absolutely continuous measure - probability theory Let $(\Omega, F, P)$ be a probability space.
My book says, that:
"In the case when $\Omega$ consists of finitely many elements and $P[{ω}] >
0$ for every $ω \in \Omega$, then for every probability measure $Q$ we have $Q << P$.
Equivalence means $Q[{ω}] > 0$ for every $ω$".
Can someone explain me why this is true?
I know that $Q << P$ means, that $P[A] = 0 \Rightarrow Q[A] = 0$, for $A \in F$, but why does this relation hold? And why must $\Omega$ be finite?
 A: The right definition is: $Q << P$ means, that $P[A] = 0 \Rightarrow \color{red}Q[A] = 0$, for $A \in F$. Since $P[A]$ is never zero except if $A$ is the empty set, the implication holds. 
As far as finiteness.
If  $\Omega$ was not finite then we couldn't say so easily that $P[\{ω\}] >
0$ for every $ω \in \Omega.$ If $\Omega$ is not finite but countable and $\sum_{\omega\in \Omega} P[\{\omega\}]=1$ then the same statement is true.
A: Proof of

In the case when $\Omega$ consists of finitely many elements and $P[{ω}] >
0$ for every $ω \in \Omega$, then for every probability measure $Q$ we have $Q << P$.

For any $A \in F$, we need to show that if $P(A)=0$ then $Q(A)=0$. So, for what $A$'s do we have that $P(A) = 0$? If we find those $A$'s, then we have to check the corresponding $Q(A)$'s.
If there's no such $A$, then, vacuously, 'If $P(A) = 0$, then $Q(A) = 0$' is true.
Actually, there is such an $A$: $P(A) = 0 \iff A=\emptyset$. (Otherwise, $A$ will contain some $\omega$'s which by definition have positive $P$-probability).
QED
Note: If you for some reason restrict your $A$'s to be the singleton's then that's fine because $P(A)$ will never be zero. Vacuously, 'If $A$ is a singleton and $P(A) = 0$, then $Q(A) = 0$' is true.

Discussion of $P[{ω}] > 0$ assumption:
The thing is if there was some sample point $\omega_z$ with zero $P$-probability, i.e. $P(\omega_z) = 0$, then it's pretty simple to find construct $Q$ with if $Q(\omega_z) > 0$. Consider the following:
$\Omega = \{\omega_n\}_{n=0}^{k}$ where $k$ is some positive integer greater than $1$.
$P(\omega_n) = \frac{1}{k} \ \forall n = 1, \dots, k$
$P(\omega_0) = 0$
$Q(\omega_n) = \frac{1}{k+1} \ \forall n = 0, 1, \dots, k$
Here, $P(\omega_0) = 0$ but $Q(\omega_0) = \frac{1}{k+1}$. You can even let $P(\omega_n)$ be something else. Convince yourself.

Proof of

Equivalence means $Q[{ω}] > 0$ for every $ω$".

$$P \ \text{and} \ Q \ \text{are equivalent}$$
$$\iff$$
$$P << Q \ \text{and} \ Q << P$$
$$\iff$$
$$(P(A) = 0 \iff Q(A) = 0) \forall A \in F$$
$$\iff$$
$$(P(A) > 0 \iff Q(A) > 0) \forall A \in F \tag{*}$$
Now since $\{\omega\} \in F$, choose $A=\{\omega\}$.
We are given that $P(\{\omega\}) > 0$. Thus, if $P$ and $Q$ are equivalent, then by $(*)$, $Q(\{\omega\}) > 0$.
QED

Discussion of finiteness assumption:
It seems that the finiteness assumption is simply so that we can talk about the individual elements of $\Omega$ having positive probability. However, we can actually relax finite to countable, as $A=\emptyset$ will still be the only $A \in F$ that gives $P(A)=0$, and we can still talk about $P(\omega)$ even with $P(\omega) > 0$.
Another explanation could be that the next part of the text, which I guess is this, is talking about atoms in a finite probability space. It looks like the text is trying to explain the concept of atoms by using a finite probability space as an example: The explanation involves decreasing/increasing $\sigma$-algebras which leads to the next definition which is on filtrations.
Also see Example 1.1.14 in the text.
