State necessary and sufficient conditions for $A$ and $B$ to be independent 
(a) Let $\mathbb P$ be the uniform measure on $[0,1]$. Define $A = (a,b)$ and $B = (c,d)$, with $a < c$. State necessary and sufficient conditions for $A$ and $B$ to be independent.
(b) Find an example of an $\omega , \mathbb P,$ and sets $A,B,C$ such that $$\mathbb P(A\cap B\cap C)=\mathbb P(A)\:\mathbb P(B)\:\mathbb P(C)$$ Where $A,B,$ and $C$ are not independent

$(a)$
For any interval $I = [a,b] ( $or $I=(a,b))$ in the set $\mathbb {R}$ of real numbers, let $\mathbb P(I)=b-a$ denote its length. I need to show the independence,
$$\mathbb P(A\cap B)=\mathbb P(A)\:\mathbb P(B)$$ but A; B; C are not independent.
I divide the whole scenario into three case:

*

*$(a,b)$ and $(c,d)$ are disjoint: then

\begin{equation}
\begin{split}
\mathbb P(A)\:\mathbb P(B) & = (b-a)(d-c) \\
\mathbb P(A\cap B) & = 0
\end{split}
\end{equation}
$\mathbb P(A\cap B)=\mathbb P(A)\:\mathbb P(B)$ either $(b-a)=0\implies a=b$ or $(d-c)=0\implies c=d$

*

*$a<c<d<b$ then $\mathbb P(A\cap B)=\mathbb P(A)\:\mathbb P(B) \implies (d-c)=(b-a)(d-c)$ which mean $(a,b)=(0,1)$


*$a<c<b<d$ then $\mathbb P(A\cap B)=\mathbb P(A)\:\mathbb P(B)$ if $(b-c)=(b-a)(d-c)$. Then I could detect any conclusion about them.
For $(b)$ I couldn't manage an example.
But I think I am not getting what they asked for. Any solution or hint will be appreciated. Sorry for the question title, I haven't any nice title for those question in my head now.
 A: For your 9-sided die example, maybe $\mathcal X=\omega$ is the set on which the measure is defined, $\omega=\{1,2,3,...,9\}$, and let the $\mathcal A=\sigma$-field be the power set, and let $\mu=\mathbb P$ be the measure $1/9 \cdot \# A$ where $A$ is any of the subsets in $\mathcal A$. Let the events $A=\{1,2,3\}$, $B=\{3,4,5\}$ and $C=\mathcal X=\omega$. $A\cap B\cap C=\{3\}$.
$$P(A\cap B\cap C)=1/9=1/3\cdot 1/3\cdot 1=P(A)P(B)P(C)$$
A: Regarding part (a)
For the disjoint case you can write:
$A$ and $B$ are independent if and only if $A=\emptyset$ or $B=\emptyset$.
Try to write precise mathematical statements for the two other cases. I don't think you can achieve anything particularly nice in the final case. What you have seems good enough.
Regarding part (b)
Edited answer after clarification:
Let $X \colon \Omega \to \mathbb{R}$ denote a random variable such that $$X \sim \text{Bernoulli}\bigg(\frac{1}{2}\bigg).$$ For instance this can model a fair coin toss. Consider now
$$
A=\{X=0\}, \quad B=\{X=1\} \quad\text{and} \quad C= \emptyset.
$$
Note that $A,B$ and $C$ are not pairwise independent (and thus not independent) as
$$
\mathbb{P}(A\cap B) = \mathbb{P}(\emptyset) = 0 \neq \frac{1}{2} \cdot \frac{1}{2} = \mathbb{P}(A)\mathbb{P}(B).
$$
However
$$
\mathbb{P}(A\cap B\cap C) = \mathbb{P}(\emptyset) = 0 = \frac{1}{2}\cdot \frac{1}{2} \cdot 0   = \mathbb{P}(A)\mathbb{P}(B)\mathbb{P}(C)
$$
as desired.
