Question about the independence definition. Why does the independence definition requires that every subfamily of events $A_1,A_2,\ldots,A_n$ satisfies $P(A_{i1}\cap \cdots \cap A_{ik})=\prod_j P(A_{ij})$ where $i_1 &lt i_2 &lt \cdots &lt i_n$ and $j &lt n$.
My doubt arose from this: Suppose $A_1,A_2$ and $A_3$ such as $P(A_1\cap A_2\cap A_3)=P(A_1)P(A_2)P(A_3)$.
Then
$$P(A_1\cap A_2)=P(A_1\cap A_2 \cap A_3) + P(A_1\cap A_2 \cap A_3^c)$$
$$=P(A_1)P(A_2)(P(A_3)+P(A_3^c))=P(A_1)P(A_2).$$
So it seems to me that if $P(A_1\cap A_2\cap A_3)=P(A_1)P(A_2)P(A_3)$ then $P(A_i\cap A_j)=P(A_i)P(A_j)$, i.e., the biggest collection independence implies the smaller ones. Why am I wrong? The calculations seems right to me, maybe my conclusion from it are wrong?

Here is a quick offhand example. Toss a fair coin once. Let $A$ be the event "head," $B$ the event "tail" and $C$ the event "head and tail." (Of course $C$ has probability $0$.)  That counterexample is trivial and boring, so we produce a non-trivial one.
Let $A$ and $B$ be events such that $P(A)=P(B)=1/2$ and $P(A\cap B)=1/5$. It is clear that $A$ and $B$ are not independent and that $P(A\cup B)=4/5$. Let $C=A\cup B$. Then 
$$P(A\cap B\cap C)=P(A\cap B)=\frac{1}{5}\qquad\text{and}\qquad P(A)P(B)P(C)=\frac{1}{5}.$$
