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While working on some probability problems I got stuck in the understanding of the following:

Question

Let $p_1, p_2, ...p_N$ be non- negative numbers s.t. $\,p_1 + p_2 + ... + p_N = 1$ and let $\Omega = \{\omega_1, \omega_2, ..., \omega_N$}, with $\mathcal{F}$ the power set of $\Omega$.

Show that the function $\mathbb{Q}$ given by:

$\mathbb{Q}(A) = \sum _{ i:\omega _{ i }\in A }^{ }{ \,p_i }\quad\,for\, A\in \mathcal{F}$

is a probability measure on $(\Omega, \mathcal{F})$.


Prove

I understand that $\mathbb{Q}(A) \geq 0$ and $\mathbb{Q}(\Omega)=1, \mathbb{Q}(\emptyset)=0$ holds.

The following should fulfill the requirement that:

if $A_1, A_2, ...$ are disjoint events in $\mathcal{F}$, then $\,\mathbb{P}(\bigcup _{ i=1 }^{ \infty}{ A_i} )\,=\,\sum_{i=1}^{\infty}\mathbb{P}(A_i)$.

For this part chose some disjoint events $A_1, ..., A_k$. Every $A_i$ contains different $\omega_j$ which indicates:

$\mathbb{Q}\,(\bigcup _{ i=1 }^{ k }{ A_ i})\quad =\quad \sum _{j:\omega_j\in\, \bigcup _{ i=1 }^{ k }{A_ i}}^{ }{ p_ i }\quad = \quad\sum_{i=1}^{k}\sum_{j:\omega_j\in \,A_i}p_i \quad = \quad \sum_{i=1}^{k}\mathbb{Q}(A_i)$

I've been trying to figure out what is going on out here for hours, but it keeps confusing me.

Is there anyone who could describe the idea behind this in a somewhat less abstract/ difficult way than shown out here?

Thank you!

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This first and the third equality follow by definition: in the first, you have the definition of $\mathbb{Q}\left(\bigcup_{i=1}^{k}A_{i}\right)$ and in the third you have the definition of $\mathbb{Q}(A_{i})$.

In the second equality, you are using the fact that the $A_{i}$'s are disjoint. That is, $w_{j}\in \bigcup_{i=1}^{k}A_{i}$ if and only if $w_{j}\in A_{i}$ for a unique $i$.

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