Independent Events are Independent of Complement/General Result

Theorem
Let $A_1, A_2, \ldots, A_m$ be events in a probability space $\struct {\Omega, \Sigma, \Pr}$.

Then $A_1, A_2, \ldots, A_m$ are independent $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_m$ are also independent.

Proof
Proof by induction:

For all $n \in \N: n \ge 2$, let $\map P n$ be the proposition:
 * $A_1, A_2, \ldots, A_n$ are independent $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_n$ are independent.

Basis for the Induction
$\map P 2$ is the case:
 * $A_1$ and $A_2$ are independent $\Omega \setminus A_1$ and $\Omega \setminus A_2$ are independent.

This is proved in Independent Events are Independent of Complement: Corollary.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $A_1, A_2, \ldots, A_k$ are independent $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_k$ are independent.

Then we need to show:
 * $A_1, A_2, \ldots, A_{k + 1}$ are independent $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_{k + 1}$ are independent.

Induction Step
This is our induction step.

Suppose $A_1, A_2, \ldots, A_{k + 1}$ are independent.

Then:

So we see that $\displaystyle \bigcap_{i \mathop = 1}^k A_i$ and $A_{k + 1}$ are independent.

So $\displaystyle \bigcap_{i \mathop = 1}^k A_i$ and $\Omega \setminus A_{k + 1}$ are independent.

So, from the above results, we can see that $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_{k + 1}$ are independent.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

The reverse implication follows directly.

Therefore:
 * $A_1, A_2, \ldots, A_n$ are independent $\Omega \setminus A_1, \Omega \setminus A_2, \ldots, \Omega \setminus A_n$ are independent.