup \dots \cup A_n)$	$=$	$P(A_1\cup A_2\cup \dots \cup A_n\cup \varnothing \cup \varnothing \cup \dots )$
	$=$	$P\left({\bigcup }_{i=1}^{\infty }{E}_i\right)$
	$=$	${\sum }_{i=1}^{\infty }P\left(E_i\right)$
	$=$	${\sum }_{i=1}^{n}P\left(E_i\right)+P(\varnothing )+P(\varnothing )+P(\varnothing )+\dots$
	$=$	${\sum }_{i=1}^{n}P\left(E_i\right)+0+0+0+\dots$
	$=$	${\sum }_{i=1}^{n}P\left(A_i\right)$

Suppose that $c\ne 0$, and note that $P\left(\Omega \right)=P\left(\left[0,1\right]\right)={\sum }_{x\in [0,1]}P\left(x\right)={\sum }_{x\in [0,1]}c$, but note that no matter how small the value of $c$, since we are summing it uncountably many times with itself the sum always goes to infinity, but at the same time, $P\left(\Omega \right)=1$ and thus we have a contradiction, so $c=0$.

Define $F_i=E_i\cap E$, we claim that $E={\bigcup }_{i=1}^{\infty }{F}_i$. Let $x\in E$, then there is some $i\in {\mathbb{N}}^1$ such that $x\in E_i$, therefore $x\in E_i\cap E:=F_i$, on the contrary suppose that $a\in F_i$ for some $i\in {\mathbb{N}}_1$, therefore $a\in E_i\cap E$, so clearly $a\in E$, therefore $E={\bigcap }_{i=1}^{\infty }{F}_i$

Define $G_1:=F_1$ and set $G_k:=F_k\cap {\left({\bigcap }_{n=1}^{k-1}{F}_i\right)}^C$ for $k\ge 2$. Firstly note that the $G$'s are pairwise disjoint, suppose that we have $G_i,G_j$ where without loss of generatlity $i<j$, then we know that $G_i\subseteq F_i$, but at the same time we know that $G_j\subseteq {\left({\bigcap }_{n=1}^{j}F\right)}^C$, since $F_i$ is part of that union, then when we take it's complement, clearly $G_j$ cannot intersect with $F_i$ and since $F_i\supseteq G_i$ it cannot intersect with $G_i$ as well.

Additionally we claim ${\bigcup }_{i=1}^{\infty }{G}_i=\bigcup F_{i=1}^{\infty }$. Let $x\in {\bigcup }_{i=1}^{\infty }{G}_i$, so there exists some $k\in {\mathbb{N}}_1$ such that $x\in G_k$, since $G_k\subseteq F_k$ then we know $x\in F_k$ that is there is some $k$ such that $x\in F_k$ so by definition $x\in {\bigcup }_{i=1}^{\infty }{F}_i$. Moving the other way suppose that $p\in {\bigcup }_{i=1}^{\infty }{F}_i$ in that case there is some $m\in {\mathbb{N}}_1$ so that $p\in F_m$, let $I\subseteq {\mathbb{N}}_1$ be the collection of indices such that for every $j\in I$, $p\in F_j$, clearly $I$ is non-empty since at least $k\in I$, therefore by the well ordering principle it has some least element $l\in I$, note that this means for any $c\in [1\dots l-1]$ we have that $p\notin F_c$, so by definition $p\in G_l$, so that we've shown $p\in {\bigcup }_{i=1}^{\infty }{G}_i$ as needed. We conclude that ${\bigcup }_{i=1}^{\infty }{G}_i={\bigcup }_{i=1}^{\infty }{F}_i=E$

$$\begin{gathered} {\displaystyle P\left(E\right)\&=P\left({\bigcup }_{i=1}^{\infty }{G}_i\right) \\ \&={\sum }_{i=1}^{\infty }P\left(G_i\right) \\ \&\le {\sum }_{i=1}^{\infty }P\left(F_i\right) \\ \&\le {\sum }_{i=1}^{\infty }P\left(E_i\right) \\ } \end{gathered}$$ The first line is justified by the paragraph above, the second by the fact that the $G_i$'s are disjoint, the third by the fact that $G_i\subseteq F_i$ and the last because $F_i\subseteq E_i$.