Independent Events are Independent of Complement

Theorem
Let $A$ and $B$ be events in a probability space $\left({\Omega, \Sigma, \Pr}\right)$.

Then $A$ and $B$ are independent iff $A$ and $\Omega \setminus B$ are independent.

Proof
For $A$ and $B$ to be independent:
 * $\Pr \left({A \cap B}\right) = \Pr \left({A}\right) \Pr \left({B}\right)$

We need to show that:
 * $\Pr \left({A \cap \left({\Omega \setminus B}\right)}\right) = \Pr \left({A}\right) \Pr \left({\Omega \setminus B}\right)$

First note that $\Omega \setminus B \equiv \mathcal C_{\Omega} \left({B}\right)$ where $\mathcal C_{\Omega}$ denotes the relative complement.

From Set Difference as Intersection with Relative Complement, we have then that $A \cap \left({\Omega \setminus B}\right) = A \setminus B$.

From Set Difference and Intersection form Partition, we have that:
 * $\left({A \setminus B}\right) \cup \left({A \cap B}\right) = A$
 * $\left({A \setminus B}\right) \cap \left({A \cap B}\right) = \varnothing$

So from the Kolmogorov axioms, we have that:
 * $\Pr \left({A}\right) = \Pr \left({A \setminus B}\right) + \Pr \left({A \cap B}\right)$

Hence:

But as $A \setminus B = A \cap \left({\Omega \setminus B}\right)$ we have:
 * $\Pr \left({A \cap \left({\Omega \setminus B}\right)}\right) = \Pr \left({A}\right) \Pr \left({\Omega \setminus B}\right)$

which is what we wanted to show.

Now, suppose $A$ and $\Omega \setminus B$ are independent.

From the above, we have that $A$ and $\Omega \setminus \left({\Omega \setminus B}\right)$ are independent.

But $\Omega \setminus \left({\Omega \setminus B}\right) = B$ from Relative Complement of Relative Complement hence the result.